- #1

snakebite

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Suppose wehave a 2x2 matrix A with one eigenvalue [tex]\lambda[/tex], but it is not a scalar matrix. Suppose [tex]\vec{v2}[/tex] is a nonzero vector which is not an eigenvector of A; show that [tex]\vec{v1}[/tex] = (A-[tex]\lambda[/tex])[tex]\vec{v2}[/tex] is an eigenvector of A. Also show that if P is the matrix with columns [tex]\vec{v1}[/tex] and [tex]\vec{v2}[/tex] then P^(-1)AP = [[tex]\lambda[/tex] 1

0 [tex]\lambda[/tex]]

## The Attempt at a Solution

I tried calculating (A-[tex]\lambda[/tex])[tex]\vec{v1}[/tex] to try and proove that it is equal to 0, however i end up with it being equal to (A-[tex]\lambda[/tex] I)^2[tex]\vec{v2}[/tex]

Thank you very much