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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?

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    @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

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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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    Isn'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
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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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    @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
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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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    Actually, eigendecomposition is defined for all diagonalizable matrices, not just normal matrices.2018-11-20
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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.