# Eigenvalues and determinant

If A is a symmetric matrix whose eigenvalues have absolute values that are less than $1$, then: $$\det(I-A)\neq0$$ where $I$ is identity matrix.

Why is that inequality correct?

• This is false, for $A=I$. Mar 28, 2017 at 11:48
• It is not correct as written. Choose $A=I$. This matrix is clearly symmetric, has some eigenvalues (equal to $1$), but $det(I-A)=0$. Mar 28, 2017 at 11:48
• ok, I edited question Mar 28, 2017 at 11:49
• This holds when $1$ is not an eigenvalue of $A$. Mar 28, 2017 at 11:51
• This is a simple consequence of the definition of eigenvalues: $\lambda I-A$ is singular iff $\lambda$ is an eigenvalue of $A$.
– amd
Mar 28, 2017 at 19:21

## 2 Answers

If you see "symmetric" (even normal is enough), diagonalize! \begin{align} \det(I - A) &= \det(UIU^* - U\Lambda U^*) \\&= \det(U(I-\Lambda) U^*) \\&= \det(U)\det(I-\Lambda)\det(U^*) \\&= \det(I-\Lambda) \\&= \prod\limits_{j=1}^n (1-\lambda_j) \\\det(I - A) &= 0 \iff \exists j\colon\lambda_j = 1 \end{align}

• Now I feel bad for not including that the last line is naturally true for any $A$, but I don't want to bump this post. Mar 28, 2017 at 17:03

Mosquito-nuking solution: $$\rho(A) < 1\implies \sum_{n=0}^{\infty}A^n\text{ is convergent}$$ (see Neumann series and spectral radius) and $$(I - A)^{-1} = \sum_{n=0}^{\infty}A^n\text{ exists}.$$