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With $$\left\Vert A \right\Vert=\max_{\mathbf{x}\ne 0}\frac{\left\Vert A\mathbf{x}\right\Vert }{\left\Vert \mathbf{x}\right\Vert }$$ and $$A=\begin{bmatrix}1 & 2\\ 2 & 4 \end{bmatrix} $$ Calculate $\Vert A \Vert$.

Not exactly homework, but an exercise from some notes on applied linear algebra that I am trying to work though. I am given the following hint:

Hint: note that $\begin{bmatrix}1 & 2\\ 2 & 4 \end{bmatrix}=\begin{bmatrix}1\\ 2 \end{bmatrix}\begin{bmatrix}1 & 2\end{bmatrix} $

I've used the hint to obtain an upper bound on $\frac{\left\Vert A\mathbf{x}\right\Vert }{\left\Vert \mathbf{x}\right\Vert }$:$$\frac{\left\Vert A\mathbf{x}\right\Vert }{\left\Vert \mathbf{x}\right\Vert } = \frac{\left\Vert \begin{bmatrix}1 & 2\\ 2 & 4 \end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert }{\left\Vert \begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert } = \frac{\left\Vert \begin{bmatrix}1\\ 2 \end{bmatrix}\begin{bmatrix}1 & 2\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert }{\left\Vert \begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert } = \frac{\left(\begin{bmatrix}1\\ 2 \end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right)\left\Vert \begin{bmatrix}1\\ 2 \end{bmatrix}\right\Vert }{\left\Vert \begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert } \leq \frac{\left\Vert \begin{bmatrix}1\\ 2 \end{bmatrix}\right\Vert \left\Vert \begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert \left\Vert \begin{bmatrix}1\\ 2 \end{bmatrix}\right\Vert }{\left\Vert \begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}\right\Vert } = \left\Vert \begin{bmatrix}1\\ 2 \end{bmatrix}\right\Vert ^{2} = 5 $$

I believe that I am headed in the right direction here but I'm rather stuck with where to go next. I suppose I need to solve $$\frac{\left\Vert A\mathbf{x}\right\Vert }{\left\Vert \mathbf{x}\right\Vert }=5$$but after some struggle, I can't figure out how to solve this equation. I tried following the through component-wise to get $$|x_1 + 2x_2|^2 + |2x_1+4x_2|^2=25[|x_1|^2+|x_2|^2]$$ but that seems no closer to what I need.

Does anyone see where I should go from here?

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Do you know the precise equality condition for your $\le$? –  Alexander Shamov Jan 8 '13 at 7:14
    
Here is a related problem. –  Mhenni Benghorbal Jan 8 '13 at 7:17
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The eigenvalues are $0$ and $5$. Since the matrix is symmetric the norm equals the largest eigenvalue. –  PAD Jan 8 '13 at 7:26
    
@PantelisDamianou ah I did not know that. That's I guess a good thing to know. If the matrix was not symmetric then this would not necessarily be the case? I guess the matrix norm by definition must at least be greater than or equal to the largest eigenvalue? –  crf Jan 8 '13 at 7:35
    
No, this is a feature specific to symmetric (or skew-symmetric, or, more generally, normal) matrices. Another good thing to know is that for any matrix $\Vert A \Vert^2 = \Vert A A^T \Vert = \Vert A^T A \Vert$, and since the latter two matrices are symmetric, their norm can be computed via eigenvalues. If the scalars are complex, the transpose should be replaced by conjugate transpose. –  Alexander Shamov Jan 8 '13 at 10:10
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2 Answers

up vote 3 down vote accepted

Consider the vector (1,2) again...

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Wow. I feel pretty stupid. I did that, and ended up finding that $\begin{bmatrix}1 & 2\\ 2 & 4 \end{bmatrix}\begin{bmatrix}1\\ 2 \end{bmatrix}=\begin{bmatrix}5\\ 8 \end{bmatrix}$. That's not correct, is it. –  crf Jan 8 '13 at 7:17
    
@crf: Indeed, the $8$ should be a $10$. Observe $\|Ax\|/\|x\|$ is the same as $\|Au\|=\|(1,2)\|\cdot|(1,2)\cdot u|$ where $u=x/\|x\|$: if $\theta$ is the angle between $u$ and $(1,2)$, then this is $5\cos\theta$ due to the formula $u\cdot v=\|u\|\|v\|\cos\theta$. This is maximized when $\theta=0$, in other words when $u$ is in the same direction as $(1,2)$, in which case we can take $x=(1,2)$. –  anon Jan 8 '13 at 7:22
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You have shown that $$\max \dfrac{\Vert Ax \Vert}{\Vert x \Vert} \leq 5$$ To show that equality is attained, choose $x = \begin{bmatrix} 1\\ 2\end{bmatrix}$. We have $$Ax = \begin{bmatrix} 1 & 2\\ 2 & 4\end{bmatrix} \begin{bmatrix}1\\2 \end{bmatrix} = \begin{bmatrix} 5 \\ 10\end{bmatrix} = 5x$$ Hence, $$\dfrac{\Vert Ax \Vert}{\Vert x \Vert} = 5$$

In general, note that if $A = uu^T$, then the reduced SVD of $A$ is $$A = \Vert u \Vert_2^2 \left(\dfrac{u}{\Vert u \Vert_2} \right) \left(\dfrac{u}{\Vert u \Vert_2} \right)^T$$ Hence, $\Vert A \Vert_2 = \Vert u \Vert_2^2$. In your case, $$A = \begin{bmatrix} 1\\ 2\end{bmatrix} \begin{bmatrix} 1 & 2\end{bmatrix}$$ Hence, $u = \begin{bmatrix} 1\\ 2\end{bmatrix}$ and thereby $$\Vert A \Vert_2 = \Vert u \Vert_2^2 = 1^2 + 2^2 = 5$$

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