If the square matrix A is positive definite, and I is the conformable identity matrix, then I − A is also positive definite.

Question

I was just wondering why this assumption is false and how to prove it

"If the squarematrix Ai spositive definite, and I is the conformable identity matrix, then I − A is also positive definite."

Answer 1

Just assume $A=I$ to see that it is not true in general.

Answer 2

To elaborate a bit, assume $A$ is a real symmetric matrix to avoid ubiquitous disputes whether or not a positive definite matrix should be symmetric by definition. Then we have the following:

• $A$ is positive definite iff $$\tag{1} x^TAx>0 \quad \text{for all} \quad x\neq 0.$$

• $I-A$ is positive definite iff $$\tag{2}x^T(I-A)x>0 \iff x^TAx<x^Tx \quad \text{for all} \quad x\neq 0.$$

If $\lambda_\min(A)$ and $\lambda_\max(A)$ are, respectively, the minimum and maximum eigenvalues of $A$, then (1) and (2) are equivalent to

• $A$ is positive definite iff $\lambda_\min(A)>0$,
• $I-A$ is positive definite iff $\lambda_\max(A)<1$.

Obviously, the two conditions are different. The only thing you can be sure of is that if $\lambda_\min(A)\geq 1$, then we cannot have $\lambda_\max(A)<1$. This is the case in the other answer with $A=I$.