$T$ 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!

  • 3
    $\begingroup$ 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

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 $c_j$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.