031 Review Part 3, Problem 10

Show that if  ${\vec {x}}$  is an eigenvector of the matrix product  $AB$  and  $B{\vec {x}}\neq {\vec {0}},$  then  $B{\vec {x}}$  is an eigenvector of  $BA.$

Foundations:
An eigenvector  ${\vec {x}}$  of a matrix  $A$  is a nonzero vector such that
$A{\vec {x}}=\lambda {\vec {x}}$
for some scalar  $\lambda .$

Solution:

Step 1:
Since  ${\vec {x}}$  is an eigenvector of  $AB,$  we know  ${\vec {x}}\neq {\vec {0}}$  and
$AB({\vec {x}})=\lambda {\vec {x}}$
for some scalar  $\lambda .$
Using associativity of matrix multiplication, we have
$A(B{\vec {x}})=\lambda {\vec {x}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {BA(B{\vec {x}})}&=&\displaystyle {B(A(B{\vec {x}}))}\\&&\\&=&\displaystyle {B(\lambda {\vec {x}})}\\&&\\&=&\displaystyle {\lambda B{\vec {x}}.}\end{array}}$
Since  $B{\vec {x}}\neq {\vec {0}},$  we can conclude that  $B{\vec {x}}$  is an eigenvector of  $AB.$

Final Answer:
See solution above.

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