Consider a vector $v$.
The magnitude of this vector (if it describes a position in euclidean space) is equal to the distance from the origin:
$$(v^Tv)^{\frac{1}{2}} = \sqrt{(v^Tv)}$$
that is, the square root of the dot product.
Suppose we compute this value for not just a vector but a matrix M, which describes an operator that transports a position vector.
What is the interpretation of the magnitude?
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$\begingroup$For a vector $v\in\mathbb{R}^n$, $\sqrt{v^Tv}$ is just the Euclidean norm $\|v\|_2$. There are infinitely many different norms for vectors and any one of them gives rise to a legitimate measure of "magnitude".
The same also holds for matrices. You can measure the "magnitudes" of matrices by using matrix norms. If you stack up the columns of an $n\times n$ matrix $A$ successively with the first column on top, you get a long vector $v$ in $\mathbb{R}^{n^2}$. The Euclidean norm of this vector (i.e. $\|v\|_2$) is called the Frobenius norm of the matrix and it is denoted by $\|A\|_F$. Again there are other kinds of matrix norms as well.
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