Show that the matrix $I-uu^T$ has rank $n-1$ where $u$ is a unit vector in $R^n$ I've tried a few examples and I know that $I-uu^T$ is symmetric but I'm stuck here. Any help will be appreciated!

Note that $u$ is in the kernel of $I-uu^\top$. Moreover, every nonzero vector orthogonal to $I-uu^\top$ in not in the kernel of $I-uu^\top$. What does this tell us about the dimension of the kernel of $I-uu^\top$?
More explicitly, note that $(I-uu^\top)u=Iu-uu^\top u=u-u1=u-u=0$ since $u$ is a unit vector so $u^\top u=1$. This means that $I-uu^\top$ is in the kernel of $I-uu^\top$. Moreover, if $u^\top v=0$, then $(I-uu^\top)v=Iv-uu^\top v=v-u0=v\neq0$.
