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


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.

  • $\begingroup$ 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

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