Prove that if $X=n\times k$ matrix of rank $k$, then $X(X^TX)^{-1}X^T$ has rank $k$

Question

Prove that if $X=n\times k$ matrix of rank $k$, then $X(X^TX)^{-1}X^T$ has rank $k$

I thought that $X^TX$ would be a full rank matrix of rank $k$, since it must have full rank to be invertable.

Then $X$ multiplied by $(k\times k)$ matrix, will be a matrix of rank $\min{k,n}$. Then multiply this by a matrix of rank $k$ we get $k$ as the total rank.

Is there a rule that says: if $A$ and $B$ have rank $k$ then so does $AB$ and $BA$?

Have you checked this post: http://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope — iamvegan, Aug 20 '16 at 12:15
Yes they prove $\mathrm{rank}AB\leq \min{k,k}$ But I want to show it is equal to — GRS, Aug 20 '16 at 12:19
Oh sorry. This might not be true. Consider $A= [0,0;0,1]$ and $B= [1,0;0,0]$ then we have $AB=[0,0;0,0]$. Am I correct? Here $A,B$ has rank 1 but the multiplication is rank 0. — iamvegan, Aug 20 '16 at 12:23
Matrix $X(X^TX)^{-1}X^T$ is the (Moore-Penrose) pseudo-inverse of $X$. — Jean Marie, Aug 20 '16 at 12:26
@Michael Hardy : error of mine, you are perfectly right, we deal here with $XX^+.$ — Jean Marie, Aug 20 '16 at 14:54

H. H. Rugh · Accepted Answer · 2016-08-20T13:04:19.970

The rank of a product does not exceed the rank of each of its factors. Multiply your expression by $X$ to the right. You get $X$. Thus, rank $X=k$ is an upper and a lower bound for the rank you look for.

When $A$ is $n\times k$ and $B$ is $k\times n$ (with $k<n$) and both have rank $k$ then $AB$ also has rank $k$. The reason is that $B$ is surjective and $A$ injective.

$BA$ may have lower rank, however. In general one can only say that the rank $\geq \max\{0,2k-n\}$ (because some of the image of $A$ may map into the kernel of $B$). But this is a bit more complicated.

Prove that if $X=n\times k$ matrix of rank $k$, then $X(X^TX)^{-1}X^T$ has rank $k$

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