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Let $V$ and $W$ be finite-dimensional vector spaces and $T$ be a linear transformation $T:V\to W$.

Let $H$ be a non-zero subspace of $V$, and let $T(H)$ be the set of images of vectors in $H$. Prove that $\dim(T(H))\leq \dim(H)$.


Proof:

Since $H\neq\{0\}$, then some subset $B$ of $H$ is a basis for $H$ (by the spanning set theorem). Let $B=\{b_1,\dots,b_p\}$ be a basis for $H$, so $\dim(H)=p$.

Then for all $x\in$Span$(H)$, there exists scalars $c_1,\dots,c_p$ such that $x=c_1b_1+\cdots+c_pb_p$.

Since $T$ is linear,

$$x=c_1b_1+\cdots+c_pb_p\implies T(x)=c_1T(b_1)+\cdots+c_pT(b_p)$$ and the vectors $T(b_1),\dots,T(b_p)$ are linearly independent. The vectors $T(b_1),\dots,T(b_p)$ also span $T(H)$, since for all $T(x)\in$Span$(T(H))$, we can find an $x\in$Span$(H)$ such that $T(x)\in$Span$(T(H))$.

So $\{T(b_1),\dots,T(b_p)\}$ is a basis for $T(H)$. Thus, $\dim(T(H))\leq\dim(H)$.


Some questions:

I'm convinced with: "The vectors $T(b_1),\dots,T(b_p)$ also span $T(H)$, since for all $T(x)\in$Span$(T(H))$, we can find an $x\in$Span$(H)$ such that $T(x)\in$Span$(T(H))$" but I am not sure if this is true.

Also, it seems to be that $\dim(T(H))=\dim(H)$, is there a case where $\dim(T(H))<\dim(H)$?

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  • $\begingroup$ what happens if $ker(T)=H$? $\endgroup$ – Chiranjeev_Kumar Jul 9 '15 at 18:51
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    $\begingroup$ I'm more concerned about the assertion that $T(b_1), \ldots, T(b_p)$ is linearly independent! $\endgroup$ – Theo Bendit Jul 9 '15 at 18:52
  • $\begingroup$ @TheoBendit Can't I simply show this by linearity? Since $T$ is linear, $0=c_1b_1+\cdots+c_pb_p\implies 0=T(0)=c_1T(b_1)+\cdots+c_pT(b_p)$ only has the trivial solution as well? $\endgroup$ – user144809 Jul 9 '15 at 18:55
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    $\begingroup$ Absolutely not! As Chiranjeev suggested (perhaps a little circuitously), if $T$ is the zero transformation, i.e. maps everything to $0$, then this is definitely not the case. The error here is that you really want your implication sign to go the other direction. Add in the assumption that $T$ is injective though, and yes, they will be linearly independent. $\endgroup$ – Theo Bendit Jul 9 '15 at 19:02
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Note that: $$T(B)=\{T(b_1),\dots,T(b_p)\}$$ is spanning set of $T(H)$ but it is not necessarily linearly independent.

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What you are saying is true only if we suppose that $\ker(T) = \{0\}$, i.e. $T$ is injective.

A counter-example: $T: \mathbb R^3 \to \mathbb R^3$, $T(x,y,z) = (x + y,x,y)$. It's linear, and $\dim(\mathbb R^3) = 3 > 2 = \dim(T(\mathbb R^3))$.

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Here is a simpler proof:

We use the fact that if $H$ is a subspace of $V$, i.e $H\subset V$, then $$\dim{H}\leqslant \dim{V}$$ Since $T(H)$ is the set of images of vectors in $H$, there is $$ T(H)=T(V)\cap H\subset H $$ So we have $$ \dim{T(H)}\leqslant \dim{H} $$

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