# Find a homogeneous system of linear equations whose space of solutions is $\operatorname{Im}T$

We have that $T:\mathbb R^3 \to \mathbb R^4$ and $T(x_1,x_2,x_3)=(3x_1+4x_2+2x_3,x_1+2x_2,2x_1+x_2+3x_3,-x_1+5x_2-7x_3)$.

I know that if $A$ is any $m \times n$ matrix over a field $\mathbb F$ viewed as a linear map $A:\mathbb F^n \to \mathbb F^m$, then $\operatorname{ker}A=\operatorname{nullsp}A$ and $\operatorname{Im}A=\operatorname{colsp}A$. Then, in this problem, $\ker A$ is described as:

$3x_1+4x_2+3x_3=0$

$x_1+2x_2=0$

$2x_1+x_2+2x_3=0$

$-x_1+5x_2-7x_3=0$

I also know that the dimension of the solution space of a homogeneous system $AX=0$ of linear equations is $s=n-r$, where $n$ is the number of unknowns and $r$ is the rank of the matrix. How would I be able to solve this problem? Thank you.

• You have to use Gauß's pivot method. You see the system has rank $3$. Feb 9, 2015 at 0:46
• So, use Gaussian elimination on the kernel system? What result will this give me? Feb 9, 2015 at 1:14
• You'll get the dimension of the kernel, and a basis. Feb 9, 2015 at 1:58
• How does this help me find the desired system of linear equations? Feb 9, 2015 at 2:03
• I'm afraid I don't quite get you. Do you want to solve the system of linear equations? Feb 9, 2015 at 2:15

Consider the images of the canonical basis of $\mathbf R^3$, written as a $3\times 4$ matrix: $$A=\begin{bmatrix}3&4&3\\1&2&0\\2&1&2\\-1&5&-7 \end{bmatrix}$$ You must under which condition on $x,y,z,t$ the system of linear equations: $$A\begin{bmatrix}\lambda\\\mu\\\nu\end{bmatrix}=\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$$ has a solution. Note there are $3$ unknowns and $4$ equations.