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

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

or in your notation,

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

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?

  • $\begingroup$ 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? $\endgroup$ – sweets Feb 20 at 7:30
  • $\begingroup$ @sweets yep thats exactly right! and hence $a_i$ are zero, what does that tell you about the vector $v$? $\endgroup$ – Andres Mejia Feb 20 at 17:23
  • $\begingroup$ v1...vn is 0? And so T(0)=0 and so null (T)=0? Thus injective? $\endgroup$ – sweets Feb 20 at 18:33
  • $\begingroup$ 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. $\endgroup$ – Andres Mejia Feb 20 at 22:09
  • $\begingroup$ Oh my bad yah that's what I mean. $\endgroup$ – 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

  • $\begingroup$ I'm pretty sure that Axler asks for linear maps. $\endgroup$ – egreg Feb 20 at 10:33
  • $\begingroup$ If he asked about linear mappings, why did he miss this essential fact in the question? $\endgroup$ – Minz Feb 21 at 0:33
  • $\begingroup$ 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. $\endgroup$ – egreg Feb 21 at 1:03

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