Is $Q( x)={x'\left(I_n-\frac{{11^\top}}{n}\right)x}$ a positive semidefinite quadratic form?

Question

I don't understand how could i demonstrate it. s=(1,1,...,1). Is Q(x)=x'Ax a semi-definite positive? Thanks I attach a photo.

Answer 1

The eigenvalues of $ss'/n$ are $1$ (with multiplicity $1$) and $0$ (with multiplicity $n-1$) since it is a rank $1$ matrix.

Thus the eigenvalues of $I - \frac{ss'}{n}$ are $0$ or $1$.

Answer 2

This quadratic form can be written:

$$\tag{1}x_2^2+\cdots+x_n^2$$

in a convenient basis (in other words $0x_1^2+1x_2^2+\cdots+1x_n^2,$ where $0,1 \cdots 1$ are certain eigenvalues).

We are going to explain it through a simple geometric interpretation.

Lemma: $P=I-NN^T$ (where $N$ is a norm 1 column vector) is the matrix of the orthogonal projection onto the hyperplane (H) orthogonal to $N$ (proof below).

As a consequence, $P$ has one eigenvector with eigenvalue $0$, namely $N$, and $n-1$ eigenvectors associated with eigenvalue 1, (take any basis of (H)) because these vectors are orthogonally projected... on themwselves.

Proof of the lemma:

• $PN=(I-NN^T)N=N-N(N^TN)=N-N1=0.$

• If $M \in (H) \ \iff \ M \perp N$ then $PN=(I-NN^T)M=M-N(N^TM)=M-0=M.$

Proof of (1): take here the norm 1 vector $N=\tfrac{1}{\sqrt{n}}{\mathbb 1}.$ (${\mathbb 1}$ being the vector with all its components equal to 1.). And take for the other components an orthonormak basis of hyperplane (H).