# Calculating norm of a $2 \times 3$ matrix

I have a $2\times 3$ matrix and am asked to calculate its norm. i was under the impression that only square matrices can have norms calculated. Is this false?

The matrix is $\quad \begin{bmatrix} 0 & 1 & -2\\ 1 & 0 & 1\\ \end{bmatrix}$

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What is the matrix norm definition? – Sigur Nov 30 '12 at 21:43
only square matrices have norms clearly related to eigenvalues. But you can still ask about the maximum length $|Ax|$ when $|x| = 1.$ Here $x \in \mathbb R^3,$ but $Ax \in \mathbb R^2.$ – Will Jagy Nov 30 '12 at 21:44
There are many different matrix norms, most of which make sense for non-square matrices. – Brian M. Scott Nov 30 '12 at 21:45
In your comment, you seem to have written $AA$, but you can't multiply a non-square matrix by itself. Look, whoever has asked you to calculate this norm must have told you at some point what she had in mind as the definition of the norm. Go over your notes, find out the precise definition you have been asked to use, and then get back to us. – Gerry Myerson Nov 30 '12 at 21:54
@Gerry: he probably meant to write $\langle A,A\rangle$ which could mean $tr(A^t A)$ – Thibaut Dumont Nov 30 '12 at 22:54

There are several norms one may calculate directly. Among the most common, mentioned in the comments are:

1. The operator norm: one can see $A$ as a linear operator $\mathbb R^3\to\mathbb R^2$. In that case its norm agrees with the biggest singular value: $$\|A\|=\lambda_\max(A^TA)^{1/2}=\lambda_\max(\begin{bmatrix}1&0&1\\0&1&-2\\1&-2&4 \end{bmatrix} )^{1/2}=\sqrt{\frac52+\frac{\sqrt{29}}2}$$

2. You can also see $A$ as an operator $\mathbb R^2\to\mathbb R^3$ (by multiplication on the right). In this case $$\|A\|=\lambda_\max(AA^T)^{1/2}=\lambda_\max(\begin{bmatrix}5&-2\\-2&2 \end{bmatrix} )^{1/2}=\sqrt{6}$$

3. You can look at the Frobenius norm, $$\|A\|_2=\text{Tr}(A^TA)^{1/2}=\text{Tr}\left(\begin{bmatrix}1&0&1\\0&1&-2\\1&-2&4 \end{bmatrix} \right)^{1/2}=\sqrt6$$

4. Or a "reverse Frobenius" norm $$\|A\|_{2'}=\text{Tr}(AA^T)^{1/2}=\text{Tr}\left(\begin{bmatrix}5&-2\\-2&2\end{bmatrix} \right)^{1/2}=\sqrt7$$

These are among the most common options, but there are of course many others.

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It's not coincidence, is it, that the 2nd and 3rd interpretations give the same numerical result? – Gerry Myerson Dec 1 '12 at 5:06
I was surprised when I got the same number. But the operator norm and the 2-norm are certainly different for square matrices. Note also that if one expects equality between 2 and 3, one would also have to expect it between 1 and 4 (since 1 and 4 are "2 and 3 for $A^T$"). – Martin Argerami Dec 1 '12 at 5:48

The product $A^{T}A$ will give you a square matrix, the norm of which will be the square of the norm of $A$.

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