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This was the proof I saw:

Simplest Proof of the Dimensions theorem

But why is T injective on $\operatorname{span}(v_{p + 1}, \dots, v_n)$?

What I understand is that because the vectors spaned by the $\operatorname{span}(v_{p + 1}, \dots, v_n)$, do not belong in the kernel of $T$, each of them should land in an unique vector in the image of $T$. My question is why not being in the kernel gives the vector this property that it lands it in a unique image vector?

Jendrik Stelzner
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Klevis Imeri
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  • This is by construction when they extend the basis vectors. – CyclotomicField Dec 10 '22 at 13:15
  • So you mean extending the bases that span the kernal to the bases that span the domain of T ensures you that the images of the vectors what are expanding the bases are linearly independent. Can you elaborate further in this topic. – Klevis Imeri Dec 10 '22 at 13:20
  • You've got the right idea. They're extending the basis to cover the orthogonal compliment of the kernel, which is isomorphic to the image of $T$. That isomorphism is why the counting argument works. – CyclotomicField Dec 10 '22 at 13:24

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Well, the wording of the linked answer is not really precise (the $\oplus$ symbol is not used correctly).

Anyway, if $v\in\operatorname{span}(v_{p+1},\dots,v_n)$ and $v\in\ker T$, then you have \begin{align} v&=\alpha_{p+1}v_{p+1}+\dots+\alpha_nv_n \\ &=\alpha_1v_1+\dots+\alpha_pv_p \end{align} by definition of span. Hence $$ \alpha_1v_1+\dots+\alpha_pv_p-\alpha_{p+1}v_{p+1}-\dots-\alpha_nv_n=0 $$ and so all those coefficients are zero. This proves $v=0$, so indeed $T$ is injective on $\operatorname{span}(v_{p+1},\dots,v_n)$.

Therefore, as $$ T(V)=\operatorname{span}(T(v_1),\dots,T(v_p),T(v_{p+1}),\dots,T(v_n)) =\operatorname{span}(T(v_{p+1}),\dots,T(v_n)) $$ you have $\dim T(V)=n-p$.

egreg
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