Given the expression of the matrix find the adjoint and the square of the matrix

Question

Let the matrix $U=\mathbb{I}-ww^* $ with $w\in\mathbb{C}^3$ being a column vector, find the expression for $U^*$ and for $U^2$.

So for the adjoint $U^* $ I gave it a try and got $U^* =\mathbb{I}-(ww^* )^* =\mathbb{I}-w^* w$ but my book says $U^* =U$, how is that possible?

And for $U^2=(\mathbb{I}-ww^* )(\mathbb{I}-ww^* )=\mathbb{I}-2ww^* +ww^* ww^* =\mathbb{I}-(2-ww^* )ww^* $ but I don't know how else to continue nor if this is right.

Answer 1

Recall the product rule for adjoints: $$ (AB)^* = B^*A^*\ . $$ Can you use this to see what you did wrong in your calculation for $U^*$? Otherwise, take a look here:

$$U^* = \Bbb I - (ww^*)^* = \Bbb I - (w^*)^*w^* =\Bbb I - ww^* = U\,. $$

Your calculation for $U^2$ seems to be correct.

EDIT: Here' why your result and the book's result are equivalent: $$ U^2 = \Bbb{I} - 2ww^* + ww^*ww^* = \Bbb{I} - 2ww^* + w(w^*w)w^* = \Bbb{I} - 2ww^* + w\|w\|^2w^* = \Bbb{I} - 2ww^* + \|w\|^2\,ww^* = \mathbb{I}-(2-\|w\|^2)ww^*\ . $$