TITLE: Calculate $\begin{Vmatrix}1&2\\2&4\end{Vmatrix}$
QUESTION [8 upvotes]: 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?

REPLY [5 votes]: 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 $$\Vert A \Vert_2 = \Vert u \Vert_2^2 = 1^2 + 2^2 = 5$$