How to Calculate Matrix Norm in a Banach Space?

[kalish1]

How can I calculate the following matrix norm in a Banach Space:

$$
A=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix}
?$$

I have tried $$\|A\|=\sup\limits_{\|x\|=1}\|Az\|$$

and then did $$Az=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5x-2y \\ x-y \end{pmatrix}.$$

Now, how can I maximize the function $\|Az\|$? I'm not sure.

The norm is not specified in the problem. Should I just go with the square root of the sum of the squares of the compoments of Az, i.e. maximize $\sqrt{(5x-2y)^2+(x-y)^2}$ given the constraint $x^2+y^2=1$? Or should I maximize $\{{(5x-2y)^p+(x-y)^p}\}^{1/p}$ given the constraint $x^2+y^2=1$?

Any help would be appreciated.

I have crossposted this question here: calculus - Matrix norm in Banach space - Mathematics Stack Exchange

 

[stainburg]

If you want to calculate 2-norm,

(a) easy way I: use software such as Matlab;)

(b) easy way II: use $$\sqrt{\text{sup}(\lambda)}$$, where $$\lambda$$ is eigenvalue of $$A^TA$$.

(c) hard way: Let $$z=[x\text{ }y]^T$$ be unit vector. You need to maximize $$\parallel Az\parallel_2^2=z^TA^TAz$$, subject to $$\parallel z\parallel_2^2=z^Tz=1$$. By Lagrange multiplier method, it is to solve equations $$\bigtriangledown_z (z^TA^TAz-\lambda (z^Tz-1))=0$$ and $$z^Tz=1$$ (actually, it's not a typical optimization problem, but we can still do it like this, why?). At last, we have $$A^TAz=\lambda z$$. $$\lambda$$ is what easy way II speaks of.

 

[Fermat1]

How can I calculate the following matrix norm in a Banach Space:

$$
A=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix}
?$$

I have tried $$\|A\|=\sup\limits_{\|x\|=1}\|Az\|$$

and then did $$Az=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5x-2y \\ x-y \end{pmatrix}.$$

Now, how can I maximize the function $\|Az\|$? I'm not sure.

The norm is not specified in the problem. Should I just go with the square root of the sum of the squares of the compoments of Az, i.e. maximize $\sqrt{(5x-2y)^2+(x-y)^2}$ given the constraint $x^2+y^2=1$? Or should I maximize $\{{(5x-2y)^p+(x-y)^p}\}^{1/p}$ given the constraint $x^2+y^2=1$?

Any help would be appreciated.

I have crossposted this question here: calculus - Matrix norm in Banach space - Mathematics Stack Exchange

Since you are taking the sup of non-negative numbers, we may as well maximise $(5x-2y)^2+(x-y)^2$ subject to $x^2+y^2=1$ and then take square root at the end. The condition implies $x=cos{\alpha}$ and $y=sin{\alpha}$ for some ${\alpha}$. Rewriting $(5x-2y)^2+(x-y)^2$ using that, its obvious that the max occurs when $cos{\alpha}$ is positive and $sin{\alpha}=-cos{\alpha}$. So $(5x-2y)^2+(x-y)^2=6.27$ to 3.s.f. Square rooting gives the norm as approximately 2.5.

 

[Opalg]

How can I calculate the following matrix norm in a Banach Space:

$$
A=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix}
?$$

I have tried $$\|A\|=\sup\limits_{\|x\|=1}\|Az\|$$

and then did $$Az=\begin{pmatrix}
5 & -2 \\
1 & -1 \\
\end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5x-2y \\ x-y \end{pmatrix}.$$

Now, how can I maximize the function $\|Az\|$? I'm not sure.

The norm is not specified in the problem. Should I just go with the square root of the sum of the squares of the compoments of Az, i.e. maximize $\sqrt{(5x-2y)^2+(x-y)^2}$ given the constraint $x^2+y^2=1$? Or should I maximize $\{{(5x-2y)^p+(x-y)^p}\}^{1/p}$ given the constraint $x^2+y^2=1$?

Any help would be appreciated.

I have crossposted this question here: calculus - Matrix norm in Banach space - Mathematics Stack Exchange

There are many different norms for a matrix considered as an operator on a Banach space. If you want the operator norm of $A$ as an operator on 2-dimensional Euclidean space then the easiest way to calculate it is by using the C*-identity $\|A\|^2 = \|A^*A\|$, where $A^*$ is the Hermitian adjoint of $A$. In this case, $A^*A = \begin{bmatrix}5&1 \\-2&-1 \end{bmatrix} \begin{bmatrix}5&-2 \\1&-1 \end{bmatrix} = \begin{bmatrix}26&-11 \\-11&5 \end{bmatrix}.$ Since $A^*A$ is a positive matrix, its norm will be its largest eigenvalue, which you can find in the usual way by solving the quadratic equation $\det(A^*A - \lambda I) = 0.$ Then take the square root to get $\|A\|.$

Edit. I hadn't noticed that stainburg already suggested this method.