Linear Algebra

If Eigenvalues of a Matrix $A$ are Less than $1$, then Determinant of $I-A$ is Positive

Problems and solutions in Linear Algebra

Problem 237

Let $A$ be an $n \times n$ matrix. Suppose that all the eigenvalues $\lambda$ of $A$ are real and satisfy $\lambda <1$.

Then show that the determinant \[ \det(I-A) >0,\] where $I$ is the $n \times n$ identity matrix.

Solution 1.

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

Then we have
\[p(t)=\det(A-tI)=\prod_{i=1}^n(\lambda_i-t)=(-1)^n\prod_{i=1}^n(t-\lambda_i),\] where $\lambda_i$ are eigenvalues of $A$.

Thus substituting $t=1$, we have
\[\det(I-A)=(-1)^n(-1)^n\prod_{i=1}^n(1-\lambda_i)=\prod_{i=1}^n(1-\lambda_i)>0.\] Since eigenvalues $\lambda_i$ are less than $1$, the product is positive, and this proves $\det(I-A)>0$.

Solution 2.

There exists an invertible matrix $P$ such that the matrix $P^{-1}AP$ is the Jordan canonical form. That is $P^{-1}AP$ is an upper triangular matrix whose diagonal entries are eigenvalues $\lambda_i$ of the matrix $A$.
Then we have
\begin{align*}
\det(I-A)=\det(I-P^{-1}AP)=\prod_{i=1}^n(1-\lambda_i).
\end{align*}

Here the last equality follows from the fact that the determinant of an upper triangular matrix is the product of its diagonal entries.

Since the eigenvalues $\lambda_i$ are less than $1$, the last product is positive. Thus we proved $\det(I-A)>0$.

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