Eigen decomposition of a particular projection matrix

Question

Let $\boldsymbol{1}$ be a $n$ dimensional all-one vector. What is the eigen decomposition of the following matrix? $$I-\frac{1}{n}\boldsymbol{1}\boldsymbol{1}^T, $$ here $I$ is an identity matrix

Answer 1

Let $V_1:=\frac{1}{\sqrt{n}}\boldsymbol{1}$ (normes vector) and $M=I-V_1V_1^T=I-\frac{1}{n}\boldsymbol{1}\boldsymbol{1}^T.$

Let $(V_2,V_3,...V_{n})$ be an orthogonal basis completing $V_1$ (an example of such a basis is given at the botto of this page.)

Consider product $$\tag{1}MV_1=V_k-\frac{1}{n}\boldsymbol{1}\boldsymbol{1}^TV_k$$

• if $k=1$: due to the fact that $\boldsymbol{1}^T\boldsymbol{1}=n$, (1) becomes $MV_1=\boldsymbol{1}-\frac{1}{n}\boldsymbol{1}n$; thus $$\tag{2}MV_1=0.$$

• if $k=2,...n$, because $V_k$ is orthogonal to $\boldsymbol{1}$, we have $\boldsymbol{1}^TV_k=0$, thus (1) becomes $$\tag{3}MV_k=V_k-0=V_k \ \ \text{for} \ \ k=2,3...n$$

Equalities (2) and (3) show

a) that $M$ is the projection operator on the hyperplane of $\mathbb{R}^n$ orthogonal to $\boldsymbol{1}.$

b) that $V_1$ is an eigenvector associated with eigenvalue $0$ and all other $V_k (k=2,3,... n)$ are eigenvectors associated with eigenvalue 1, which is thus or order $(n-1).$

In a concrete way, one can choose :

$$V_2=\pmatrix{-1\\1\\0\\0\\ \vdots \\ 0\\0}, \ \ V_3=\pmatrix{0\\-1\\1\\0\\ \vdots \\ 0\\0}, \ \ V_4=\pmatrix{0\\0\\-1\\1\\ \vdots \\ 0\\0}, \ \ \cdots \ \ V_n=\pmatrix{0\\0\\0\\0\\ \vdots \\ -1\\1}$$