Spectral radius

Question

Suppose I have a matrix $M$ and $||M||_2$ denotes the spectral radius of the matrix.

I came across a note which says $||M||_2 \leq \sqrt{||M||_1||M||_\infty}$.

Could someone explain to me how this inequality holds? I don't understand it.

Answer 1

Let $v$ be a vector of length $1$.

$$||Mv||^2 = \sum_{i=1}^n \left(\sum_{j=1}^n m_{ij}v_j \right)^2 \\ \leq \sum_{i=1}^n \left(\sum_{j=1}^n m_{ij}^2 \right) \left(\sum_{j=1}^n v_j^2 \right) \\ = \sum_{i=1}^n \sum_{j=1}^n m_{ij}^2 = \sum_{i=1}^n \sum_{j=1}^n |m_{ij}| |m_{ij}| \\ \leq \sum_{i=1}^n \sum_{j=1}^n |m_{ij}| ||M||_{\infty} \\ = ||M||_1 ||M||_{\infty} $$

• In the first inequality we use Cauchy–Schwarz.
• In the last inequality we use the fact $|m_{ij}| \leq ||M||_{\infty}$ for all $i,j$.
• The last equality is the definition of $||M||_1$, since the constant $||M||_{\infty}$ can be pulled out of the sum.

Answer 2

$$||M||_2^2=\rho(M^*M)\leq||M^*M||_1\leq||M^*||_1||M||_1=||M||_{\infty}||M||_1$$