# Operator norm is equal to max eigenvalue

Take a matrix $$A \in M_{2 \times 2}(\mathbb{R})$$ and consider the norm $$\vert\vert A\vert\vert = \sup\limits_{x \in \mathbb{R}^2} \frac{ \vert\vert Ax\vert\vert}{\vert\vert x \vert\vert} = \sup\limits_{x, \vert\vert x \vert\vert = 1} \vert\vert Ax\vert\vert$$.

I am unable to see that the norm must be less than or equal to the maximum eigenvalue of $$A$$:

$$\vert\vert A\vert\vert \le \max\limits_{\lambda \in \sigma(A)} \lambda$$

and I am also unable of characterizing the type of $$A$$ such that:

$$\vert\vert A\vert\vert = \max\limits_{\lambda \in \sigma(A)} \lambda$$

• What is your norm $\|\cdot \|$? – Jacky Chong Apr 3 '18 at 5:32
• It is the euclidean norm/ 2-norm – blanchey Apr 3 '18 at 5:33

1. If $x$ is a unit eigenvector corresponding to eigenvalue $\lambda$, then $\|Ax\| = |\lambda|$, so $|\lambda| \le \|A\|$. So the inequality that you stated should be reversed.

2. If $A$ is diagonalizable with respect to an orthonormal basis $\{v_1, v_2\}$ with eigenvalues $\lambda_1$ and $\lambda_2$, then for any $x=c_1 v_1 + c_2 v_2$ we have $Ax = c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2$ so $\frac{\|Ax\|^2}{\|x\|^2} = \frac{c_1^2 \lambda_1^2 + c_2 \lambda_2^2}{c_1^2 + c_2^2} \le \max\{\lambda_1^2, \lambda_2^2\}$, with equality if $x$ is an eigenvector corresponding to the largest eigenvalue (in absolute value). This situation happens whenever $A$ is symmetric, or more generally, normal ($A^\top A = A A^\top$).

An example for strict inequality is if $A$ is similar to a matrix of the form $\begin{bmatrix} 0 & 1 \\ & 0\end{bmatrix}$, in which case the eigenvalues are all zero, but the operator norm is $1$.

These facts can be generalized to arbitrarily sized square matrices as well.

It is not true that $||A|| \leq \max \{|\lambda|\}$ in general. The reverse inequality always holds: $Ax=\lambda x,x \neq 0$ implies $|\lambda| ||x|| \leq ||A|| ||x||$ so $|\lambda| \leq ||A||$ for any eigen value $\lambda$. Equality holds if $A$ is symmetric.

Observe we have that \begin{align} \|Ax\|_2^2= x^TA^TAx \end{align} where $A^TA$ is diagonalizable since it's symmetric. Then we see that \begin{align} x^TA^TAx = y^TDy \end{align} where $D$ consists of the square of the singular values. Hence, we get that \begin{align} \|Ax\|_2^2 = \lambda_1^2y_1^2+\lambda_2^2y_2^2. \end{align} In the case when $\|x\|_2=1$ then we also have that $\|y\|_2=1$. Hence it follows \begin{align} \|Ax\|_2^2 \leq \max_{1\le i \le 2}\lambda_i^2 \ \ \implies \ \ \|Ax\|_2 \leq \max_{1 \le i \le 2}\lambda_i. \end{align}

Note that I say singular value, not eigenvalue.

Additional: Consider \begin{align} A= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \end{align} then we see that \begin{align} \sup_{\|x\|_2=1}\|Ax\|_2 = 1 \end{align} but the eigenvalues of $A$ are $0$. Moreover, observe that \begin{align} A^TA= \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} \end{align} which means the singular values of $A$ are equal to $1$ and $0$.

• I think this only works when $A$ is itself symmetric. In particular, the singular values of $A$ are not the same as its eigenvalues. – angryavian Apr 3 '18 at 5:46
• @angryavian I know. I assume the OP meant singular value. – Jacky Chong Apr 3 '18 at 5:49
• @Jacky Chong If $A^{2}=0$ then the inequality you tried to prove fails. – Kavi Rama Murthy Apr 3 '18 at 5:52
• @KaviRamaMurthy I think my proof is correct. – Jacky Chong Apr 3 '18 at 5:55