Let $m \in \Bbb N$ and $v_1,\dots,v_m \in V$ be distinct vectors. Furthermore let $A\colon =\{v_1,\dots,v_m\} $ and $$T: \Bbb K^m \to V \\ T(x_1,\dots,x_m) = x_1v_1+\dots+x_mv_m$$ be a linear map.

Under what conditions on $A$ is $T$ injective and surjective?

My Solution:

$A$ must be a basis for $V$, because:

  • Let $A$ be linearly independent: Then $\lambda_1v_1+\dots\lambda_mv_m = 0$ has only the trivial solution.

    $T$ is injective iff $\ker T = \{0\}$. Let $k \in \ker T$, so $k_1v_1+\dots+k_mv_m = 0$. Now using equating the coefficients, we get $$\lambda_1v_1+\dots+\lambda_mv_m = k_1v_1+\dots+k_mv_m \Rightarrow \lambda_1 = k_1 = \dots = \lambda_m = k_m = 0$$ because of the linearly independence of $A$.

    Therefore $\ker T = \{0\}$.

  • Let $A$ be a spanning set of $V$:

    $T$ is surjective iff $im T = V$. Because $A$ is a spanning set and $T$ can be interpreted as mapping all linear combinations of $A$, we get $im T = V$.

Is this correct?


1 Answer 1


Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.

In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.

As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!


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