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Is there an intuitive meaning for the spectral norm of a matrix? Why would an algorithm calculate the relative recovery in spectral norm between two images (i.e. one before the algorithm and the other after)? Thanks


2 Answers 2

up vote 4 down vote accepted

The spectral norm is the maximum singular value of a matrix. Intuitively, you can think of it as the maximum 'scale', by which the matrix can 'stretch' a vector.

thanks - this helps me conceptually. –  val Aug 29 '12 at 6:56

Let us consider the singular value decomposition (SVD) of a matrix $X = U S V^T$, where $U$ and $V$ are matrices containing the left and right singular vectors of $X$ in their columns. $S$ is a diagonal matrix containing the singular values. A intuitive way to think of the norm of $X$ is in terms of the norm of the singular value vector in the diagonal of $S$. This is because the singular values measure the energy of the matrix in various principal directions.

One can now extend the $p$-norm for a finite-dimensional vector to a $m\times n$ matrix by working on this singular value vector:

\begin{align} ||X||_p &= {\Big(} \sum_{i=1}^{\text{min}(m,n)} \sigma_i^p {\Big)}^{1/p} \end{align}

This is called the Schatten norm of $X$. Specific choices of $p$ yield commonly used matrix norms:

  1. $p=0$: Gives the rank of the matrix (number of non-zero singular values).
  2. $p=1$: Gives the nuclear norm (sum of absolute singular values). This is the tightest convex relaxation of the rank.
  3. $p=2$: Gives the Frobenius norm (square root of the sum of squares of singular values).
  4. $p=\infty$: Gives the spectral norm (max. singular value).
thanks - can we really talk about Schatten p=0 for matrices? since we would be looking at 1/0...? –  val Aug 29 '12 at 7:02
Yes we can. And it is equal to the rank of the matrix. Can you clarify your question a bit more? –  Kartik Audhkhasi Aug 29 '12 at 13:41
Hi - i was referring to the Schatten norm equation above: the exponent is 1/p. If p=0 we have 1/0. thanks.. p must > 1. –  val Aug 29 '12 at 17:31
Think of $p \rightarrow 0$. Then any nonzero singular values will lead to $1$, while $0$ singular values will give $0$ anyway. Hence, you will end up with the number of non-zero singular values. However, for $p \in [0,1]$, the Schatten norm is not a "norm" since it does not satisfy all the properties. However, it is still common practice to call it a "norm". –  Kartik Audhkhasi Aug 29 '12 at 18:10
ok, i understand better - thanks again. –  val Aug 30 '12 at 16:17

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