Suppose that I have an $n\times n$ real matrix $A$ such that $PA^T = AP$ for every invertible matrix $P$. Does this imply that $A$ is a multiple of the identity matrix (i.e., $A= c\, I_{n\times n}$ for some $c\in \mathbb{R}$.) ?
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The following proof works over any field, not just $\mathbb R$:
Put $P=I$, we see that $A$ is real symmetric. Hence $PA=AP$ for every invertible $P$. Since the set of all invertible matrices spans the whole matrix space, we in turn get $XA=AX$ for every square matrix $X$. Now it is well known (and also an easy exercise) that a matrix that commutes with every other square matrix must be a scalar multiple of $I$.
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Can you clarify please. Why invertible matrices are dense in the $\mathbb{R}^{n\times n}$?. If so, how does this imply that $AX=XA$ for each $X\in \mathbb{R}^{n\times n}$ ? – Hussein Eid Apr 05 '17 at 17:31
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@HusseinEid This is because for every matrix $X$, the perturbed matrix $P_t = X+tI$ is invertible when $t>0$ is sufficiently small. As $P_tA=AP_t$, we get $XA=AX$ if we take the limit $t\to0$. – user1551 Apr 05 '17 at 18:33
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Excuse me sir, why $P_t$ is invertible for sufficiently small $t > 0$ ? I tried to prove that by checking that the determinant is nonzero but I failed . – Hussein Eid Apr 05 '17 at 20:03
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@HusseinEid $X$ has finitely many eigenvalues. So, $-t$ will eventually become not an eigenvalue when $t$ is small enough. It follows that $X+tI$ is invertible. – user1551 Apr 05 '17 at 20:12
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Yes, indeed we have $A^T= P^{-1}AP$ for every invertible matrix $P$. In particular if $P$ is the identity, then $A^T=A$. Hence $A$ is symmetric and thus have a real eigenvector, say $v$, such that $Av = \lambda v$. Now, note that for any invertible matrix $P$, we have $$ A Pv = PA^T v = PAv = P\lambda v = \lambda Pv,$$ i.e. $Pv$ is also an eigenvector of $A$. Since for every $w\neq 0$, there exists an invertible matrix $P$ such that $Pv = w$, it follows that $Au = \lambda u$ for every $u \neq 0$. Hence $A$ must be a multiple of the identity, i.e. $A=\lambda I$.
Surb
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I appreciate your help. But there is a point that worth discussing. If $w \neq 0$, then there is an $n\times n$ matrix $P = (w v^T)/ (v^T v)$ ($v^T v$ is a $1\times 1$ matrix or simply a scalar). such that $Pv=w$. This is what you stated. But there is no guarantee that $P$ is invertible ! – Hussein Eid Apr 05 '17 at 17:15
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@HusseinEid The matrix you stated is indeed not invertible. However, you could take the matrix $P = I-2\frac{zz^T}{z^Tz}$ with $z=v-w$. Then $P$ is orthogonal, i.e. $P^{-1}=P^T$ (it is a Householder matrix) and $Pv = w$. Another way to obtain such a matrix is the following: complete $v$ into a basis $v_1=v,v_2,\ldots,v_n$ and $w$ into another basis $w_1=w,w_2,\ldots,w_n$ and consider the basis change matrix that maps $v_i$ to $w_i$ for every $i$. – Surb Apr 05 '17 at 18:09
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It is not clear at all then $P$ is invertible ! . It is clear that $P^T = P$. But, $PP^T = PP \neq I_{n\times n}$ . – Hussein Eid Apr 05 '17 at 20:31
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@HusseinEid This is a well known property: $$PP = (I-2\frac{zz^T}{z^Tz})(I-2\frac{zz^T}{z^Tz})= I-4\frac{zz^T}{z^Tz}+4\frac{z(z^Tz)z^T}{(z^Tz)^2}= I-4\frac{zz^T}{z^Tz}+4\frac{(z^Tz)zz^T}{(z^Tz)^2}=I.$$ – Surb Apr 05 '17 at 20:40
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@HusseinEid No problem, sorry from my side for the bit unnice tone. Anyway, I think the Householder matrix is a good one to remember as it has plenty of nice properties and is easy to build. – Surb Apr 05 '17 at 21:43
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Here's a very elementary proof.
First, as the other answers already stated, if you set $P=I$, you directly get that $A^T=A$. Therefore in all the following we can just omit the transposition.
Next, take for $P$ the diagonal matrix that has $-1$ as the $k$-th diagonal element and $1$ for the others. Then $PA=AP$ means the matrix that you get when reversing all signs in the $k$-th row is equal to the matrix that you bet by reversing the signs in the $k$-th column. This is only possible if all nondiagonal elements in the $k$-th row and $k$-th column are $0$, and since this is true for all $k$, $A$ must be diagonal.
Finally, by using permutation matrices as $P$ you find that all diagonal elements must be equal. Therefore $A$ is a multiple of the unit matrix.
On the other hand, for any multiple of the unit matrix it is easily verified that $PA^T=AP$ for any $P$, so there are no further conditions on $A$.
celtschk
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