# Lower bound for norm of matrix

I have the following problem: $A$ is a positive definite, symmetric matrix.

Firstly I was required to find a matrix $B$ such that $B^n = A$. I believe this to be $C(D^{\frac1n}) C'$ where C is the orthogonal matrix of eigenvectors of $A$, and $A = CDC'$.

After this I am asked to find a lower bound for the norm of $B$ as a function of the norm of $A$. It is not specified which norm to take, but by default I took the spectral norm which gave me an equality rather than a bound, because the eigenvalues of $A$ correspond to those of $B$.

Is there something I am missing here? Thanks in advance!

• No, I don't think you are missing anything. – s.harp May 25 '16 at 13:48
• As you have noticed, there is a correspondence between the Eigenvalues of $A$ and $B$, and hence of their norms. Now just write out what this equivalence is, i.e. as a function of $n$ – b00n heT May 25 '16 at 13:49
• If they mean the operator norm, then you have $\|A\| = \|B^n\| \leq \|B\|^n$, hence $\|B\| \geq \|A\|^{1/n}$. – Simon May 25 '16 at 14:04

## 1 Answer

For the spectral norm, you can write a direct relation between the norm of $B$ and the norm of $A$. Since $A$ is symmetric and positive-definite, the spectral norm of $A$ is just the maximal eigenvalue of $A$. Your $B$ is also symmetric and positive-definite and so its norm also equals to the maximal eigenvalue which will be $||A||^{\frac{1}{n}}$.

• Great, thanks so much! – patter2809 May 25 '16 at 13:58