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?
What is the difference between Singular Value and Eigenvalue?
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3@AlexBecker : The DO NOT agree in finite dimensions! Clearly you're not familiar with the singular value decomposition. All real matrices have singular values, but non-square matrices don't have eigenvalues. – 2013-01-23
4 Answers
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!
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1Isn't it sufficient for the square matrix to be diagonalizable, rather than symmetric (ie orthogonally diagonalizable) for the singular values to be the same as the eigenvalues? – 2014-05-21
Given a matrix $A$, if the eigenvalues of $A^HA$ are $\lambda_i \geq 0$, then $\sqrt{\lambda_i}$ are the singular values of $A$. If $t$ is an eigenvalue of $A$, then $|t|$ is a singular value of $A$. And here is an example should be noticed, $A = \begin{pmatrix}1&0&1\\0&1&1\\0&0&0\end{pmatrix},$ the eigenvalues of $A$ are $1,1,0$ while the singular values of $A$ are $\sqrt{3},1,0$.
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0@EricKightley The problem with your argument is that if $t$ is a right eigenvalue for $A$ ($Av=tv$ for some $v$) then $\overline t$ is a _left_ eigenvalue for $A^*$ ($\overline{Av})=\overline{v}A^*=\overline{t}\overline{v}$) – 2018-01-31
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.
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0Actually, eigendecomposition is defined for all diagonalizable matrices, not just normal matrices. – 2018-11-20
Singular values of the SVD decomposition of the matrix A is the square root of the eigenvalues of the matrix (A multiplied by A transpose) or(A transpose multplied by A), the two ar identical with positive eigenvalues.