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**Deveno** for any 0-vector (nx1 matrix) u (= 0), and any other row-vector $\displaystyle w^T$ (1xn matrix),

$\displaystyle w^Tu = w^T0 = 0$.

we are given that the vector $\displaystyle (\lambda I - A)^2v = 0$ for some $\displaystyle \lambda$.

so multiplying it by $\displaystyle v^T$ will also give a 0 1x1 matrix, that is, the number 0.

remember, A is symmetric so λI - A is symmetric too: $\displaystyle (\lambda I - A)^T = \lambda I - A$.

thus $\displaystyle 0 = v^T(\lambda I - A)^2v = v^T(\lambda I - A)^T(\lambda I - A)v$

$\displaystyle = ((\lambda I - A)v)^T(\lambda I - A)v) =\ <(\lambda I - A)v,(\lambda I - A)v>\ = ||(\lambda I - A)v||^2$.

you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.