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Eigenvalues of $2$ symmetric $4times 4$ matrices: why is one negative of the other?
Congruence of invertible skew symmetric matricesEigenvalues of a general block hermitian matrixEigenvalues of Overlapping block diagonal matricesBuilding matrices from eigenvalueseigenvalues for certain hermitian matricesEigenvalues and eigenspaces in a symmetric matrixThe matrix of an endomorphismProve 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:
$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
and a very similar matrix:
$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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 $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ 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.
$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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:
$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
and a very similar matrix:
$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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 $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ 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.
$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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
31 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
29 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
15 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago
add a comment |
$begingroup$
Consider the following symmetric matrix:
$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
and a very similar matrix:
$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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 $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ 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.
$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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:
$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
and a very similar matrix:
$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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 $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ 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.
$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$
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 2 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
31 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
29 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
15 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago
add a comment |
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
31 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
29 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
15 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
31 mins ago
$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
31 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
29 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
29 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
15 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
15 mins ago
2
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$-M_1=D^-1M_0D$$
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,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$
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
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs 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 $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
beginalign*
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).
endalign*
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
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 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
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
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
endalign*
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
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)
endalign*
This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
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
endalign*
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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$-M_1=D^-1M_0D$$
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,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$
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
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs ago
add a comment |
$begingroup$
$$-M_1=D^-1M_0D$$
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,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$
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
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs ago
add a comment |
$begingroup$
$$-M_1=D^-1M_0D$$
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,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$
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.
$endgroup$
$$-M_1=D^-1M_0D$$
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,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$
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 9 mins ago
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs ago
add a comment |
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs ago
1
1
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
6 mins ago
$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
6 mins ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs ago
$begingroup$
okay, this is amazing..
$endgroup$
– Troy
43 secs 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 $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
beginalign*
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).
endalign*
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
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 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 $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
beginalign*
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).
endalign*
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
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 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 $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
beginalign*
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).
endalign*
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 $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$
Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.
For,
beginalign*
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).
endalign*
And the cases of the third and fourth rows are obviously similar.
edited 4 mins ago
answered 18 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
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 mins ago
add a comment |
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 mins ago
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
15 mins ago
$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
15 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 mins ago
$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
12 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
10 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
10 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
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
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
endalign*
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
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)
endalign*
This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
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
endalign*
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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 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
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
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
endalign*
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
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)
endalign*
This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
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
endalign*
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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 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
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
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
endalign*
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
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)
endalign*
This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
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
endalign*
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
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
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
endalign*
Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
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)
endalign*
This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$
In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
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
endalign*
A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.
edited 17 mins ago
answered 27 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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 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
19 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
19 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
16 mins ago
$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
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
12 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
12 mins ago
add a comment |
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$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
31 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
29 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
15 mins ago
2
$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
14 mins ago
$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
12 mins ago