[Math] Show that the range of a linear transformation is a subspace

linear algebralinear-transformationsmatricesvector-spacesvectors

Let $T : \mathbb V \to \mathbb W$ be a linear transformation from a vector space $\mathbb V$ into a vector space $\mathbb W$. Prove that the range of $T$ is a subspace of $\mathbb W$.

OK here is my attempt…

If we let $x$ and $y$ be vectors in $\mathbb V$, then the transformation of these vectors will look like this… $T(x)$ and $T(y)$.

If we let $\mathbb V$ be a vector space in $\mathbb R^3$ and $\mathbb W$ be a vector space in $\mathbb R^2$, then

$$
T \begin{pmatrix} x_1\\ x_2 \\ x_3
\end{pmatrix} = T\begin{pmatrix} x_1 + 2x_2 \\ 3x_3 + 4 \end{pmatrix}.
$$

Now if we tried to row reduce the matrix $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

we would get $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} .$

SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.

This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $\mathbb R^2$, and not in the subspace of $\mathbb W$.

Any help will be appreciated.

Best Answer

Given any two vector spaces $\mathbb{V, W}$ over $F$, and $T:\mathbb{V}\to\mathbb{W}$ is a linear transformation.

  1. $0_\mathbb{V}\in \mathbb{V}$. Since $T$ is linear, $T(0_\mathbb{V})=0_\mathbb{W},$ which implies $0_\mathbb{W}\in R(T).$
  2. Given any $x,y\in R(T),$ which implies there exist $v_x,v_y\in \mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $c\in F, T(cv_x+v_y)=^\dagger cT(v_x)+T(v_y)=cx+y,$ so $cx+y\in R(T).$

The 1. and 2. complete the proof.

$\dagger:$ linearity of $T:\mathbb{V}\to\mathbb{W}$.

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