Given a $n \times n$ matrix and a vector $b \neq 0$, the solution of a non-homogeneous system of linear equations $AX = b$ is $X = -b$. It follows that $-b$ is an Eigenvector of $A$ with Eigenwert $1$

Is this statement True or False?

I would answer False, since $b$ is the Eigenvector with Eigenwert $-1$. $\text{Span}(b)$ is invariant subspace under $A$ but that only follows from $b$ being the Eigenvector.



The eigenvalue is suppose to be $-1$.

Suppose on the contrary that we have $A(-b)=(-b)$, this would imply that $b=-b \implies b=0$ which is a contradiction.


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