# Eigenvalue perturbation of singular matrix

Consider a Hermitian matrix $\mathbf{A_0} \in \mathbb{C}^{N \times N}$ with one singularity, i.e. its eigenvalues in increasing order are: $$0 < \lambda_2 \leq \lambda_3 \leq \cdots \leq \lambda_N$$ The eigenvector corresponding to $\lambda_k$ is $\mathbf{v_k}$. Now consider a matrix with small entries $\mathbf{\delta A}$ and $\mathbf{A} = \mathbf{A_0} + \mathbf{\delta A}$, let the eigenvalues of $\mathbf{A}$ be $$\gamma_1 \leq \gamma_2 \leq \cdots \leq \gamma_N$$ The eigenvector corresponding to $\gamma_k$ is $\mathbf{u_k}$. It is well-known that a $1^{st}$ order perturbation of eigenvalues is $$\gamma_k = \lambda_k +\mathbf{v_k^H} \mathbf{\delta A} \mathbf{v_k} + O(\delta^2)$$ There is also a similar expression of $\mathbf{u_k}$ as a function of $\mathbf{v_1} \ldots \mathbf{v_k}$.

The questions are:

1) Is this true for $k = 1$, i.e. can we say $\gamma_1 = \mathbf{v_1^H} \mathbf{\delta A} \mathbf{v_1} + O(\delta^2)$ ? The reason i'm asking this is that i see repeatedly that the matrix $\mathbf{A_0}$ should be positive definite so that the perturbation on eigenvalues hold true.

2) If (1) is true, can we approximate $\mathbf{A}^{-1}$ by $\frac{1}{\gamma_1}\mathbf{u_1}\mathbf{u_1^H}$ ? Is there any formal theory behind this apprxoimation ?

Thank you very much in advance.

• Do you mean by "one singularity" that $\lambda_1=0$ and $Ker(A)=\mathbb{v_1}$ for a certain $v_1 \neq 0$ ? And that when you say, at the end that $A^{-1}$ is approximated by... it means that the pseudo-inverse $A^+$ IS $\frac{1}{\gamma_1}\mathbf{u_1}\mathbf{u_1^H}$? – Jean Marie Jul 23 '16 at 10:07
• Hi JeanMarie, For the first part of your question, yes "one singularity" is what you just said. For the second part, no i mean $\mathbf{A}^{-1}$, what happens if i say pseudo-inverse ? – Ahmad Bazzi Jul 23 '16 at 10:15
• But the problem is that $\mathbf{A}^{-1}$ doesn't exist because of this eigenvalue $0$... – Jean Marie Jul 23 '16 at 10:58
• Hi again, well 0 is an eigenvalue of $\mathbf{A_0}$ and not $\mathbf{A}$. – Ahmad Bazzi Jul 23 '16 at 11:07
• Sorry, I should have watched... – Jean Marie Jul 23 '16 at 11:55