A matrix is similar to its transpose without jordan form
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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
13112
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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
13112
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
13112
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
13112
asked Nov 18 at 12:13
Jimmy
13112
asked Nov 18 at 12:13
Jimmy
13112
asked Nov 18 at 12:13
Jimmy
13112
asked Nov 18 at 12:13
Jimmy
13112
13112
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
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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 |
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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