When does $y^T (A^{-1})^T x = y^T (A^{-1}) x$ for invertible $A$
$begingroup$
Background: Trying to do a Cauchy expansion. I want to show that for $M = begin{bmatrix}k & x^T\y & Aend{bmatrix}$, $det(M) = det(A) cdot (k - y^T A^{-1} x)$ for invertible A. Note, M is a 2-by-2 block matrix.
So I take the transpose of M, giving:
$$det begin{bmatrix}k & x^T\y & Aend{bmatrix} = det begin{bmatrix}k & y^T\x & A^Tend{bmatrix}$$
Then using the Schur complement, I get $det begin{bmatrix}k & y^T\x & A^Tend{bmatrix} = det(A^T) cdot det(k - y^T (A^T)^{-1} x) = det(A) cdot (k - y^T (A^{-1})^T x)$.
Close, but not quite what I want.
When $A$ is invertible, are there any rules relating $y^T (A^{-1})^T x$ and $y^T (A^{-1}) x$?
linear-algebra matrices
asked Jan 22 at 7:20
T. FoT. Fo
466311
$endgroup$
add a comment |
$begingroup$
Background: Trying to do a Cauchy expansion. I want to show that for $M = begin{bmatrix}k & x^T\y & Aend{bmatrix}$, $det(M) = det(A) cdot (k - y^T A^{-1} x)$ for invertible A. Note, M is a 2-by-2 block matrix.
So I take the transpose of M, giving:
$$det begin{bmatrix}k & x^T\y & Aend{bmatrix} = det begin{bmatrix}k & y^T\x & A^Tend{bmatrix}$$
Then using the Schur complement, I get $det begin{bmatrix}k & y^T\x & A^Tend{bmatrix} = det(A^T) cdot det(k - y^T (A^T)^{-1} x) = det(A) cdot (k - y^T (A^{-1})^T x)$.
Close, but not quite what I want.
When $A$ is invertible, are there any rules relating $y^T (A^{-1})^T x$ and $y^T (A^{-1}) x$?
linear-algebra matrices
asked Jan 22 at 7:20
T. FoT. Fo
466311
$endgroup$
add a comment |
$begingroup$
Background: Trying to do a Cauchy expansion. I want to show that for $M = begin{bmatrix}k & x^T\y & Aend{bmatrix}$, $det(M) = det(A) cdot (k - y^T A^{-1} x)$ for invertible A. Note, M is a 2-by-2 block matrix.
So I take the transpose of M, giving:
$$det begin{bmatrix}k & x^T\y & Aend{bmatrix} = det begin{bmatrix}k & y^T\x & A^Tend{bmatrix}$$
Then using the Schur complement, I get $det begin{bmatrix}k & y^T\x & A^Tend{bmatrix} = det(A^T) cdot det(k - y^T (A^T)^{-1} x) = det(A) cdot (k - y^T (A^{-1})^T x)$.
Close, but not quite what I want.
When $A$ is invertible, are there any rules relating $y^T (A^{-1})^T x$ and $y^T (A^{-1}) x$?
linear-algebra matrices
asked Jan 22 at 7:20
T. FoT. Fo
466311
$endgroup$
Background: Trying to do a Cauchy expansion. I want to show that for $M = begin{bmatrix}k & x^T\y & Aend{bmatrix}$, $det(M) = det(A) cdot (k - y^T A^{-1} x)$ for invertible A. Note, M is a 2-by-2 block matrix.
So I take the transpose of M, giving:
$$det begin{bmatrix}k & x^T\y & Aend{bmatrix} = det begin{bmatrix}k & y^T\x & A^Tend{bmatrix}$$
Then using the Schur complement, I get $det begin{bmatrix}k & y^T\x & A^Tend{bmatrix} = det(A^T) cdot det(k - y^T (A^T)^{-1} x) = det(A) cdot (k - y^T (A^{-1})^T x)$.
Close, but not quite what I want.
When $A$ is invertible, are there any rules relating $y^T (A^{-1})^T x$ and $y^T (A^{-1}) x$?
linear-algebra matrices
linear-algebra matrices
asked Jan 22 at 7:20
T. FoT. Fo
466311
asked Jan 22 at 7:20
T. FoT. Fo
466311
asked Jan 22 at 7:20
T. FoT. Fo
466311
asked Jan 22 at 7:20
T. FoT. Fo
466311
asked Jan 22 at 7:20
T. FoT. Fo
466311
466311
add a comment |
add a comment |
1 Answer
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The equation $y^{T}(A^{-1})^{T}x=y^{T}A^{-1}x$ holds for all $x$ and $y$ iff $(A^{-1})^{T}=A^{-1}$.
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The equation $y^{T}(A^{-1})^{T}x=y^{T}A^{-1}x$ holds for all $x$ and $y$ iff $(A^{-1})^{T}=A^{-1}$.
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
add a comment |
$begingroup$
The equation $y^{T}(A^{-1})^{T}x=y^{T}A^{-1}x$ holds for all $x$ and $y$ iff $(A^{-1})^{T}=A^{-1}$.
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
add a comment |
$begingroup$
The equation $y^{T}(A^{-1})^{T}x=y^{T}A^{-1}x$ holds for all $x$ and $y$ iff $(A^{-1})^{T}=A^{-1}$.
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
The equation $y^{T}(A^{-1})^{T}x=y^{T}A^{-1}x$ holds for all $x$ and $y$ iff $(A^{-1})^{T}=A^{-1}$.
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 7:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
add a comment |
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
$begingroup$
Just take $x=e_{i}, y=e_{j}$ for $i,j=1, ldots, n$. Then its clear. Here $e_i$ is a coloumn vector with 1 on $i^{th}$ position and 0 everywhere elses.
$endgroup$
– Shubham Namdeo
Jan 22 at 8:08
add a comment |
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