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Suppose that there is matrix $A = \begin{bmatrix} 3 & 2 \\ 2 & 0 \end{bmatrix}$. We want to figure out what $$\max_{||\hat{\mathbb{x}}|| = 1}||A\hat{\mathbb{x}}||$$ is with the Euclidean norm. (where $\hat{\mathbb{x}}$ is $\frac{1}{||\mathbb{x}||}\mathbb{x}$.)

Answer is stated to be 4, but I am unsure why it is.

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Hint: for a normal matrix, so certainly for a Hermitian complex, or real symmetric, matrix, this norm is the maximum absolute value of an eigenvalue. – Geoff Robinson Aug 30 '12 at 17:51
I get it now. Thanks! – Linear niner Aug 30 '12 at 18:02
@GeoffRobinson: I think your comment answers this question entirely, or in any case, it provides the answer the OP was looking for. I encourage you to post this hint so that the question does not remain unanswered. – Eric Naslund Aug 30 '12 at 18:41
up vote 0 down vote accepted

To complete this question we look for the eigenvalues since diagonlisation will give us a matrix with maximum $a_{ii}$ and taking $x_i=1$ and $0$ otherwise will give us maximum euclidean norm. We have

$$ |A-\lambda I|=(3-\lambda)(-\lambda)-4=\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1) $$

So $\lambda=4$ is the largest eigenvalue so $ \max_{||\hat{\mathbb{x}}|| = 1}||A\hat{\mathbb{x}}||=4$.

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