For given vector space $V$ and Linear Transf. $T$ satisfying certain property. Show that $\{ x,y \}$ basis for $V$.

Question

Suppose $V$ is a vector space over $\mathbb R$ with dimension $2$ and we have linear transformation $T:V \to V$ satisfying for some non-zero $x$ and $x,y \in V,$ $Tx =y$ and $Ty = -x$

Question is the following: For given linear transformation $T$, show that $\{ x,y \}$ is basis for $V$ and also find matrix representation of $T$.

My attemps: To show basis we need to check two things: linearly independency and they must span the set $V$.

For linearly independency pick $c_1,c_2 \in \mathbb R$ and look at $c_1 x+c_2y=0$. To being linearly independent $c_1$ and $c_2$ must be $0$ as $T(c_1 x+c_2y)=T(0) \implies c_1T(x)+c_2T(y)=c_1 (y) + c_2 (-x)=0$. After that I say scalars must be $0$ but I am not satisfied.

I have a trouble about showing they span $V$ because there is no specific rule for $T$.

For matrix representation: $[T]_{B}= [ [T(x)]_B \mid [T(y)]_B] = [[y]_B \mid [-x]_B] = \quad \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Answer 1

We know that $V$ has dimension $2$, so if we can show the linear independence of $\{x,y\}$, we automatically prove that it is a basis.

To show linear independence, we note that: $$ \begin{align*} c_1x + c_2y &= 0 \tag{1} \\ -c_2x + c_1y &= 0 \tag{2} \end{align*} $$ From $(2)$ we have $c_1y=c_2x$. Multiply $(1)$ by $c_1$ and substitute: $$ (c_1^2+c_2^2)x=0 $$ Since $x\neq 0$, we have $c_1^2+c_2^2=0$, which implies $c_1=0$ and $c_2=0$ since they are real numbers. This then shows the linear independence of $\{x,y\}$.

The matrix you obtained is correct.

Answer 2

Assume $c_1x+c_2y=0$. Therefore: $$T(c_1x+c_2y)=c_1T(x)+c_2T(y)=c_1y-c_2x=0$$ So we have a system of equations: $$\cases{c_1x+c_2y=0\\c_1y-c_2x=0}$$ The only solution is $c_1=c_2=0$ - this is because the vectors $(c_1,c_2), (-c_2,c_1)$ are linearly independent over $\mathbb{R}$. Indeed if not, we have some $\lambda\neq 0$ such that: $$\cases{\lambda c_1=-c_2\\\lambda c_2=c_1}$$ Meaning: $$\lambda^2 c_2=-c_2$$ which is impossible for $c_2\neq 0$. Similarly $c_1=0$. So you have two linearly indepedent vectors in $V$ so they must be a basis.