# Proof: A bijective linear map cannot send a line to a point

Let $$T\colon\mathbb R^3 \rightarrow\mathbb R^3$$ be a bijective linear map. Prove that $$T$$ can not send a line to a point.

I started off thinking that $$T(t\vec{m}+\vec{b}) = \vec{x}$$ should have infinitely many solutions, but then I don't know how to go on with the equation.

• Do you know about kernels/null spaces? Then translate the point back to the origin... Commented Mar 1, 2021 at 2:47
• If it sends a line to a point, it is not bijective, is it? Commented Mar 1, 2021 at 2:56

## 2 Answers

Assume that $$T(x_0 + tx_1) = x_2$$ for all $$t\in\mathbb R$$ with some $$x_1\neq 0$$ (otherwise this wouldn't be a line). By linearity, $$x_2 = Tx_0 + tTx_1$$, that is, $$t\cdot Tx_1 = x_2 - Tx_0$$ for all $$t\in\mathbb R$$. Setting $$t=0$$ yields $$x_2 - Tx_0 = 0$$ and then $$t=1$$ shows that $$Tx_1 = 0$$ and so $$x_1 = T^{-1}(0) = 0$$. Contradiction!

If it sends a line to a point, it is not one-one. Thus it is not bijective. QED