Show that $A= B^3$ has a solution if $A$ is symmetric?

Question

Let $A$ be a symmetric $n \times n$ matrix. Show that the equation $B^3 = A$ has a solution.

I tried solving it with the help of an answer sheet but there is something I don't understand. Anyways here is how I've come so far.

Since $A$ is symmetric, it can be written as $A = PDP^T$. And then what they did was somehow manipulate the diagonal. As in let $X$ be equal to a new diagonal matrix where each diagonal element is raised to $\dfrac{1}{3}$.

Then they did: $PXP^T$= $B$.

My question is that how come $PXP^T$ is even defined when $X$ is not made of the eigenvalues. I mean why does it work? Why aren't the eigenvectors affected by this change?

Answer 1

I'll explain why the math works, then explain the intuition. Once you know that $A=PDP^T$ where $D$ is a diagonal matrix and $P$ is an orthogonal matrix (so that $P^TP$ is the identity), then by setting $X=D^{1/3}$ we can produce a "candidate" matrix $PXP^T$ which we will show satisfies the property: $$ (PXP^T)^3=A,\qquad (\star) $$ and therefore $PXP^T$ can be used as one choice for a $B$ which satisfies $B^3=A$. How do we show $(\star)$? Simply by calculating it out: $$ (PXP^T)^3=(PXP^T)(PXP^T)(PXP^T)=PX(P^TP)X(P^TP)XP^T. $$ But since $P$ is orthogonal, this means $P^TP$ is the identity, so we get $$ (PXP^T)^3=PX^3P^T=PDP^T=A. $$

Now for the intuition: by introducing the $P$ matrix, we are effectively changing basis, from the original basis in which $A$ is just some arbitrary symmetric basis, to a new basis where it is diagonal. And then we can just do the calculations with a diagonal matrix, which is much easier.