SVD square and invertible matrix

Question

Let $A=U \Sigma V^T$ be a $n \times n$ square and invertible matrix. Matrices $U$ and $V$ are orthogonal and $\Sigma$ is a diagonal matrix.

I've read in some lecture notes that if matrix $A$ is square and invertible then $U$, $V$, and $\Sigma$ are also square matrices. How is this statement possible?

Isn't it possible that $U$ and $V$ are $n \times m$ and $\Sigma$ is $m \times m$ where $n > m$? In this case, we can only say $U^TU=I$ and $V^TV=I$.

Answer 1

It is possible to make the matrices smaller, yes. But then $U\Sigma V^T$ wouldn't be the SVD of $A$.

If $A$ is singular, then some of the diagonal entries of $\Sigma$ are zero, and could, in principle, be cut out, along with the corresponding columns of $U$ and $V$. But that's not the true SVD. Also, $A$ is assumed to be invertible, and a simple rank argument then shows that $U, \Sigma$ and $V$ must all be at least $n\times n$.

Answer 2

When $A$ is a square matrix, SVD just becomes the diagonalization. In that Case $A$ can be written as $P^{-1}DP$ where $P$ is the matrix with orthonormal eigen vectors of $A$ as columns. In such a case $P^{-1} = P^{T}$. Since $A$ is a square matrix, it has $n$ eigen values, and $n$ eigen vectors. So, all the matrices on the r.h.s are square.