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Would you help me to prove the norm of 2 by 2 matrix

\begin{equation} A=\begin{bmatrix} a & b \\ c & d% \end{bmatrix} \end{equation}

is $$\def\trc{\operatorname{tr}}\frac{\trc(A^*A)+\sqrt{\trc(A^*A)^2-4\det(A^*A)}}{2}.$$ I just try by maximizing $h(x,y)=||A\begin{bmatrix} x \\ y% \end{bmatrix}||$ where $x^2+y^2=1$ but not get this result.

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As I mentioned in my answer, your formula does not give a norm. You need to take the square root of that. For example, if you let $a=2$, $b=c=0$, $d=1$, then for $A+A$ your formula gives $16$, while for $A$ it gives $4$: so it doesn't satisfy the triangle inequality. –  Martin Argerami Oct 3 '12 at 1:36
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2 Answers

up vote 2 down vote accepted

I think you are talking about the operator norm induced by the euclidean norm, and not the euclidean norm per se. In that case, the norm will be the square root of the number you mention.

The operator norm of $A$ can be characterized as the square root of the biggest eigenvalue of $A^*A$. The eigenvalues of $A^*A$ are the roots of its characteristic polynomial. As $A^*A$ is $2\times 2$, its characteristic polynomial is $p(t)=t^2-\mbox{tr}(A^*A)\,t+\det(A^*A)$. So its biggest eigenvalue is $$ \frac{\mbox{tr}(A^*A)+\sqrt{\mbox{tr}(A^*A)^2-4\det(A^*A)}}2 $$ (the polynomial $p$ has always non-negative roots, so the formula above certainly gives the biggest eigenvalue). So the norm of $A$ is $$ \left(\frac{\mbox{tr}(A^*A)+\sqrt{\mbox{tr}(A^*A)^2-4\det(A^*A)}}2\right)^{1/2}. $$

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Couldnt be solve by calculus only? –  beginner Oct 1 '12 at 15:04
    
Probably. It will be certainly more convoluted, because if you try to do Lagrange multipliers to your $h$, what you are really doing is first finding a vector where the maximum occurs, and then evaluating $h$ at that vector. –  Martin Argerami Oct 1 '12 at 15:19
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It's the spectral radius of $A^*A$, that is, the largest modulus of the eigenvalues of $A^*A$. We can express them with a second degree equation satisfied by the eigenvalues, and knowing they are non-negative, we can see that the given expression is the norm of the matrix.

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The problem cant be solved only by multivariable calculus (using Lagrange Multiplier)? –  beginner Oct 1 '12 at 14:50
    
It's possible, but I don't know whether it is simpler. You could include your attempt with this method in the OP, then we will try to see where you think there is a problem. –  Davide Giraudo Oct 1 '12 at 14:53
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