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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.

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    $\begingroup$ Do you know about kernels/null spaces? Then translate the point back to the origin... $\endgroup$
    – Randall
    Commented Mar 1, 2021 at 2:47
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    $\begingroup$ If it sends a line to a point, it is not bijective, is it? $\endgroup$
    – Kenta S
    Commented Mar 1, 2021 at 2:56

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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!

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

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