Lin. Alg. Invertility

[hydralisks]

To find out if 4 in an eigenvalue of A, decide if A-4I is invertible...

So, if A-4I is invertible, then its cols are lin ind by IMT, and also there is only the trivial solution to A-4I=0, so thus 4 is not an eigenvalue of A

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[HallsofIvy]

? The definition of "eigenvalue" is that $\lambda$ is an eigenvalue if and only if $Av= \lambda v$ has non-trivial solutions. That is the same as saying that $Av- \lambda v= (A- \lambda I)v= 0$ has non-trivial solutions. Since v=0 is obviously a solution, saying it has non-trivial solutions means it does NOT have a "unique" solution. If the matrix M has an inverse, then the equation Mv= u has the unique solution \(\displaystyle v= M^{-1}u\).