# Proving an injective map exists from $V$ to $W \iff$ dim $V$≤ dim$W$

I'm reading Axler's Linear Alg Done Right and working through problems. For this one, I proved the forward direction already using Fundamental Theorem of Linear Maps.

And for the backwards direction I assumed dim $$V$$≤ dim $$W$$. Let $$v_1,\ldots, v_n$$ be a basis for $$V$$ and $$w_1,\ldots, w_n$$ be a basis for $$W$$. Then $$T(a_1v_1 \ldots a_nv_n)=(a_1w_1\ldots a_mw_m)$$.

I know I first I have to prove this is indeed a linear map, which I can do. Then I prove this is injective. I know an injective map null($$T$$)=$$0$$.

I'm stuck on proving this map is injective. Any help/guidance is appreciated.

I'm not sure how your map is defined.

Take $$v_1, \dots,v_n$$ a basis for $$V$$ and $$w_1, \dots, w_n, w_{n+1}, \dots w_m$$ a basis for $$W$$.

Try your idea again by sending $$T(v_i)= w_i$$ in other words

$$T(a_1v_1+ \dots +a_n v_n)=a_1w_1+\dots +a_nw_n$$

$$T(a_1, \dots,a_n)=(a_1, \dots, a_n,0, \dots ,0)$$

My hint would be take a vector $$a_1v_1+ \dots +a_n v_n=v \in V$$

and suppose that $$T(v)=a_1w_1+\dots +a_nw_n=0$$.

Now what do we know about sums likes this when we have basis vectors?

• So your second line is bc assumption dimV<=dimW. So mine was incorrectly defined. And then T(v)=a1w1+⋯+anwn=0, basis vectors are linearly independent? – sweets Feb 20 at 7:30
• @sweets yep thats exactly right! and hence $a_i$ are zero, what does that tell you about the vector $v$? – Andres Mejia Feb 20 at 17:23
• v1...vn is 0? And so T(0)=0 and so null (T)=0? Thus injective? – sweets Feb 20 at 18:33
• well $T(0)=0$ is always true. What you want to say is that the preimage of zero is zero, i.e: $\ker T= T^{-1}(\{0\})=0$ and hence it is injective. – Andres Mejia Feb 20 at 22:09
• Oh my bad yah that's what I mean. – sweets Feb 20 at 23:36

The map $$T$$ is defined by $$T(a_1v_1+a_2v_2+\dots+a_nv_n)=a_1w_1+a_2w_2+\dots+a_nw_n$$ (you forgot the + signs). It is injective by the fundamental theorem, because the image is spanned by $$\{w_1,\dots,w_n\}$$, which is linearly independent. Thus the image has dimension $$n$$.

1) The statement is wrong because there are no restriction on map exept injectivity. So we can find injective map from $$\mathbb R^2$$ to $$\mathbb R$$ because they are have the same cardinality.

2) The statement will be true if a linear injective mapping is implied

• I'm pretty sure that Axler asks for linear maps. – egreg Feb 20 at 10:33
• If he asked about linear mappings, why did he miss this essential fact in the question? – Minz Feb 21 at 0:33
• I think that the OP was a bit sloppy in reporting the exercise. Axler is usually very precise; but it could be that he considers “linear” implicit. – egreg Feb 21 at 1:03