Proving that the inverse of a matrix $H$ is itself

Question

Let $u$ be a column vector in $\mathbb{R}^n$, and let $H = I_n - uu^T$ where $I_n$ is the $n \times n$ identity matrix. Also, it is known that $H$ is symmetric. Prove that if $u^Tu = 2$, then the inverse of $H$ is itself.

Theoretically speaking, I know the easiest way to approach this is to show that $HH = I_n$. Currently I'm stuck at the following and can't get past the last equation: $$HH = (I_n - uu^T)(I_n - uu^T) = I_n^2 - 2uu^T + (uu^T)^2 = \cdots$$

How can I solve this equation using $u^Tu = 2$? I just can't see where this piece of information can assist me in proving the desired equation. I can't manipulate the last part to suit my needs.

Tips and tricks are greatly appreciated!

Answer 1

HINT

Notice that $(uu^T)^2 = u(u^Tu)u^T = 2uu^T$.

Answer 2

You're almost there. Just use these facts:

• ${I_n}^2=I_n$;
• $(uu^T)^2=u\overbrace{u^Tu}^{\phantom{2}=2}u^T=2uu^T$.