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

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?

• Do you understand it if $P=I$? I.e. do you understand why cube rooting the entries of a diagonal matrix gives you a cube root to the diagonal matrix? This is the same thing, just in a different basis. You wouldn't want to change the eigenvectors.
– anon
Commented Dec 28, 2022 at 21:23

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.