# Conditions of a set of vectors, so that a specific linear map is injective/surjective

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?

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.