$A$ is a symmetric positive-definite matrix it has square root using SVD 
What I want:
If $A$ is SPD then there is a matrix $X$ such that $A = X^2$

I am able to prove this using the Schur Decomposition, but I was asked to prove it using SVD decomposition.
I was trying to prove that if $A = UDV^T = VDU^T = A^T$ then $U = V$, but I'm failing to do that.
 A: $A$ is symmetric. $A$ possesses complete set of orthogonal eigenvectors. So we have $Ax = \lambda x$, where $x$ a right eigenvector corresponding to eigenvalue $\lambda$. Now, it is also true that $x^TA = \lambda x^T$, and so $x$ is also a left eigenvector. Therefore we have in the SVD, the orthonormal matrices $U$ and $V$ are the same. Now
\begin{align}
A &= UDV^T \\ &\stackrel{a}= UD^{\frac{1}{2}}D^{\frac{1}{2}}V^T \\ &\stackrel{b}= UD^{\frac{1}{2}}D^{\frac{1}{2}}U^T \\ &\stackrel{c}= UD^{\frac{1}{2}}V^TVD^{\frac{1}{2}}U^T \\ &\stackrel{d}= X.X  \\ &= X^2
\end{align}

*

*a: Since A is PD, the eigenvalues are positive so a halving operation is allowed.

*b: $U=V$

*c: $V^TV$ = I

*d: Defining $X$ to be a matrix with eigenvalues half power of the corresponding eigenvalues of $A$, eigenvectors same as that of $A$, and the fact that $U=V$
For every symmetric matrix $\in \mathbb{R}^{n \times n}$ there exist $n$-linearly independent eigenvectors. And for any matrix, for any eigenvalue, we have algebraic multiplicity $\geq$ geometric multiplicity of the. Using the above two statements, one can conclude that a symmetric matrix has $n$ eigenvalues.
One can represent any symmetric PD matrix in its eigenvalue decomposition form. The way one would derive this is take the idea $Ax = \lambda x$ to a matrix level to get $AX = XD$, where $D$ is a diagonal matrix with eigenvalues in its diagonal. These eigenvalues are arranged so that they get multiplied with their corresponding eigenvector in $X$ (columns of $X$). Since $X$ is orthonormal, $X^T = X^{-1}$ and one can write $A =XDX^T$, which also happens to be the SVD of $A$ with $U=V=X$.
