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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.

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  • $\begingroup$ You have to use Gauß's pivot method. You see the system has rank $3$. $\endgroup$
    – Bernard
    Commented Feb 9, 2015 at 0:46
  • $\begingroup$ So, use Gaussian elimination on the kernel system? What result will this give me? $\endgroup$
    – user208614
    Commented Feb 9, 2015 at 1:14
  • $\begingroup$ You'll get the dimension of the kernel, and a basis. $\endgroup$
    – Bernard
    Commented Feb 9, 2015 at 1:58
  • $\begingroup$ How does this help me find the desired system of linear equations? $\endgroup$
    – user208614
    Commented Feb 9, 2015 at 2:03
  • $\begingroup$ I'm afraid I don't quite get you. Do you want to solve the system of linear equations? $\endgroup$
    – Bernard
    Commented Feb 9, 2015 at 2:15

1 Answer 1

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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.

Running Gauß's elimination will give you the condition(s) of compatibilty for this system: these conditions are the equations of the image.

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