Calculate ${\|A\|}_2$ for $A = \begin{bmatrix} - 4&0\\0&1\end{bmatrix}$

Question

Let $A = \begin{bmatrix} - 4&0\\0&1\end{bmatrix}$. Calculate ${\|A\|}_2$.

What I've tried: I know that ${\|A\|}_2 = \sup_{x\neq 0}\frac{{\|Ax\|}_2}{{\|x\|}_2}$. So I get $${\|A\|}_2 = \sup_{x\neq 0} \dfrac{\sqrt{16x_1^2+ x_2^2}}{\sqrt{x_1^2 + x_2^2}} = \sup_{x\neq 0} \sqrt{\frac{15x_1^2}{x_1^2+x_2^2}+1}.$$ I'm not sure how to proceed here, since I would assume that I can increase $x_1^2$ as much as I want so that I would get ${\|A\|}_2 = \infty$.

Question: Is my answer correct? If not; what am I doing wrong?

Thanks!

Answer 1

If you increase $x_1^2$, then you also increase $x_1^2+x_2^2$, and since that number appears in the denominator, you have no reason to believe $\|A\|_2=\infty$.

So no, your answer is not correct.

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To get the corect answer, you have two options:

Option $1$:

Use the alternative definition of $\|A\|_2$ which states $$\|A\|_2=\max_{\|x\|=1}\|Ax\|$$

which allows you to assume $x_1^2+x_2^2=1$

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Option $2$

Since you are calculating the supremum of the expression $$\sqrt{\frac{15x_1^2}{x_1^2+x_2^2} + 1}$$

you need to see how big that expression gets. You can see, for example, that the expression is (at a fixed value of $x_1$) the biggest when $x_2=0$. What's the value of the expression at that point?

Answer 2

5xum already posted the correct answer but I started typing this already so I will finish it.

Recall that $\|A\|_2$ is also equal to $\sup\limits_{x:\|x\|\leq 1}\frac{\|Ax\|_2}{\|x\|_2}$. This leads to the same expression you had, $\sqrt{\frac{15x_1^2}{x_1^2+x_2^2}+1}$ but now we can use $x_1^2+x_2^2\leq 1$ to replace $x_2^2$ by $1-x_1^2$ to get the inequality, $\sqrt{\frac{15x_1^2}{x_1^2+x_2^2}+1}\leq \sqrt{15x_1^2+1}$. Now take the $\sup$ over the unit ball...

Answer 3

$ \dfrac{15x_1^2}{x_1^2+x_2^2}+1 \le 15+1=16$, hence

$\dfrac{\sqrt{16x_1^2+ x_2^2}}{\sqrt{x_1^2 + x_2^2}} \le 4.$

With $x=(1,0)$ we have $\dfrac{{\|Ax\|}_2}{{\|x\|}_2}=4$.

Conclusion ?