Answer to Question #150753 in Linear Algebra for Ashweta Padhan

Question

Find a linear mapping T: R^3 tends to R^3, whose image is spanned by (1, 2, 3),
(4, 5, 6)

Answer 1

Since $\begin{pmatrix} 1 \\ 2 \\ 3\\ \end{pmatrix}$ and $\begin{pmatrix} 4\\ 5\\6\\ \end{pmatrix}$ spans $Img(T),$ then for any $y \in Img(T), y$ can be written as a linear combination of $\begin{pmatrix} 1 \\ 2 \\ 3\\ \end{pmatrix}$ and $\begin{pmatrix} 4\\ 5\\6\\ \end{pmatrix}$ .

i.e.

$T(x)=y=\begin{pmatrix} 1 \\ 2 \\ 3\\ \end{pmatrix}\alpha + \begin{pmatrix} 4\\ 5\\6\\ \end{pmatrix}\beta$

So, the two vectors and any vector will form the linear mapping $T$.

Since only the two vectors span the space, the third column is not linearly independent. Any vector that is a linear combination of the other two, will do. So any matrix of the form $T=\begin{pmatrix} 1&4&x+4y\\ 2&5&2x+5y\\ 3&6&3x+6y\\ \end{pmatrix}$

has the designed properties of the linear mapping for any $x, y\in \R.$

A trivial example is

$T=\begin{pmatrix} 1&4&0\\ 2&5&0\\ 3&6&0\\ \end{pmatrix}$