Is this a correct abreviation of the singular value decomposition algorithm?

Question

I want to describe the process as:

For an $m\times n$ matrix A:

• Compute $A^TA$ and $AA^T$

• $U=$ orthogonally diagonalize $AA^T$

• $V=$ orthgononally diagonalize $A^TA$

• $\Sigma =$ "Diagonal" $m\times n$ matrix of the singular values

• $A=U\Sigma V^T$

Answer 1

Yes, you're right.

You get the singular values by looking at the square roots of the eigenvalues of $AA^T$ or $A^TA$. (They are the same. They are all real and positive because both matrices are positive semi-definite.)

The matrix $\Sigma$ inherits the shape of $A$, and it looks like this: $$ \left[ \begin{array}{ccc} \sigma_1 &0 & \cdots\\ 0 & \ddots & 0 \\ 0 & \cdots & \sigma_n\\ 0 & \cdots & 0 \\ \end{array} \right] $$ or its transpose. The point is that there is a block of zero on the bottom or on the right.