I'm just getting to know matrix computations can someone please help with this two proofs?
Prove that for semidefinite $A \succeq 0$, $(A \in R^{nxn})$ and for arbitrary $U \in R^{nxm}$, the matrix $B=U^{T}AU$ is semidefinite too: $B \succeq 0$
Prove that for semidefinite matrix, all its powers are also semidefinite:
$A \succeq 0 \Rightarrow A^k \succeq 0$ for all positive integers $k \in Z_{+}$
A matrix $A\in\mathbb{R}^{n\times n}$ is positive semidefinite if and only if for all $x\in\mathbb{R}^{n},$ $x^{T}Ax\geq0.$ Then in particular, this is true for $x$ of the form $Uy$ for some $y\in\mathbb{R}^{m},$ which is to say that $(Uy)^{T}A(Uy)\geq0$ for all $y\in\mathbb{R}^{m},$ which means that $y^{T}By\geq0$ for all $y\in\mathbb{R}^{m},$ proving that $B$ is positive semidefinite.
Another characterization for positive semidefiniteness is that $A\in\mathbb{R}^{n\times n}$ is positive semidefinite if (and only if) it is Hermitian and all of its eigenvalues are nonnegative. If this is the case, then $(A^{k})^{T}=(A^{T})^{k}=A^{k},$ so $A^{k}$ is Hermitian, and the eigenvalues of $A^{k}$ are all of the form $\lambda^{k}$ for $\lambda\in\sigma(A).$ Since all of the eigenvalues of $A$ satisfy $\lambda\geq0$, $\lambda^{k}\geq0$ and therefore all of the eigenvalues of $A^{k}$ are nonnegative. Thus, $A^{k}$ is positive semidefinite for all $k\geq0.$