031 Review Part 3, Problem 11

Suppose  $\{{\vec {u}},{\vec {v}}\}$  is a basis of the eigenspace corresponding to the eigenvalue 0 of a  $5\times 5$  matrix  $A.$

(a) Is  ${\vec {w}}={\vec {u}}-2{\vec {v}}$  an eigenvector of  $A?$  If so, find the corresponding eigenvalue.

If not, explain why.

(b) Find the dimension of  ${\text{Col }}A.$

Foundations:
1. An eigenvector  ${\vec {x}}$  of a matrix  $A$  corresponding to the eigenvalue  $\lambda$  is a nonzero vector such that
$A{\vec {x}}=\lambda {\vec {x}}.$
2. By the Rank Theorem, if  $A$  is a  $m\times n$  matrix, then
${\text{rank }}A+{\text{dim Col }}A=n.$

Solution:

(a)

Step 1:
First, notice
${\vec {w}}\neq {\vec {0}}$
since  $\{{\vec {u}},{\vec {v}}\}$  is a basis of the eigenspace corresponding to the eigenvalue 0 of  $A.$
Also, we have
$A{\vec {u}}={\vec {0}}$  and  $A{\vec {v}}={\vec {0}}$
since  ${\vec {u}}$  and  ${\vec {v}}$  are eigenvectors of  $A$  corresponding to the eigenvalue 0.

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {A{\vec {w}}}&=&\displaystyle {A({\vec {u}}-2{\vec {v}})}\\&&\\&=&\displaystyle {A{\vec {u}}-2A{\vec {v}}}\\&&\\&=&\displaystyle {{\vec {0}}-2{\vec {0}}}\\&&\\&=&\displaystyle {{\vec {0}}.}\end{array}}$
Hence,  ${\vec {w}}$  is an eigenvector of  $A$  corresponding to the eigenvalue  $0.$

(b)

Step 1:
Since  $\{{\vec {u}},{\vec {v}}\}$  is a basis for the eigenspace of  $A$  corresponding to the eigenvalue 0, we know that
${\text{dim Nul }}A=2.$

Step 2:
Then, by the Rank Theorem, we have
${\begin{array}{rcl}\displaystyle {5}&=&\displaystyle {{\text{dim Col }}A+{\text{dim Nul }}A}\\&&\\&=&\displaystyle {{\text{dim Col }}A+2.}\end{array}}$
Hence, we have
${\text{dim Col }}A=3.$

Final Answer:
(a)     See solution above.
(b)     ${\text{dim Col }}A=3.$

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