Product of a matrix and its transpose is equal to its square.

Question

If $UU^T = UU^TUU^T$, what conditions must be true for $U^TU=I$?

Answer 1

Let $U\in\mathbb R^{n\times d}$, $n\le d$. Apply a singular value decomposition (SVD), i.e. $U=A\Sigma B$ with $A, B$ orthonormal and $\Sigma = (D \mid 0)$ where $D=\operatorname{diag}(\sigma_i)$, $\sigma_i\ge 0$ are the singular values. Then

• On the one hand $UU^T = UU^TUU^T \iff \Sigma\Sigma^T = \Sigma\Sigma^T\Sigma\Sigma^T \iff D^2=D^4\iff \sigma_i^2=\sigma_i^4$ for all $i$.

• On the other hand $U^TU=I \iff \Sigma^T\Sigma = I \iff D^2=I\iff \sigma_i^2=1$ for all $i$.

As $\sigma_i\ge 0$, the former has exactly two solutions $\sigma_i\in\{0,1\}$ whereas the latter only has $\sigma_i=1$. Hence the implication $UU^T=UU^TUU^T \implies U^TU=I$ only holds true when $U$ has full row rank.

Answer 2

As it is worded now, the statement is clearly false.

Suppose $U$ is a zero matrix. Then $U^T$ is also a zero matrix. And $UU^T$ is zero and thus indeed idempotent. But $U^TU$ is a zero matrix (and thus not equal to the identity matrix).