Show that the matrix $I-uu^T$ has rank $n-1$ where $u$ is a unit vector in $R^n$

Question

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!

Answer 1

Let $A=I-uu^t$. Then $Au=u-u(u^tu)=u-u=0$ ($u^tu=1$ as $u$ is a unit vector).

If $v$ is a vector orthogonal to $u$ then $u^tv=0$ and $Av=v-u(u^tv)=v$.

If we take an orthogonal basis $v_1=u$, $v_2,\ldots,v_n$, then with respect to this basis $A$ has matrix $$\pmatrix{0&0&0&\cdots&0\\0&1&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\\}$$ so surely has rank $n-1$.

Answer 2

Note that $\rm P_u := u u^\top$ is the (rank-$1$) projection matrix that projects onto the line spanned by vector $\rm u$. Hence, ${\rm I}_n - {\rm P_u}$ is the projection matrix that projects onto the $(n-1)$-dimensional orthogonal complement of the line. Since the rank of a projection matrix is equal to its trace,

$$\mbox{rank} \left( {\rm I}_n - {\rm P_u} \right) = \mbox{tr} \left( {\rm I}_n - {\rm P_u} \right) = n - 1$$

Answer 3

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 $(I-uu^\top)v=Iv-uu^\top v=v-u0=v\neq0$.

Answer 4

Another approach, once you show that $A = I - u u^{\mathsf T}$ satisfies $Au = 0$, is to use the rank inequality $\text{rank}(A+B) \le \text{rank}(A) + \text{rank}(B)$.

In our case, if we take $B = uu^{\mathsf T}$, then $\text{rank}(B) =1$: all the columns of $B$ are multiples of $u$. But $\text{rank}(A+B) = \text{rank}(I) = n$. Therefore $\text{rank}(A)$ needs to be at least $n-1$. It is not $n$ (because $A$ has a nontrivial kernel), so it must be $n-1$ exactly.