# Relation between rank of a symmetric positive semi-definite matrix and its number of non-zero eigen values (or singular values)

Is there any relation between the rank of a symmetric positive semi-definite matrix and its number of non-zero eigenvalues (or singular values)?

For a matrix $\mathbb{P}$ Can we find the relationship between the rank and the eigenspace of the matrix?

Yes. For any square matrix, the rank is equal to the dimension of the space minus the dimension of the zero eigenspace (that is, the "geometric multiplicity" of zero). The rank is also the number of non-zero singular values (this one works for non-square matrices).

For a symmetric positive semi-definite matrix (which is symmetric and therefore diagonalizable), the rank is simply the total multiplicity of non-zero eigenvalues (or the dimension of the space minus the multiplicity of the zero eigenvalues).

In a sense, this follows from the fact that (orthogonally/unitarily) diagonalizing a positive semidefinite matrix gives you a singular value decomposition.

• not all PSD matrices are symmetric. – Sother Aug 28 '17 at 15:49
• The "multiplicity of the zero eigenvalues" is always 1, right? Because 0 would be the only eigenvalue of the nullity(A) (the nullspace of $A$). – Sother Aug 28 '17 at 16:03
• @Sother it is common to include symmetry as part of the definition of "positive definite", but I agree I should reword this answer. The multiplicity is not always $1$; for example, $$\pmatrix{1&1&1\\1&1&1\\1&1&1}$$ has a $0$-eigenvalue of multiplicity $2$. – Omnomnomnom Aug 28 '17 at 19:30
• multiplicity technically means the number of unique values, which is why I said the multiplicity is 1. – Sother Aug 30 '17 at 4:43
• @Sother that's not what the "multiplicity of an eigenvalue" refers to. Similarly, the term "multiplicity" is used in this way to describe the root of a polynomial. With respect, I'd rather not have an endless interview here; if you'd like more answers I suggest you ask your own new question. – Omnomnomnom Aug 30 '17 at 12:01