What is the difference between "singular value" and "eigenvalue"?

Question

I am trying to prove some statements about singular value decomposition, but I am not sure what the difference between singular value and eigenvalue is.

Is "singular value" just another name for eigenvalue?

Answer 1

is singular value just another name for eigenvalue?

No, singular values & eigenvalues are different.

What is the difference between Singular Value and Eigenvalue?

There are many possible answers to this question. Since I don't know what you're trying to prove, I'd recommend carefully comparing definitions between the two: eigendecomposition, singular value decomposition

[EDIT: You might find the first several chapters of the book "Numerical Linear Algebra" by Trefethen and Bau more useful than the Wikipedia article. They're available here.]

Two important points:

• Notice in particular that the SVD is defined for any matrix, while the eigendecomposition is defined only for square matrices (and more specifically, normal matrices).

• Notice that singular values are always real, while eigenvalues need not be real.

Answer 2

The singular values of a $M\times N$ matrix $X$ are the square roots of the eigenvalues of the $N\times N$ matrix $X^*\,X$ (where $^*$ stands for the transpose-conjugate matrix if it has complex coefficients, or the transpose if it has real coefficients).

Thus, if $X$ is $N\times N$ real symmetric matrix with non-negative eigenvalues, then eigenvalues and singular values coincide, but it is not generally the case!

Answer 3

Consider the comparison between largest singular value and largest eigen value of a matrix with real entries, $A\in \mathbb{R}^{m \times n}$:

\begin{align*} \sigma_{\max}(A)&= \sup_{x \in S^{n-1}} ||Ax||_2 =\sup_{x \in S^{n-1}} \sqrt{||Ax||_2^2}= \sup_{x \in S^{n-1}} \sqrt{x^\top A^\top A x} = \sqrt{\lambda_{\max}(A^\top A)} \end{align*}

For $A^\top =A$, we have, $\sigma_{\max}(A)=\sqrt{\lambda_{\max}(A^2)}=\sqrt{\lambda^2_{\max}(A)}=|\lambda_{\max}(A)|=\sup\limits_{x \in S^{n-1}}| x^\top A x|$