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Consider a $2\times 2$ symmetric matrix, in this case, is there some closed formula for its spectral norm ?

By spectral norm I mean the induced 2-norm, there is a definition here.


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up vote 3 down vote accepted

It's the square root of the largest (in magnitude) eigenvalue of $A^2$. Since $A^2$ is $2 \times 2$, the eigenvalues are given by the roots of a quadratic, which you can work out. You might not like the "max", in which case you can avoid it by replacing it with

$$ max(x, y) = \frac{x+y+ |y-x|}{2} $$ in your formula.

If $$ A = \begin{bmatrix} a & b \\ b & c \end{bmatrix} $$ then $A^2$ is $$ A^2 = \begin{bmatrix} a^2 + b^2 & b(a+c) \\ b(a+c) & c^2 + b^2 \end{bmatrix} $$ and its characteristic polynomial is $$ p(x) = x^2 - (a^2 + 2b^2 + c^2) x + (ac-b^2)^2. $$ The roots are then $$ x = \frac{- (a^2 + 2b^2 + c^2) \pm \sqrt{ (a^2 + 2b^2 + c^2)^2 - 4(ac-b^2)^2}}{2} $$ The greater root (in absolute value) is the one with the $-$ sign. (No need for the max trick in this case, because the coefficient of $x$ in the quadratic is so obviously negative). So that leads to your answer: the 2-norm is $$ \sqrt{\frac{ (a^2 + 2b^2 + c^2) + \sqrt{ (a^2 + 2b^2 + c^2)^2 - 4(ac-b^2)^2}}{2}} $$

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I was looking for a formula that would work without having to choose one of two values. Your formula looks exactly what I need, thanks. – Integral Apr 21 '14 at 17:22

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