Show that rank $(I - uu^*) = n-1$ where $\|u\| = 1$. Does the following proof go through?
Row reduce $uu^*$ as follows:
$$ \begin{bmatrix} u_1\bar{u}_1 & \cdots & u_1 \bar{u}_n \\ \vdots & \ddots & \vdots \\ u_n\bar{u}_1 & \cdots & u_n\bar{u}_n \end{bmatrix} \rightarrow \begin{bmatrix} 1 & (\bar{u}_1)^{-1}\bar{u}_2 & \cdots & (\bar{u}_1)^{-1} \bar{u}_n \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}. $$ Compute $I - $ rref $uu^*$ and row reduce the result, obtaining $$ \begin{bmatrix} 0 & -(\bar{u}_1)^{-1}\bar{u}_2 & -(\bar{u}_1)^{-1}\bar{u}_3 & \cdots & -(\bar{u}_1)^{-1} \bar{u}_n \\ 0 & 1 & 0 &\cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}, $$ which has $n-1$ pivots, and therefore $I - uu^*$ has rank $n - 1$.
My suspicion is that in general NO, you cannot just row reduce two matrices, add them (or subtract them), row reduce the result, and expect that to mean something. But maybe in this case YES because one of the matrices is the identity matrix and we're not doing any row interchanges?
Thanks in advance! I always appreciate this community.
As you may know, the usual geometric proof is to observe that if $u^{*}v = 0$ (i.e., $v$ is orthogonal to $u$ with respect to the standard Hermitian inner product), then $(I - uu^{*})v = v$. One therefore gets not only the rank, but the eigenvalues and eigenspaces: Since $|u| = 1$, the matrix $I - uu^{*}$ is orthogonal projection to the orthogonal complement of $u$.
This viewpoint confirms Enrico's comment that row reduction does not block-diagonalize $I - uu^{*}$: The columns span the orthogonal complement of $u$, and row operations effect change of basis in that hyperplane, which generally does not contain (a subset of) the standard basis.