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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $\{u_1, \dots, u_r\}$ is an orthonormal set in $\Bbb C^n$ ($r \leq n$), and consider the matrix $R$ whose columns are these vectors: $$R = [u_1, \dots, u_r].$$ Show that $R^\ast R$ and $RR^\ast$ are projectors, one of them being $I$.

What I did is this: Since $u_1, \dots, u_r$ are orthonormal vectors in $\Bbb C^n$, then:

$$(R^\ast R)^2 = (R^\ast R)(R^\ast R) = R^\ast RR^\ast R = R^\ast (RR^\ast)R = R^\ast(I)R = R^\ast R$$ and $$(RR^\ast)^2 = (RR^\ast)(RR^\ast) = RR^\ast RR^\ast = R(R^\ast R)R^\ast = R(I)R^\ast = RR^\ast.$$

Is this right? But how can I show that one of them is $I$? Well, in my opinion, I may think both of them are $I$.

share|cite|improve this question
up vote 1 down vote accepted

Since $u_1, \cdots, u_r$ are orthonormal, they are linearly independent, and so $R$ has full column rank. Now $R^* R=I_r$ because $(R^*R)_{ij}=u_i^*u_j=\delta_{ij}$, by the orthonormality of the $u_i$. However, notice that $R R^*$ is an $n \times n$ matrix of rank $r$. So unless $r=n$, $R R^*$ can not be invertible and hence can not be equal to $I_n$. What is true is that $R R^*$ is an orthogonal projection operator. It takes a vector $x$ and maps it to its projection on the space spanned by $u_1, \cdots, u_r$, i.e. $x \mapsto R (R^*x)=\sum_{i=1}^n <u_i,x> u_i$. To show that $R R^*$ is a projection operator, you need to show that $\mathbb{C}^n= \mathcal{R}(RR^*) \oplus \mathcal{N}(RR^*)$, i.e. $\mathbb{C}^n$ is the direct sum of the range space and the nullspace of $R R^*$.

share|cite|improve this answer

Your Answer

 
discard

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.