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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 $I$.

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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 $R R^*$.

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