Let $T\colon\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation such that $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}$. I am asked to find the image of $T$.
So, let $p=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}$ be in the image of $T$.
$$
T\begin{pmatrix}x\\y\\z\end{pmatrix}=p \implies \begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}=p \implies \begin{matrix}2x+2y-z=p_1\\2x+5y-2z=p_2\\4x+4y-2z=p_3\end{matrix}
$$
Clearly, $p_3=4x+4y-2z=2p_1\implies p_3=2p_1$. Now, I think if I can show that $p_1$ and $p_2$ does not depend on each other I can prove that the plane $p_3=2p_1$ is the image of $T$.
I do not know how to proceed from here.
Best Answer
You can write $$T\left(\begin{array}{c} x\\ y\\ z\end{array}\right)=\left(\begin{array}{rrr} 2 & 2 & -1\\ 2 & 5 & -2\\ 4 & 4 & -2\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\end{array}\right).$$ Take a basis of $\mathbb{R}^3$. For example, the canonical basis $B=\{\vec u_1(1,0,0),\vec u_2(0,1,0),\vec u_3(0,0,1)\}$, and compute: $$T\left(\begin{array}{c} 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c} 2\\ 2\\ 4\end{array}\right),\quad T\left(\begin{array}{c} 0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c} 2\\ 5\\ 4\end{array}\right),\quad T\left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c} -1\\ -2\\ -2\end{array}\right).$$ As every vector in $T(\mathbb{R}^3)$ can be written as a linear combination of $T(\vec u_1)=(2,2,4)$, $T(\vec u_2)=(2,5,4)$ and $T(\vec u_3)=(-1,-2,-2)$, you know that $T(\mathbb{R}^3)$ is the linear span of these vectors: $$T(\mathbb{R}^3)=\big\langle(2,2,4),(2,5,4),(-1,-2,-2)\big\rangle.$$ However, the above set is not a basis of $T(\mathbb{R}^3)$ because the vectors are dependents (you can check that with Gauss or the determinant). You can remove any one of the vectors and then you'll have a basis of the image.