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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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    Apologies for delay, it was operator norm that I was after. Now I see it was just the largest singular value.2012-12-02

2 Answers 2

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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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    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 $A^T$").2012-12-01
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The product $A^{T}A$ will give you a square matrix, the norm of which will be the square of the norm of $A$.