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$\boldsymbol{a} = [2, 1]$ can be regarded as a vector or a $1$x$2$ matrix. Compute the 1-norm, 2-norm, infinity norm for $a$ as a vector, also compute the matrix norm. Is the matrix norm equal to any of the vector norms? Which one? Repeat the question for $\boldsymbol{b} = [2, 1]^{T}$

The definition for the matrix norm is $$\Vert a \Vert = \max_{x:\Vert x \Vert = 1} \Vert ax \Vert \>$$

Hint:Use Cauchy-Schwarz inequalities

Here is what I have so far:

For both vectors $a$ and $b$, $1$-norm = $3$, $2$-norm = $\sqrt{5}$, infinity-norm = $2$

I'm having trouble calculating the matrix norm. The hint is using Cauchy-Schwarz inequalities,

$\Vert a \cdot x \Vert \leq \Vert a \Vert \Vert x \Vert \>$

Since $\Vert x\Vert=1$, I'm thinking there is an upper bound on $\Vert a \cdot x \Vert $, but how should I proceed after this?

Thanks for any help

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2 Answers 2

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Let $\mathrm A := \begin{bmatrix} 2 & 1\end{bmatrix}$. Hence,

$$\|\mathrm A\|_2 = \sigma_{\max} (\mathrm A) = \sqrt{\lambda_{\max} (\mathrm A \mathrm A^{\top})} = \sqrt{\lambda_{\max} (5)} = \sqrt{5}$$

Let $\mathrm B := \mathrm A^{\top}$. Hence,

$$\|\mathrm B\|_2 = \sigma_{\max} (\mathrm B) = \sqrt{\lambda_{\max} (\mathrm B^{\top} \mathrm B)} = \sqrt{\lambda_{\max} (\mathrm A \mathrm A^{\top})} = \|\mathrm A\|_2 = \sqrt{5}$$

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There are many possible matrix norms, depending upon the choice of norms in your definition. Consider the 2-norm:

So we view $a$ as a map from ${\Bbb R}^2$ to ${\Bbb R}$ both equipped with the $2$-norm $|\cdot|_2$. For ${\Bbb R}$ this just mean the absolute abs value. For $x\in {\Bbb R}^2$ we have $ax=a_1 x_1+a_2 x_2\in {\Bbb R}$ which is just like a scalar product. So
$$ \|a\|_2 =\sup_{x\in {\Bbb R}^2, |x|_2=1} |a\cdot x| \leq \sup_{x\in {\Bbb R}^2, |x|_2=1} |a|_2\; |x|_2 =|a|_2 =\sqrt{5} $$ by the Cauchy-Schwarz inequality. The inequality in the middle is, however, an equality as it is attained for $x=a/|a|_2$ (when $a\neq 0$). So for the matrix 2-norm we have $\|a\|_2=\sqrt{5}$.

For the matrix $\infty$-norm we get: $$ \|a\|_\infty =\sup_{x\in {\Bbb R}^2, |x|_\infty=1} |a_1x_1+a_2x_2| \leq |a_1|+|a_2|=|a|_1.$$ Again equality may be attained by choosing $x_i$ to be the sign of $a_i$. So $\|a\|_\infty = |a|_1=3$.

For the matrix 1-norm you get in a similar way $\|a\|_1=|a|_\infty=2$.

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  • $\begingroup$ Could you elaborate more? Why the dot product of $a$ and $x$? and how did you get the equality? by the Cauchy-Schwarz inequality? $\endgroup$
    – user59036
    Sep 15, 2016 at 23:43

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