SVD(Singular Value Decomposition) I have several questions regarding SVD. I know that we have to find eigen vectors of A^T A and go from there but what I don't understand is how do you know that the eigen vectors of A^T A will also be eigen vectors of A? Also, the book says if there are n eigen vectors for A^T A then, for some r in 1<= r <= n, Av1... Avr will form a orthogonal basis for columns of A. Again, how do we know for certain that A^TA will have more eigen vectors than A? A might have rank of 15 and A^TA might only have 4 eigen vectors what then?
 A: 
what I don't understand is how do you know that the eigen vectors of A^T A will also be eigen vectors of A?

The eigenvalues of $A^TA$ will generally not be the eigenvalues of $A$; they will be the singular values of $A$.  Similarly, the eigenvectors of $A^TA$ are singular vectors of $A$, which need not coincide with the eigenvectors of $A$.  Notably: if $A$ isn't square, then $A$ doesn't have eigenvalues/eigenvectors, but it will still have singular values/singular vectors.

how do we know for certain that A^TA will have more eigen vectors than A?

$A^TA$ is a symmetric matrix.  By the spectral theorem, it is orthogonally diagonalizable, which is to say that it necessarily has an orthonormal basis of eigenvectors.  This is true regardless of the properties of $A$.

A might have rank of 15 and A^TA might only have 4 eigen vectors what then?

The rank of a matrix generally doesn't tell you whether that matrix is diagonalizable (i.e. has enough eigenvectors).
A: The book doesn't say "more eigenvectors than $A$".
If $A$ is $m \times n$, then $A^T A$ is $n \times n$.  Being a real symmetric matrix, $A^T A$ has $n$ linearly independent eigenvectors.   The rank of $A$ is at most $n$ (which is the number of columns), and also at most $m$ (the number of rows).  In particular, it's impossible for $A$ to have rank $15$ if $A^T A$ has only $4$ linearly independent eigenvectors.
