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I'm struggling to make any progress with this question. I have defined C as the standard n-dimensional identity matrix. As A is semidefinite, I believe the diagonal matrix D must have positive values? How can I show this? Further, how does this help answer the original question?

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By the finite-dimensional spectral theorem there exists an orthogonal matrix $C$ (i.e. $C^-1 = C^T$ such that $C^TAC = D = diag(d_1,\ldots, d_n)$ is a diagonal matrix. Let $e_k \in \mathbb{R}^n$ denote the vector, which $k$-th coordinate is $1$ and all others are $0$. Then $$d_k = e_k^T D e_k = e_k^T C^T A Ce_k = (Ce_k)^T A (Ce_k) \geq 0.$$ Hence, all entries of $D$ are $\geq 0$. This means we can consider the square root $\sqrt{d_k}$ of $d_k$ and define $E := diag(\sqrt{d_1},\ldots, \sqrt{d_k})$. It follows that $$E^2 = diag(\sqrt{d_1},\ldots, \sqrt{d_k}) \cdot diag(\sqrt{d_1},\ldots, \sqrt{d_k}) = diag(d_1,\ldots, d_k) = D = C^T A C.$$ Using that $C^T = C^{-1}$, we see that $$A = CE \cdot EC^T = CEC^T \cdot CEC^T.$$ Now, if we define $B = CEC^T$, then $$B^T = (CEC^T)^T = (C^T)^T E^T C^T = CEC^T = B$$ and $A = B^2$.

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I have defined C as the standard n-dimensional identity matrix.

Don't do that; choose the real unitary (orthogonal) $C$ that diagonalizes $A$. That is, let $C$ be a matrix whose columns are an orthonormal eigenbasis of $A$, so that $C^T = C^{-1}$ and $D = C^TAC$. Note that $A = CDC^T$.

As A is semidefinite, I believe the diagonal matrix D must have positive values? How can I show this?

Let $v = Ce_i$, where $e_1,\dots,e_n$ are the standard basis vectors. Compute $v^TAv$, noting that $A = CDC^T$.

Further, how does this help answer the original question?

Try to find a matrix $E$ so that $E^2 = D$, then define $B = C^TEC$. In particular, make your choice of $E$ diagonal, with positive entries on the diagonal.

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