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 $\quad \begin{bmatrix} 0 & 1 & -2\\ 1 & 0 & 1\\ \end{bmatrix}$
 A: There are several norms one may calculate directly. Among the most common, mentioned in the comments are:


*

*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:
$$
\|A\|=\lambda_\max(A^TA)^{1/2}=\lambda_\max(\begin{bmatrix}1&0&1\\0&1&-2\\1&-2&4 \end{bmatrix} )^{1/2}=\sqrt{\frac52+\frac{\sqrt{29}}2}
$$

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

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

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