# Perturbation of a Singular Matrix is Nonsingular ## Problem 56

Suppose that $A$ is an $n\times n$ singular matrix.
Prove that for sufficiently small $\epsilon>0$, the matrix $A-\epsilon I$ is nonsingular, where $I$ is the $n \times n$ identity matrix. Add to solve later

Contents

## Hint.

Consider the characteristic polynomial $p(t)$ of the matrix $A$.

Note that the eigenvalues of $A$ are the roots of $p(t)$.

Thus if $\epsilon$ is not an eigenvalue, then $p(\epsilon)\neq 0$.

## Proof.

Let $p(t)=\prod_{i=}^n(\lambda_i-t)$ be the characteristic polynomial for $A$, where $\lambda_i$ are eigenvalues of $A$.

Let $\lambda_{i_0}$ be the nonzero eigenvalue of $A$ of the smallest absolute value.
That is $|\lambda_{i_0}|\leq |\lambda_i|$ for any nonzero eigenvalue $\lambda_i$.
Then for any $0< \epsilon <|\lambda_{i_0}|$, we have $\det(A-\epsilon I)=p(\epsilon)\neq 0$, otherwise $\epsilon$ would be an eigenvalue of $A$ but it is impossible because of the minimality of $\lambda_{i_0}$.

Therefore the matrix $A-\epsilon I$ is nonsingular for all $0<\epsilon<|\lambda_{i_0}|$. Add to solve later

### 1 Response

1. 08/07/2016

[…] note that $A-epsilon I$ is invertible matrix for sufficiently small $epsilon$. (See Problem Perturbation of a singular matrix is nonsingular for a proof of this […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### Simple Commutative Relation on Matrices

Let $A$ and $B$ are $n \times n$ matrices with real entries. Assume that $A+B$ is invertible. Then show that...

Close