# operator norm of a matrix and the largest eigenvalue

Is it true that for any $n \times n$ matrix $A$, with real entries, and where all eigenvalues are real, and non zero, the operator norm ( where$||A|| = \max_{|x|=1} |Ax|$ with |.| is the standard length in $\mathbb{R}^{n}) is the same as the largest eigenvalue in absolute value? If it is true, can someone please give me a reference for this result? If it is not true, what if we assume that eigenvalues are also distinct, would that make it true? Thank you for any help. ## 1 Answer No, it's not true. Let$A = \begin{bmatrix} 2 & 1000 \\ 0 & 1 \end{bmatrix}$. The eigenvalues of$A$are$1$and$2$, but the norm of$A$is not$2$. (The norm of$A$is larger than$2$, because$\begin{bmatrix} 2 & 1000 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1000 \\ 1 \end{bmatrix}\$.)

• Sorry, I edited to assume eigenvalues are not zero. – Jessica Jun 2 '15 at 9:06