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.
 A: 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.
A: 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. 
A: 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 $\ker(A)$.
