Eigenvalues of $2$ symmetric $4times 4$ matrices: why is one negative of the other?If the eigenvalues are...
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Eigenvalues of $2$ symmetric $4times 4$ matrices: why is one negative of the other?
If the eigenvalues are distinct then the eigenspaces are all one dimensionalCongruence of invertible skew symmetric matricesEigenvalues of a general block hermitian matrixEigenvalues of Overlapping block diagonal matricesHow to find a symmetric matrix that transforms one ellipsoid to another?The matrix of an endomorphismA conjecture regarding the eigenvalues of real symmetric matricesProve that the span of ${M_1, M_2, M_3}$ is the set of all symmetric $2times2$ matrices.Looking for properties of, or formulae for eigenvalues of a symmetric matrix reminiscent of Toeplitz matricesDo hermitian matrices commute when they occupy they same elements but have different values?
$begingroup$
Consider the following symmetric matrix:
$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.
Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
$endgroup$
add a comment |
$begingroup$
Consider the following symmetric matrix:
$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.
Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
$endgroup$
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
1
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago
add a comment |
$begingroup$
Consider the following symmetric matrix:
$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.
Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
$endgroup$
Consider the following symmetric matrix:
$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.
Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
edited 7 mins ago
YuiTo Cheng
2,2734937
2,2734937
asked 1 hour ago
TroyTroy
4231519
4231519
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
1
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago
add a comment |
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
1
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
1
1
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
2
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
$endgroup$
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
add a comment |
$begingroup$
This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}
And the cases of the third and fourth rows are obviously similar.
$endgroup$
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
add a comment |
$begingroup$
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
$endgroup$
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
$endgroup$
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
add a comment |
$begingroup$
$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
$endgroup$
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
add a comment |
$begingroup$
$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
$endgroup$
$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
answered 14 mins ago
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
add a comment |
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
1
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
11 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
5 mins ago
add a comment |
$begingroup$
This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}
And the cases of the third and fourth rows are obviously similar.
$endgroup$
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
add a comment |
$begingroup$
This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}
And the cases of the third and fourth rows are obviously similar.
$endgroup$
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
add a comment |
$begingroup$
This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}
And the cases of the third and fourth rows are obviously similar.
$endgroup$
This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}
And the cases of the third and fourth rows are obviously similar.
edited 8 mins ago
answered 23 mins ago
M. VinayM. Vinay
7,33322136
7,33322136
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
add a comment |
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
19 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
17 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
14 mins ago
add a comment |
$begingroup$
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
$endgroup$
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
add a comment |
$begingroup$
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
$endgroup$
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
add a comment |
$begingroup$
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
$endgroup$
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
edited 22 mins ago
answered 31 mins ago
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
21.8k42959
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
add a comment |
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
24 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
20 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
16 mins ago
add a comment |
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$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
36 mins ago
$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
33 mins ago
1
$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
19 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
19 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
17 mins ago