A matrix is similar to its transpose without jordan form
up vote
0
down vote
favorite
Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then
$(A^∗)^{−1} = (A^{−1})^∗$
Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).
I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.
Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.
matrices numerical-linear-algebra matrix-decomposition transpose
asked Nov 18 at 12:13
Jimmy
15312
add a comment |
up vote
0
down vote
favorite
Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then
$(A^∗)^{−1} = (A^{−1})^∗$
Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).
I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.
Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.
matrices numerical-linear-algebra matrix-decomposition transpose
asked Nov 18 at 12:13
Jimmy
15312
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then
$(A^∗)^{−1} = (A^{−1})^∗$
Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).
I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.
Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.
matrices numerical-linear-algebra matrix-decomposition transpose
asked Nov 18 at 12:13
Jimmy
15312
Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then
$(A^∗)^{−1} = (A^{−1})^∗$
Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).
I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.
Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.
matrices numerical-linear-algebra matrix-decomposition transpose
matrices numerical-linear-algebra matrix-decomposition transpose
asked Nov 18 at 12:13
Jimmy
15312
asked Nov 18 at 12:13
Jimmy
15312
asked Nov 18 at 12:13
Jimmy
15312
asked Nov 18 at 12:13
Jimmy
15312
asked Nov 18 at 12:13
Jimmy
15312
15312
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08
add a comment |
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
add a comment |
up vote
2
down vote
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
add a comment |
up vote
2
down vote
up vote
2
down vote
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
edited Nov 22 at 14:28
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
answered Nov 22 at 13:41
Algebraic Pavel
16.1k31839
16.1k31839
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003460%2fa-matrix-is-similar-to-its-transpose-without-jordan-form%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35
"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08