Is $r.b*=b*A$ true?
$begingroup$
Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?
vectors matrix-equations
edited Jan 2 at 18:27
Omnomnomnom
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
$endgroup$
add a comment |
$begingroup$
Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?
vectors matrix-equations
edited Jan 2 at 18:27
Omnomnomnom
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
$endgroup$
$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41
add a comment |
$begingroup$
Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?
vectors matrix-equations
edited Jan 2 at 18:27
Omnomnomnom
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
$endgroup$
Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?
vectors matrix-equations
vectors matrix-equations
edited Jan 2 at 18:27
Omnomnomnom
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
edited Jan 2 at 18:27
Omnomnomnom
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
edited Jan 2 at 18:27
Omnomnomnom
127k790178
edited Jan 2 at 18:27
Omnomnomnom
127k790178
edited Jan 2 at 18:27
Omnomnomnom
127k790178
127k790178
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
asked Jan 2 at 18:09
Şenay ErdinçŞenay Erdinç
11
11
$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41
add a comment |
$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41
$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
Taking the conjugate transpose, we get
$$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
Which we need to be equal to
$$(b^*A)^* = A^*b$$
Since matrix multiplication is associative, we can explicitly write
$$ ba^*A^*b = ((ba^*)A^*)b$$
Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
$endgroup$
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
answered Jan 2 at 19:06
OmnomnomnomOmnomnomnom
127k790178
127k790178
add a comment |
add a comment |
$begingroup$
Taking the conjugate transpose, we get
$$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
Which we need to be equal to
$$(b^*A)^* = A^*b$$
Since matrix multiplication is associative, we can explicitly write
$$ ba^*A^*b = ((ba^*)A^*)b$$
Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
$endgroup$
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
add a comment |
$begingroup$
Taking the conjugate transpose, we get
$$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
Which we need to be equal to
$$(b^*A)^* = A^*b$$
Since matrix multiplication is associative, we can explicitly write
$$ ba^*A^*b = ((ba^*)A^*)b$$
Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
$endgroup$
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
add a comment |
$begingroup$
Taking the conjugate transpose, we get
$$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
Which we need to be equal to
$$(b^*A)^* = A^*b$$
Since matrix multiplication is associative, we can explicitly write
$$ ba^*A^*b = ((ba^*)A^*)b$$
Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
$endgroup$
Taking the conjugate transpose, we get
$$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
Which we need to be equal to
$$(b^*A)^* = A^*b$$
Since matrix multiplication is associative, we can explicitly write
$$ ba^*A^*b = ((ba^*)A^*)b$$
Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
answered Jan 2 at 19:09
AlexanderJ93AlexanderJ93
6,158823
6,158823
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
add a comment |
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
$begingroup$
Тhanks a lot for quick answers.
$endgroup$
– Şenay Erdinç
Jan 2 at 19:22
add a comment |
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$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28
$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37
$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41