Let $T$ be a linear transformation from the vector space $\R^3$ to $\R^3$.
Suppose that $k=3$ is the smallest positive integer such that $T^k=\mathbf{0}$ (the zero linear transformation) and suppose that we have $\mathbf{x}\in \R^3$ such that $T^2\mathbf{x}\neq \mathbf{0}$.

Show that the vectors $\mathbf{x}, T\mathbf{x}, T^2\mathbf{x}$ form a basis for $\R^3$.

(The Ohio State University Linear Algebra Exam Problem)

We first recall that $T(\mathbf{0})=\mathbf{0}$. To see this, we compute
\[T(\mathbf{0})=T(0\cdot \mathbf{0})=0\cdot T(\mathbf{0})=\mathbf{0}.\]

To show that $\mathbf{x}, T\mathbf{x}, T^2\mathbf{x}$ form a basis, it suffices to prove that these vectors are linearly independent since the dimension of the vector space $\R^3$ is three and any three linearly independent vectors form a basis in such a vector space.
(For a different proof, see the post A linear transformation maps the zero vector to the zero vector.)

Let us consider a linear combination
\[a_1\mathbf{x}+a_2(T\mathbf{x})+a_3(T^2\mathbf{x})=\mathbf{0}, \tag{*}\]
where $a_1, a_2, a_3$ are real scalars.

If the only solution is $a_1=a_2=a_3=0$, then the vectors are linearly independent.

Applying the linear transformation $T$ on both sides, we obtain
\begin{align*}
T\left (a_1\mathbf{x}+a_2(T\mathbf{x})+a_3(T^2\mathbf{x})\right)=T(\mathbf{0})=\mathbf{0}.
\end{align*}

Using the linearity of $T$, we have
\[a_1(T\mathbf{x})+a_2(T^2\mathbf{x})+a_3(T^3\mathbf{x})=\mathbf{0}\]
and since $T^3=\mathbf{0}$, we have
\[a_1(T\mathbf{x})+a_2(T^2\mathbf{x})=\mathbf{0} \tag{**}\]

r

We apply the linear transformation $T$ again, and by using the linearity as above we obtain
\[a_1(T^2\mathbf{x})+a_2(T^3\mathbf{x})=T(\mathbf{0})=\mathbf{0}\]
and since $T^3=\mathbf{0}$, we have
\[a_1(T^2\mathbf{x})=\mathbf{0}.\]
Since $T^2\mathbf{x}\neq \mathbf{0}$, we must have $a_1=0$.

Then from (**), we have $a_2(T^2\mathbf{x})=\mathbf{0}$, and hence $a_2=0$ since $T^2\mathbf{x}\neq \mathbf{0}$.
Substituting $a_1=a_2=0$ in (*), we have $a_3(T^2\mathbf{x})=\mathbf{0}$, and thus $a_3=0$ since $T^2\mathbf{x}\neq \mathbf{0}$.

Therefore the solution to (*) is $a_1=a_2=a_3=0$, and hence the vectors
\[\mathbf{x}, T\mathbf{x}, T^2\mathbf{x}\]
are linearly independent.

We conclude, hence, that these vectors form a basis for $\R^3$.

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