Geometric representation of range and kernel of a linear transformation Let $T$ be a linear operator on $\Bbb{R}^3$ defined by $TX$ = $AX$ where $X$ is a $3$$\times$$1$ column vector and $A$ is
$$\begin{bmatrix}1 & 3 & 4\\3 & 4 & 7\\-2 & 2 & 0\end{bmatrix}$$
I need to show that the range and kernel of $T$ are a plane and a line passing through the origin respectively.
I have found that the dimension of range and kernel are $2$ and $1$ respectively. So all elements in the kernel are of the form $ax$ where $a$ is a real number and $x$ is a fixed element in $\Bbb{R}^3$.From here how can I conclude the above.
 A: First, X is not a "3 x 3 column vector"!  "3 x 3" would mean a matrix with three rows and three columns.  X is a 3 dimensional column vector.
The most direct way (not necessarily the simplest) to solve this is to actually find the kernel and range.
A vector $v= \begin{bmatrix}a \\ b \\ c \end{bmatrix}$ is in the range of A if and only if there exist a vector $u= \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ such that Au= v.
That is $\begin{bmatrix} 1 & 3 & 4 \\ 3 & 4 & 7 \\ -2 & 2 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 3y+ 4z \\ 3x+ 4y+ 7z \\ -2x+ 2y\end{bmatrix}= \begin{bmatrix}a \\ b \\ c \end{bmatrix}$.
Given any $a, b, c$ there must exist $x, y, z$ such that  $x+ 3y+ 4z= a, 3x+ 4y+ 7z= b$, and $=-2x+ 2y= c$.  We can determine for which a, b, c that is true by finding $x, y, z$ in terms of $a, b, c$.  From the last equation, $-2x+ 2y= c, y= x+ c/2$. Replacing y with that in the first two equations, $x+ 3(x+ c/2)+ 4z= 4x+ c/2+ 4z= a$ or $4x+ 4z= a- c/2$, and $3x+ 4(x+ c/2)+ 7z= 7x+ 2c+ 7z= b$ or $7x+ 7z= b- 2c$.  Dividing the first of those two equations by $ =4, x+ z=  a/4- c/8$.  Dividing the second of those two equations by  $7, x+ z= b/7- 2c/7$.  Since they are both equal to $x+ z$, we must have $a/4- c/8= b/7- 2c/7$.  So $a/4= b/7- 2c/7+ c/8= b/7- 9c/56,  a= 4b/7- 9c/14= (8b- 9c)/14$.  So any vector in the range must be of the form (assuming my arithmetic is correct) $\begin{bmatrix}\frac{8b- 9c}{14} \\ b \\ c\end{bmatrix}$.  That is a two dimensional subspace (i.e. a plane). Taking $b= 7, c= 0$ we have $\begin{bmatrix} 4 \\ 7 \\ 0\end{bmatrix}$.  Taking b= 0, c= 14 we have $\begin{bmatrix}-9 \\ 0 \\ 14\end{bmatrix}$.  Those two vectors can be taken as basis vectors for the range.
The kernel is actually easier to find.  It is the set of all vectors, u, such that Au= 0.  If we take $u= \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ then we must have
$Av= \begin{bmatrix} 1 & 3 & 4 \\ 3 & 4 & 7 \\ -2 & 2 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 3y+ 4z \\ 3x+ 4y+ 7z \\ -2x+ 2y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$.
So now we have the three equations $x+ 3y+ 4z= 0, 3x+ 4y+ 7z= 0$, and $-2x+ 2y= 0$.  The last equation immediately gives y= x.  Replacing y with x in the first two equations we get $4x+ 4z= 0 and 7x+ 7z= 0$.  Both of those give z= -x.  So every vector in the kernel is of the form $\begin{bmatrix} x \\ x \\ -x\end{bmatrix}$ for x any number.  Taking x= 1 the single vector $\begin{bmatrix}1 \\ 1 \\ -1 \end{bmatrix}$ spans that one dimensional subspace.
