Using the trace to calculate determinant I've been working on the following problem for my analysis class for quite a while now, but can't seem to find a neat solution to the following question:
Let $B$ be a symmetric, positive definite, $n\times n$  matrix. Then I've already shown the existence of a symmetric, positive definite matrix $A$ such that $B^3 = A^2$.
We've been given a function $$f(x,y) := X^2 - Y^3$$ where both $X$ and $Y$ are symmetric, positive definite matrices. I've calculated $DF(A,B)(H,0)$ (which is $AH+HA$ if I'm not mistaken) for two arbitrary symmetric, positive definite matrices. Next we were supposed to show that, if $M = DF(A,B)(H,0)$, then $$\mathrm{tr}(H^{\mathsf{T}}M) = \mathrm{tr}(H^{\mathsf{T}}AH) + \mathrm{tr}(HAH^{\mathsf{T}}).$$ Using the fact that $A$ is positive definite, we can show that  $\mathrm{tr}(H^{\mathsf{T}}AH) > 0$ and $\mathrm{tr}(HAH^{\mathsf{T}}) > 0$.
The next question was to show that $DF(A,B)(H,0)$ is an invertible linear map from $\mathrm{Mat}(n, \mathbb{R}$) to itself. Now we can do this by showing that the determinant is not zero. Now my guess is that I have to use that $\mathrm{tr}(H^{\mathsf{T}}M) > 0$ but I don't see how to do this.
 A: Forget about using determinant.  Hint: if $e_j$ are the standard unit vectors, $\text{trace}(H A H^T) = \sum_j (H e_j)^T A (H e_j)$.  Since $A$ is positive definite,   $(H e_j)^T A (H e_j)\ge 0$, and $=0$ only if $He_j = 0$. 
A: Let $\varphi : M_{n\times n}(\mathbb{R}) \to M_{n\times n}(\mathbb{R})$ be the linear map:
$$M_{n\times n}(\mathbb{R}) \ni H \mapsto DF(A,B)(H,0) \in M_{n\times n}(\mathbb{R})$$
You have shown $\text{tr}(H^T\varphi(H)) > 0$ whenever $H \ne 0$. For any $X \in \ker(\varphi)$, one have:
$$\varphi(X) = 0 \implies \text{tr}(X^T\varphi(X)) = 0 \implies X = 0$$
This means $\ker(\varphi) = 0$ and $\varphi$ is injective. Because $\dim M_{n\times n}(\mathbb{R}) < \infty$, $\varphi$ is bijective.
Alternatively, you can verify for any $X,Y \in M_{n\times n}(\mathbb{R})$, one have:
$$\text{tr}(X^T\varphi(Y)) = \text{tr}(Y^T\varphi(X))$$
Combine with what you have $\text{tr}(X^T\varphi(X)) > 0$ for $X \ne 0$. 
The matrix corresponds to $\varphi$ in $\color{red}{M_{n^2\times n^2}(\mathbb{R})} \approx \text{Hom}(M_{n\times n}(\mathbb{R}), M_{n\times n}(\mathbb{R}) )$ is positive definite and hence invertible! This again implies $\varphi$ is invertible.
