In singular value decomposition why is $V$ orthogonal matrix?

Question

I have studied SVD from multiple resources and what I have understood is that in ($AV = \Sigma U$), A is a transformation matrix, $V$ is the matrix on which transformation is applied. $\Sigma$ is scaling matrix and $U$ is rotation matrix. What I don't understand is how $V$ can be an orthogonal matrix. It could be any matrix, why are we assuming that it's an orthogonal matrix ?

Answer 1

In the SVD of a real matrix $A$ of order $m\times n$, the eigenvectors of $A^TA$ make up the columns of $V$, while the eigenvectors of $AA^T$ make up the columns of $U$. Now, as $A^TA$ is a symmetric matrix, i.e., $(A^TA)^T = A^TA$, and every symmetric real matrix of order $n$, has $n$ linearly independent eigenvectors. By normalizing them you can construct the orthogonal matrix $V$ such that $V^TV = I$. In other words, the orghotonoality of $V$ and $U$ stems from the very construction of the SVD, and it is not assumed apriori.