Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


I am trying to solve the following problem, but I couldn't. The problem is:

Let $U$ be unitary matrix. Let $P$ and $UP$ be orthogonal projections. Is it true that $U^{2}=P$? If yes, please show me how to prove it.

share|improve this question
What if $P=0$? Or what if the nullspace of $P$ has dimension at least $2$, and $U$ exchanges two orthonormal vectors in a given basis for $N(P)$? –  Arturo Magidin Apr 24 '12 at 3:49
add comment

2 Answers 2

up vote 5 down vote accepted

A very strange question. $U^2$ is also a unitary matrix, so it can't be $P$ unless $P = I$ (in which case $UP=U$ must also be $I$, because that's the only orthogonal projection that is unitary). For any other orthogonal projection $P$, you can get counterexamples by taking $U$ to be a unitary matrix for which $\text{Ran}(P)$ and $\text{Ker}(P)$ are invariant subspaces, with $Ux=x$ for $x \in \text{Ran}(P)$.

share|improve this answer
add comment

Note that if $P=U^2$, then $P$ would be invertible; but since $P^2=P$ (being a projection) that would imply that $P=I$.

So the answer would be "yes" if and only if the only way for $P$ and $UP$ to both be orthogonal projections when $U$ is unitary would be for $P$ to be the identity.

This is never true, since we can always take $U=I$ no matter what $P$ is. And we can always take $P=0$, no matter what $U$ is. In neither case will it follow necessarily that $U^2=P$.

So let us make this a bit more interesting and add the assumption that $U\neq I$ and that $P\neq 0$. Will the implication hold then?

If $\dim(V)\geq 3$, this is not the case: let $P$ be an orthogonal projection with nullspace of dimension $2$. Let $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3,\ldots,\mathbf{e}_n$ be an orthonormal basis, where $\mathbf{e}_1,\mathbf{e}_2$ are basis for the nullspace, and $\mathbf{e}_3,\ldots,\mathbf{e}_n$ are a basis for the range. Then let $U$ be the linear transformation that exchanges $\mathbf{e}_1$ and $\mathbf{e}_2$, and leaves the rest of the vectors invariant. Then $UP=P$, but $P\neq U^2$, since $P$ is not invertible.

If $\dim(V)= 2$, then the result is still false: if $P\neq 0$, then $P$ is either similar to $P(x,y) = (x,0)$, or is the identity. If $P$ is (essentially) $P(x,y)=(x,0)$, then $UP$ must send $(0,y)$ to $(0,0)$ hence $\{(0,y)\}^{\perp} = \{(x,0)\}$ to itself; that is, $U$ must send $(1,0)$ to $(1,0)$, and must send $(0,1)$ to $(0,u)$ where $u$ is a unit of the underlying field. Either way, $U^2\neq P$.

If $\dim(V)=1$, then the only nonzero projection is the identity; the result will hold since $U$ must act like the identity on the image of $P$, hence if $P\neq 0$ then $U=P=I$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.