The linear system $Bx = b$ is consistent $\iff$ $BB^Tx = b$ is consistent

Question

I want to prove the following:

The linear system $Bx = b$ is consistent $\iff$ $BB^Tx = b$ is consistent ($B$ is $m \times n$ matrix)

$\Leftarrow$ is trivial, OK. But I have no idea, how to prove $\Rightarrow$.

I know, that if $Bx = b$ is consistent then if $p$ is a string of properly length such as $pB = 0$ then $pb = 0$. And I think that the idea of the proof should be: $pBB^T = 0 \rightarrow pB=0 \rightarrow pb = 0$ but I failed to prove the first $\rightarrow$.

If rank of $B$ is $n$ then indeed $pBB^T = 0 \rightarrow pB = 0$ (otherwise the non trivial linear combination of columns gives a zero vector which isn't possible).

But if $\operatorname{rank}B \neq n$?

Could you please give me any hints?

Thanks in advance!

Answer 1

Oh, I seem to have managed to think up a proof:

Assume $\exists \ x_0 : Bx_0 = b$. $\operatorname{rank}[\ BB^T |b\ ] \ge \operatorname{rank} BB^T$ as it has more columns, but $[\ BB^T | \ b\ ] = [BB^T \ | \ Bx_0] = B\ [B^T \ | \ x_0]$ and $\operatorname{rank} (B \ [B^T\ |\ x_0]) \le \operatorname{rank} B$.

Hence, from the fact $\operatorname{rank} BB^T = \operatorname{rank} B \rightarrow \operatorname{rank} [BB^T\ |\ b] = \operatorname{rank} BB^T$ therefore the system is consistent!