031 Review Part 3, Problem 9

Assume  $A^{2}=I.$  Find  ${\text{Nul }}A.$

Foundations:
Recall that the subspace  ${\text{Nul }}A$  is the set of all solutions to  $A{\vec {x}}={\vec {0}}.$

Solution:

Step 1:
Since  $A\cdot A=I,$  we know that  $A$  is invertible.
Additionally, since  $A$  is invertible, we know that  $A$  is row equivalent to the identity matrix.

Step 2:
Since  $A$  is row equivalent to the identity matrix, the only solution to  $A{\vec {x}}={\vec {0}}$  is the trivial solution.
Hence,
${\text{Nul }}A={\Bigg \{}{\begin{bmatrix}0\\0\end{bmatrix}}{\Bigg \}}.$

Final Answer:
${\text{Nul }}A={\Bigg \{}{\begin{bmatrix}0\\0\end{bmatrix}}{\Bigg \}}$

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