What is the difference between "singular value" and "eigenvalue"? 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?
 A: 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 $\sqrt{3},1,0$.
A: 
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. 
A: 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!
A: A very clear explanation from Cleve Moler's text book 
https://www.mathworks.com/content/dam/mathworks/mathworks-dot-com/moler/eigs.pdf
An eigenvalue and eigenvector of a square matrix A are a scalar λ and a nonzero
vector x so that
$$Ax = λx.$$
A singular value and pair of singular vectors of a square or rectangular matrix A
are a nonnegative scalar σ and two nonzero vectors u and v so that
$$Av = σu,$$
$$A^Hu = σv.$$
Eigenvectors and singular vectors are them same if $A$ is a real symmetric matrix (so $A^H = A$).
A: Singular values of the SVD decomposition of the matrix A is the square root of the eigenvalues of the matrix ($A$ multiplied by $A^T$) or ($A^T$ multiplied by $A$), the two are identical with positive eigenvalues.
A: 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, $\sigma_{\max}(A)=\sqrt{\lambda_{\max}(A^2)}=\sqrt{\lambda^2_{\max}(A)}=|\lambda_{\max}(A)|=\sup\limits_{x \in S^{n-1}}| x^\top A x|$
A: The eigenvalues $\mathbf{\Lambda}$ of the sample matrix $\mathbf{X}$ can be obtained applying SVD to the covariance of $\mathbf{X}$:
$$\mathbf{S} = cov(\mathbf{X}) \mapsto \mathbf{S}=\mathbf{U}\mathbf{\Lambda}\mathbf{V}^{T} $$
