The spectral norm of a positive definite matrix is its largest eigenvalue

Question

Let $A$ be a symmetric positive definite matrix. Show that $$\lambda_n=\max \left\{\frac{\|Ax\|}{\|x\|}: x\ne 0 \right\}$$ is the largest eigenvalue of $A$.

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My try is $\frac{\|Ax\|}{\|x\|}\le \lambda,\forall \lambda$ as a not eigenvalue of $A$, and the equality occurs when $\lambda$ is an eigenvalue. So $\lambda$ is maximum? May be I am not even understanding the question.

Answer 1

By the spectral theorem, there is an orthonormal basis of eigenvectors of $A$, say $x_1,\dots,x_n$. WLOG, assume $0 < \lambda_1 \leq \dots \leq \lambda_n$, where $Ax_i = \lambda_i x_i$. For any non-zero $x$, write $x = c_1x_1 + \cdots + c_nx_n$. Then $$ \frac{\lVert{Ax\rVert}^2}{\lVert{x\rVert}^2} = \frac{\lVert{c_1\lambda_1x_1 + \cdots + c_n\lambda_nx_n\rVert}^2}{c_1^2 + \cdots + c_n^2} = \frac{c_1^2\lambda_1^2 + \cdots + c_n^2\lambda_n^2}{c_1^2 + \cdots + c_n^2} \leq \lambda_n^2 \frac{c_1^2 + \cdots + c_n^2}{c_1^2 + \cdots + c_n^2} = \lambda_n^2,$$ with equality achieved when $x = x_n$ (whence $\lVert{Ax\rVert}=\lVert{\lambda_n x_n\rVert}=\lambda_n$). You may wish to read about Rayleigh quotients or see Problem 37 in Brezis.

Answer 2

Let $\bf A$ be symmetric and positive semidefinite.

$$ \| {\bf A} \|_2 = \sigma_{\max} ({\bf A}) = \sqrt{\lambda_{\max} \left( {\bf A}^\top {\bf A} \right)} = \sqrt{\lambda_{\max} \left( {\bf A}^2 \right)} = \sqrt{\lambda_{\max}^2 \left( {\bf A} \right)} = \left| \lambda_{\max} \left( {\bf A} \right) \right| = \color{blue}{\lambda_{\max} \left( {\bf A} \right)} $$

Answer 3

It is all of the answer first you can prove that for each eigenvalue of positive definite matrix $M$ you have $$ \operatorname{eig}(M^2)=\operatorname{eig}(M)^2.$$ In the other words if $s$ is the eigenvalue of $M$ then $s^2$ is the eigenvalue of $M^2$ since $s$ is the eigenvalue of $M$: $|s\cdot I-M|=0$ now by means of determinant properties we can write |s^2*I-M^2|=|(sI-M)(sI+M)|=|sI-M||sI+M|=0 we proved that s^2 is the eigenvalue of M^2 we now that 2 norm of a matrix M is derived from this equation ||M||=sqrt(max(eig(M'*M))) you should now that your claim is only true for symmetric matrix M so we can write ||M||=sqrt(max(eig(M^2))) according to our proof we can write ||M||=sqrt(max(s^2))) since M is positive definite s is positive now we can write ||M||=max(s)=max(eig(M))