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$T$ is injective if and only if the columns of the matrix representing the $T$ are linearly independent

Let $T: V \to W$ be an injective linear transformation and let $B_V = (v_1,\ldots,v_n)$ be a basis for $V$ and $B_W = (w_1,\ldots,w_m)$ a basis for $W.$ Then $A = \mathcal{M}_T(B_V,B_W) = (x_1, x_2,\ldots,x_n) = \left([T(v_1)]_{B_W},[T(v_2)]_{B_W},\ldots, [T(v_n)]_{B_W} \right)$ where $[T(v_i)]_{B_W} = \pmatrix{a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi}}.$

Pretty stuck on this, how do I get started? Do I use the fact that $\ker T = \{0\}?$ If I could get a hint, I would appreciate it. Thank you!

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    $\begingroup$ Hint: If $A$ is an $m\times n$ matrix, how do you write $Ax$ in terms of the column vectors of $A$? $\endgroup$ – Ted Shifrin Oct 16 '15 at 2:56
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Suppose the columns are LI.

Write $v \in V$ as $v = a_1v_1 + \cdots + a_nv_n$.

Then if $T(v) = 0$, we have $0 = T(v) = T(a_1v_1 + \cdots + a_nv_n) = a_1T(v_1) +\cdots + a_nT(v_n)$.

By the LI of the $T(v_j)$, it follows that $a_1 = \cdots = a_n = 0$. Hence $v = 0v_1 +\cdots + 0v_n = 0$.

You do the other direction, suppose $\text{ker }T = \{0\}$, and:

$c_1T(v_1) +\cdots + c_nT(v_n) = 0$. What can you deduce about the $c_j$?

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