Operator norm of matrices

Suppose X is a vector space that can be endowed with two different norms, say $\|\cdot\|_1$ and $\|\cdot\|_2$. Consider a linear map $\varphi_1 : (X, \|\cdot\|_1) \rightarrow (Y, \|\cdot\|)$ as well as the linear map $\varphi_2 : (X, \|\cdot\|_2) \rightarrow (Y, \|\cdot\|)$ defined by $\varphi_2(x) := \varphi_1(x)$.

Clearly, the operator norms of $\varphi_1$ and $\varphi_2$ differ in general, i.e. $\|\varphi_1\| \neq \|\varphi_2\|$.

My question concerns the particular case where both $X$ and $Y$ are finite dimensional vector spaces, in which case $\varphi_1$ and $\varphi_2$ are matrices. If I'm not mistaken, the operator norm of a matrix is the square root of the largest eigenvalue. I'm a bit confused as to what $\|\varphi_1\| \neq \|\varphi_2\|$ means in this case. Does this mean that the matrices associated to $\varphi_1$ and $\varphi_2$ will differ? Or does it mean that the norm of one of (or possibly both) $\varphi_1$ and $\varphi_2$ is not the square root of the largest eigenvalue?

• It means that the norm of one of (or possibly both) $\varphi_1$ and $\varphi_2$ is not the square root of the largest eigenvalue. – Omnomnomnom Jul 19 '17 at 17:10
• To add to what @Omnomnomnom already said, the operator norm with respect to the Euclidean norm equals the absolute value of the largest singular value of the matrix, which could be larger than the absolute value of the largest eigenvalue. – fourierwho Jul 19 '17 at 17:37

You're mistaken. If $\|\cdot\|$ denotes the $2$-norm (i.e. the Euclidean norm), then the induced operator norm on matrices is given by $$\|A\| = \max_{\|x\| = 1}\|Ax\| = \sqrt{\lambda_{max}(A^*A)}$$ On the other hand, if $\|\cdot \|_1$ denotes $1$-norm (i.e. the "taxicab" norm), then the induced operator norm is given by $$\|A\|_1 = \max_{\|x\|_1 = 1} \|Ax\|_1 = \max_{j=1,\dots,n} \sum_{i=1}^m |a_{ij}|$$ And it is certainly possible that $\|A\| \neq \|A\|_1$