031 Review Part 3, Problem 6

(a) Show that if  ${\vec {x}}$  is an eigenvector of the matrix  $A$  corresponding to the eigenvalue 2, then  ${\vec {x}}$  is an eigenvector of  $A^{3}-A^{2}+I.$  What is the corresponding eigenvalue?

(b) Show that if  ${\vec {y}}$  is an eigenvector of the matrix  $A$  corresponding to the eigenvalue 3 and  $A$  is invertible, then  ${\vec {y}}$  is an eigenvector of  $A^{-1}.$  What is the corresponding eigenvalue?

Foundations:
An eigenvector  ${\vec {x}}$  of a matrix  $A$  corresponding to the eigenvalue  $\lambda$  is a nonzero vector such that
$A{\vec {x}}=\lambda {\vec {x}}.$

Solution:

(a)

Step 1:
Since  ${\vec {x}}$  is an eigenvector of  $A$  corresponding to the eigenvalue  $2,$  we know  ${\vec {x}}\neq {\vec {0}}$  and
$A{\vec {x}}=2{\vec {x}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {(A^{3}-A^{2}+I){\vec {x}}}&=&\displaystyle {A^{3}{\vec {x}}-A^{2}{\vec {x}}+I{\vec {x}}}\\&&\\&=&\displaystyle {A\cdot A\cdot A{\vec {x}}-A\cdot A{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {A\cdot A\cdot 2{\vec {x}}-A\cdot 2{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {2A\cdot A{\vec {x}}-2A{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {2A\cdot 2{\vec {x}}-2\cdot 2{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {(2\cdot 2)A{\vec {x}}-4{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {(4)\cdot 2{\vec {x}}-4{\vec {x}}+{\vec {x}}}\\&&\\&=&\displaystyle {5{\vec {x}}}.\end{array}}$
Hence, since  ${\vec {x}}\neq {\vec {0}},$  we conclude that  ${\vec {x}}$  is an eigenvector of  $A^{3}-A^{2}+I$  corresponding to the eigenvalue  $5.$

(b)

Step 1:
Since  ${\vec {y}}$  is an eigenvector of  $A$  corresponding to the eigenvalue  $3,$  we know  ${\vec {y}}\neq {\vec {0}}$  and
$A{\vec {y}}=3{\vec {y}}.$
Also, since  $A$  is invertible,  $A^{-1}$  exists.

Step 2:
Now, we multiply the equation from Step 1 on the left by  $A^{-1}$ to obtain
${\begin{array}{rcl}\displaystyle {A^{-1}(A{\vec {y}})}&=&\displaystyle {A^{-1}(3{\vec {y}}}\\&&\\&=&\displaystyle {3(A^{-1}{\vec {y}}).}\end{array}}$
Now, we have
${\begin{array}{rcl}\displaystyle {3(A^{-1}{\vec {y}})}&=&\displaystyle {A^{-1}(A{\vec {y}})}\\&&\\&=&\displaystyle {(A^{-1}A){\vec {y}}}\\&&\\&=&\displaystyle {I{\vec {y}}}\\&&\\&=&\displaystyle {{\vec {y}}.}\end{array}}$
Hence,  $A^{-1}{\vec {y}}={\frac {1}{3}}{\vec {y}}.$
Therefore,  ${\vec {y}}$  is an eigenvector of  $A^{-1}$ corresponding to the eigenvalue  ${\frac {1}{3}}.$

Final Answer:
(a)     See solution above.
(b)     See solution above.

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