How many orthogonal eigenvectors does a symmetric and positive semidefinite matrix $A_{n\times n}$ has?

Suppose $A_{n\times n}$ is a symmetric and positive semidefinite matrix, and Rank(A)=k. I know that $A$ has k nonzero eigenvalues and corresponding orthogonal eigenvectors $v_1,\ldots,v_k$. I have two questions:

(1) I wonder whether eigenvectors corresponding to remaining zero eigenvalues is orthogonal to $v_1,\ldots,v_k$.

(2) I wonder whether eigenvectors corresponding to remaining zero eigenvalues is distinct or same.

The eigenvectors corresponding to the zero eigenvalues form a basis for $\mathcal{N}(A)$. Now, this basis is independent of the basis for $R(A)\setminus\mathcal{N}(A)$ which can be easily verified. Thus, this two sets of vectors can together form a basis for $R(A)$ and from that using Gram-Schmidt orthogonalization, one can form an orthogonal set of vectors.

• I wonder whether eigenvectors corresponding to remaining zero eigenvalues is distinct or same. – Dajiang Lei Feb 27 '15 at 16:13
• Well this is exactly what is said in my answer (5 min earlier).... – Surb Feb 27 '15 at 16:51
• Oh, Sorry @Surb, I did not notice your answer while I was writing mine. I upvoted yours by the way. – Samrat Mukhopadhyay Feb 28 '15 at 9:07
• @DajiangLei, the eigenvectors corresponding to the zero eigenvalues form a basis for $\mathcal{N}(A)$, so they are definitely distinct. – Samrat Mukhopadhyay Feb 28 '15 at 9:09
• I want to know that if I put the eigenvectors corresponding to nonzero eigenvalues and the eigenvectors corresponding to zero eigenvalues together to form one matrix $U$, whether U is an orthogonal matrix. – Dajiang Lei Feb 28 '15 at 17:02

The spectral theorem tells us that every $n \times n$ symmetric matrix has $n$ mutually orthogonal (or, if you prefer, orthonormal) eigenvectors. A positive semidefinite symmetric matrix is no exception.

• If I just employ SVD on $A$ to get all eigenvectors corresponding to zero eigenvalues and nonzero eigenvalues, after normalizing all the eigenvectors, then put all the eigenvectors together to form one matrix $U$, whether $U$ will be orthogonal matrix. – Dajiang Lei Mar 2 '15 at 10:18
• But SVD already gives you that matrix $U$. And yes, it must be orthogonal. – Omnomnomnom Mar 2 '15 at 11:24

If $v_i,v_i$ are eigenvectors corresponding to distinct eigenvalues $\lambda_i,\lambda_j$, then $$\lambda_i \langle v_i,v_j\rangle = \langle \lambda_iv_i,v_j\rangle =\langle Av_i,v_j\rangle = \langle v_i,Av_j\rangle = \lambda_j\langle v_i,v_j\rangle,$$ and thus we must have $\langle v_i,v_j\rangle = 0$. The space of eigenvectors corresponding to the zero eigenvalue is the kernel of $A$. Now from the Gram-schmidt process, if you have an orthogonal basis of $V\setminus \ker(A)$, then you can complete it into an orthogonal basis of $V$ and the additional basis vectors forms a basis of $\ker(A)$.

• How many basis vectors does $V$ have? – Dajiang Lei Feb 27 '15 at 16:04
• Do you mean V include all eigenvectors corresponding to nonzero eigenvalues? – Dajiang Lei Feb 27 '15 at 16:06
• I wonder whether eigenvectors corresponding to remaining zero eigenvalues is distinct or same. – Dajiang Lei Feb 27 '15 at 16:06
• How many basis vectors does $ker(A)$ have? – Dajiang Lei Feb 27 '15 at 16:08
• @DajiangLei So, since $A$ is of dimension $n\times n$, here $V$ is the space of dimension $n$ such that $A\colon V \to V$. So there are $n$ basis vectors in $V$. The number of basis vectors in $\ker(A)$ depends on $A$ (e.g. if $A=0$, then there are $n$ and if $A = Id$ then there are $0$). The eigenvectors corresponding to the eigenvalue $0$ are orthogonal yes (follows from my comment with gram-schmidt process). – Surb Feb 27 '15 at 16:17