Question from worksheet, I don't fully understand the solution the teacher gave.
Question: let $S$ be the set of symmetric positive definite matrices of dimension $n$x$n$.
Let $T: S \to S$, $T(A)=A^2$
Show that $T$ is locally invertible at every point.
Teacher's proof:
The differential $D_TH=AH+HA$, this is true,I checked it using the definition $Df(a)H=\lim_{t \to 0}\frac{f(a+th)-f(a)}{t}$.
We want to show that $D_TH$ is invertible, it is the same as showing that $AH+HA$ is not degenerate, meaning $Ker(AH+HA)=0$, meaning if $AH+HA=0$ then $H=0$. that is what we will show.
and indeed, if $AH=-HA$, then since $A$ is symmetric positive definite, then $A=PDP^{-1}$ where $P$ is orthogonal and $D$ is diagonal with positive values only.
(Here is the part I don't understand) if $H=PQP^{-1}$ where $Q$ is diagonal, then we get $DQ=-QD$ this implies that $d_iq_i=-d_iq_i$ for all $i$. since $d_i >0$, this must mean that $q_i=0$ and so $H=0$.
Since $AH=-HA$ implies $H=0$, this means that $AH+HA$ is not degenerate, and so the differential at any point is invertible, from the inverse function theorem, we can deduct that indeed $T$ is locally invertible at any point.
What I don't understand in this proof is, how can we assume we can write $H=PQP^{-1}$? We don't know if $H$ is diagonlizable, and certainly we can't say that the matrix that diagonlizes it is the same as the one that diagonalizes $A$. Why is this proof valid?
The matrices A and H are not necessary diagonalizable by the same matrix. But we can still prove what we need. Since $A$ is symmetric and positive definite, it follows that there exists a basis $\{f_j\}$ of eigenvectors of $A$, $Af_j=\lambda_jf_j$, $\lambda_j>0$, $j=1,\dots,n$. Now if $AH+HA=0$, then $$ 0=\langle f_j,(AH+HA)f_k\rangle=(\lambda_j+\lambda_k)\langle e_j,He_k\rangle, $$ because both $A$ and $H$ are symmetric. Hence $\langle e_j,He_k\rangle=0$ for all $j$ and $k$, and hence $H=0$.