1
$\begingroup$

So I'm trying to prove the classic dimension theorem for linear transformations. Here is what I have so far:

Given a linear transformation $T:V\rightarrow W$

Let the basis of $\mathrm{Im}T$ be $\{u_1,...,u_m\}$ where $\forall u\in V:T(u)$.

Let the basis of $\ker T$ be $\{v_1,..,v_k\}$ where $\forall v:T(v)=0$.

Lets prove $B=\{u_1,...,u_m,v_1,..,v_k\}$ is a basis of $V$:

(1) $B$ is linearly independent: Assume (falsely) that $B$ isn't linearly independent. Since $\{u_1,...,u_m\}$ and $\{v_1,..,v_k\}$ are both the basis of $\ker T$ and $\mathrm{Im}T$ repectively, they must each be linearly independent.If this is the case, $\exists b\in V:b\in \{v_1,..,v_k\}\cap \{v_1,..,v_k\}$ however this contradicts $\mathrm{Im}T\cap \ker T=\emptyset$ so $B$ must be linearly independent.

(2)$\mathrm{span}(B)=V$: Assume (falsely) that $b_2\in V| b_2\notin \mathrm{span}(B)$. If this is the case, then $b_2\notin \{\forall u\in V:T(u)\}\cap\{\forall v:T(v)=0\}$. This is impossible because $b_2\in V$ so $\mathrm{span}(B)=V$

Since (1) and (2) are true, according to basis identities, $B$ is a basis of $V$. This means that $\dim V=|B|=\dim\ker T+\dim\mathrm{Im}T$ QED

$\endgroup$
1
$\begingroup$

Your proof doesn't work. For instance, consider the transformation $$ T = \pmatrix{0&1\\0&0} $$ Both the image and kernel are spanned by $(1,0)^T$, so there is no pair of bases for your initial step whose union will form a linearly independent set.

If you look up a correct proof, you'll see that the usual idea is to take the basis of the kernel and extend it to a basis for $V$, without regard for what happens with the image.

$\endgroup$
3
  • $\begingroup$ Some of the proof is outlined in this recent question $\endgroup$ – Ben Grossmann Feb 4 '18 at 1:34
  • $\begingroup$ So my proof only applies if the transformation is isomorphic? $\endgroup$ – CluelessButCurious Feb 4 '18 at 1:41
  • $\begingroup$ That's not how the word "isomorphic" is used. Really, your proof "applies" if and only if $\ker T \cap \operatorname{Im}T = \{0\}$ $\endgroup$ – Ben Grossmann Feb 4 '18 at 3:17

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.