Mixing
![What exactly is a function? [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Eigen values of this block matrix?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose that $M$ is an $mtimes m$ matrix.
Can we find the eigen values of the block matrix
beginequation
B=
beginbmatrix
-I&M\
M^T&-I
endbmatrix
endequation
in terms of the eigenvlaues of $M$?
linear-algebra eigenvalues-eigenvectors
asked Jul 19 at 0:30
user35616
294
add a comment |Â
up vote
1
down vote
favorite
Suppose that $M$ is an $mtimes m$ matrix.
Can we find the eigen values of the block matrix
beginequation
B=
beginbmatrix
-I&M\
M^T&-I
endbmatrix
endequation
in terms of the eigenvlaues of $M$?
linear-algebra eigenvalues-eigenvectors
asked Jul 19 at 0:30
user35616
294
4
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
1
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $M$ is an $mtimes m$ matrix.
Can we find the eigen values of the block matrix
beginequation
B=
beginbmatrix
-I&M\
M^T&-I
endbmatrix
endequation
in terms of the eigenvlaues of $M$?
linear-algebra eigenvalues-eigenvectors
asked Jul 19 at 0:30
user35616
294
Suppose that $M$ is an $mtimes m$ matrix.
Can we find the eigen values of the block matrix
beginequation
B=
beginbmatrix
-I&M\
M^T&-I
endbmatrix
endequation
in terms of the eigenvlaues of $M$?
linear-algebra eigenvalues-eigenvectors
asked Jul 19 at 0:30
user35616
294
edited Jul 28 at 19:47
asked Jul 19 at 0:30
user35616
294
asked Jul 19 at 0:30
user35616
294
asked Jul 19 at 0:30
user35616
294
294
4
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
1
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38
add a comment |Â
4
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
1
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38
4
4
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
1
1
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
You can relate the eigenvalues of $B$ to those of $M^TM$. First of all, we have
$$
(B-lambda)binom x y = binom 0 0quadLongleftrightarrowquad My = (1+lambda)xquadtextandquad M^Tx = (1+lambda)y.
$$
From here, it is quite elementary to show that $lambda$ is an eigenvalue of $B$ if and only if $(1+lambda)^2$ is an eigenvalue of $M^TM$. This also shows you that if $B$ has an eigenvalue $lambda$, then also $-2-lambda$ is an eigenvalue of $B$. That is, the eigenvalues of $B$ are symmetric with respect to the point $-1$.
Even if $M$ is square, you can in general not relate the eigenvalues of $B$ to those of $M$ (except the zero eigenvalue). Only if $M$ is symmetric and positive semi-definite, then $lambda$ is an eigenvalue of $B$ if and only if $|1+lambda|$ (i.e., the distance of $lambda$ to $-1$) is an eigenvalue of $M$.
answered Jul 19 at 3:05
amsmath
1,613114
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can relate the eigenvalues of $B$ to those of $M^TM$. First of all, we have
$$
(B-lambda)binom x y = binom 0 0quadLongleftrightarrowquad My = (1+lambda)xquadtextandquad M^Tx = (1+lambda)y.
$$
From here, it is quite elementary to show that $lambda$ is an eigenvalue of $B$ if and only if $(1+lambda)^2$ is an eigenvalue of $M^TM$. This also shows you that if $B$ has an eigenvalue $lambda$, then also $-2-lambda$ is an eigenvalue of $B$. That is, the eigenvalues of $B$ are symmetric with respect to the point $-1$.
Even if $M$ is square, you can in general not relate the eigenvalues of $B$ to those of $M$ (except the zero eigenvalue). Only if $M$ is symmetric and positive semi-definite, then $lambda$ is an eigenvalue of $B$ if and only if $|1+lambda|$ (i.e., the distance of $lambda$ to $-1$) is an eigenvalue of $M$.
answered Jul 19 at 3:05
amsmath
1,613114
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
add a comment |Â
up vote
0
down vote
You can relate the eigenvalues of $B$ to those of $M^TM$. First of all, we have
$$
(B-lambda)binom x y = binom 0 0quadLongleftrightarrowquad My = (1+lambda)xquadtextandquad M^Tx = (1+lambda)y.
$$
From here, it is quite elementary to show that $lambda$ is an eigenvalue of $B$ if and only if $(1+lambda)^2$ is an eigenvalue of $M^TM$. This also shows you that if $B$ has an eigenvalue $lambda$, then also $-2-lambda$ is an eigenvalue of $B$. That is, the eigenvalues of $B$ are symmetric with respect to the point $-1$.
Even if $M$ is square, you can in general not relate the eigenvalues of $B$ to those of $M$ (except the zero eigenvalue). Only if $M$ is symmetric and positive semi-definite, then $lambda$ is an eigenvalue of $B$ if and only if $|1+lambda|$ (i.e., the distance of $lambda$ to $-1$) is an eigenvalue of $M$.
answered Jul 19 at 3:05
amsmath
1,613114
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can relate the eigenvalues of $B$ to those of $M^TM$. First of all, we have
$$
(B-lambda)binom x y = binom 0 0quadLongleftrightarrowquad My = (1+lambda)xquadtextandquad M^Tx = (1+lambda)y.
$$
From here, it is quite elementary to show that $lambda$ is an eigenvalue of $B$ if and only if $(1+lambda)^2$ is an eigenvalue of $M^TM$. This also shows you that if $B$ has an eigenvalue $lambda$, then also $-2-lambda$ is an eigenvalue of $B$. That is, the eigenvalues of $B$ are symmetric with respect to the point $-1$.
Even if $M$ is square, you can in general not relate the eigenvalues of $B$ to those of $M$ (except the zero eigenvalue). Only if $M$ is symmetric and positive semi-definite, then $lambda$ is an eigenvalue of $B$ if and only if $|1+lambda|$ (i.e., the distance of $lambda$ to $-1$) is an eigenvalue of $M$.
answered Jul 19 at 3:05
amsmath
1,613114
You can relate the eigenvalues of $B$ to those of $M^TM$. First of all, we have
$$
(B-lambda)binom x y = binom 0 0quadLongleftrightarrowquad My = (1+lambda)xquadtextandquad M^Tx = (1+lambda)y.
$$
From here, it is quite elementary to show that $lambda$ is an eigenvalue of $B$ if and only if $(1+lambda)^2$ is an eigenvalue of $M^TM$. This also shows you that if $B$ has an eigenvalue $lambda$, then also $-2-lambda$ is an eigenvalue of $B$. That is, the eigenvalues of $B$ are symmetric with respect to the point $-1$.
Even if $M$ is square, you can in general not relate the eigenvalues of $B$ to those of $M$ (except the zero eigenvalue). Only if $M$ is symmetric and positive semi-definite, then $lambda$ is an eigenvalue of $B$ if and only if $|1+lambda|$ (i.e., the distance of $lambda$ to $-1$) is an eigenvalue of $M$.
answered Jul 19 at 3:05
amsmath
1,613114
answered Jul 19 at 3:05
amsmath
1,613114
answered Jul 19 at 3:05
amsmath
1,613114
answered Jul 19 at 3:05
amsmath
1,613114
1,613114
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
add a comment |Â
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
I get how $lambda$ is an eigenvalue of $B$ implies that $(lambda + 1)^2$ is an eigenvalue of $M^TM$ and $MM^T$; but I don't get how the reverse implication works . . .
– Robert Lewis
Jul 19 at 4:25
1
1
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
If $MM^Tx = (1+lambda)^2x$, put $y = tfrac 11+lambdaM^Tx$.
– amsmath
Jul 19 at 10:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856145%2feigen-values-of-this-block-matrix%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
$M$ does not have eigenvalues if $mneq n$.
– amsmath
Jul 19 at 0:38
You can use the Schur complement to relate the eigenvalues $neq -1$ of $B$ to the eigenvalues of $M^TM$ (or, equivalently, the singular values of $M$) which are not $0$. Also, $-1$ is an eigenvalue of $B$ if and only if $0$ is an eigenvalue of $M^TM$.
– amsmath
Jul 19 at 0:45
1
When $M$ is a real square matrix, the eigenvalues of $B$ are $-1pm s_i$, where $s_1,ldots,s_n$ are the singular values of $M$. They are functions of singular values rather than eigenvalues of $M$.
– user1551
Jul 19 at 2:59
How does this relate to linear programming?
– Erwin Kalvelagen
Jul 20 at 14:38