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I am a little bit confused about such fundamental problems:

Suppose
1. $Ax=\lambda x$.
2. $A \in \mathbb{R^{n \times n}} $.

Case I: $$A^TAx = \lambda A^Tx=\lambda \lambda x=\lambda^2x$$

Case II: $$AAx = \lambda Ax=\lambda \lambda x=\lambda^2x$$

From above, it seems that $A^TA$ and $AA$ have the same eigenvalues.

However, consider $$A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$$

If checking the eigenvalues of $A^TA$ and $AA$ by matlab, they are not the same.

Why is that?

sleeve chen
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    $A^Tx \neq \lambda x$ in genreal. –  May 27 '15 at 00:29
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    Consider the following: http://math.stackexchange.com/questions/123923/a-matrix-and-its-transpose-have-the-same-set-of-eigenvalues – sleeve chen May 27 '15 at 00:31
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    The eigenvalues are the same yet they might corresponds to different eigenvectors (not the same $x$). –  May 27 '15 at 00:33
  • How about this: $||A||2 = \sigma{max}(A) = (\lambda_{max}(A^TA))^{\frac{1}{2}}$, the second part is the singular value. Does it mean, the definition of singular value comes from $A^TA$ instead of $AA$? – sleeve chen May 27 '15 at 00:36
  • It seems that the singular value in general is not the eigenvalue. (They are the same when $A = A^T$). –  May 27 '15 at 00:40
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    So you can say that singular values of $A$ are square root of the eigenvalues of $AA^T$, not that of $AA$. –  May 27 '15 at 00:47

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