# $A$ 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.

• hint: also use the positive defineted-ness of the matrix Sep 9 '20 at 19:13
• @learner I don't see why it's useful to prove $U = V$ since $D$ for the SVD has positive values anyway Sep 9 '20 at 19:19
• @learner tried a lot, but all I can get is that $A^2 = A^TA = VD^2V^T = U D^2 U^T = AA^T$... But I don't know how to define $X$ :( Sep 9 '20 at 19:49
• there is no merit in finding $A^2$. If at all you should be squaring anything it should be $X$ ;) Sep 9 '20 at 20:12

$$A$$ 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 $$U=V=X$$.

• I still can't understand why $U = V$. I have never used this argument of right and left eigenvectors for solving SVD problems. Could you detail more this part, please? The conclusion after $U = V$ I can understand. :) Sep 10 '20 at 13:59
• @Figurinha I have added more explanation. Let me know if it is clear. Like we have a definition for right eigenvector (or "eigenvector") which is $Ax = \lambda x$, we have a similar definition for left eigenvector which is, $x$ is left eigenvector if $x \neq 0, x^TA = \lambda x$. In the definition of SVD, the columns of $U$ are left eigenvalues. Sep 10 '20 at 15:29