Linear transformation, show that points on a plane mapped back on the same plane.

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I need help with the second part of this question

Question

The eigenvalues, with their corresponding eigenvectors are as follows (answers to the first part):

λ = 1 , $e_1$ = –2j + k

λ = 2 , $e_2$ = i + j

λ = 3 , $e_3$ = 2i + 2j + k

The answer scheme gives these steps (for the second part):

r = se + tf

A(se + tf) = sAe + tAf = (sλ)e + (tμ)f

I do not understand how λ and μ come about, a step by step guide would be appreciated.

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By parametric equation of the plane $\Pi$, your question means the set of all points in the plane $r \in \Pi$ written as linear combinations of the two eigenvectors $e$ and $f$. The question asks you to show that for any $r \in \Pi$, $Ar \in \Pi$. The scalars $\lambda, \mu$ are the eigenvalues of $e$ and $f$, respectively.

Let $r = se + tf$. Then, clearly, $r \in \Pi$. Consider $Ar$. Your job is to show that $Ar \in \Pi$ by illustrating that $Ar$ can be written as a linear combination of $e$ and $f$.

$Ar = A(se + tf) = A(se) + A(tf) = sAe + tAf = s(\lambda e) + t(\mu f) = s\lambda e + t\mu f = \beta_1 e + \beta_2 f$.

The $\lambda$ and $\mu$ come from applying the fact that $e$ and $f$ are eigenvectors of $A$ (i.e., $Ae = \lambda e$, etc.).

Therefore, you have shown that $Ar$ can be written as a linear combination of $e$ and $f$, as desired to show that $T$ maps points in $\Pi$ to points in $\Pi$.