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Let $T:R^3 \to R^4$, 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 set up a system of equations with these four equations and solved it for $x_1=-2, x_2=1, x_3=1$. I don't know what to do with this though. Honestly, I don't even know what I even have to do. If anyone has any hints or anything thanks in advance!
Generally, given a matrix $A$ the set $$ Ker(A):=\{v\mid Av=0\} $$
is a subvector space and it can be represented in the forms of a system of linear equations naturally.
For example if $$ A=\begin{pmatrix}1 & 2\\ 3 & 4 \end{pmatrix} $$
then $Ker(A)$ is given by $$ Av=0 $$
$$ \begin{pmatrix}1 & 2\\ 3 & 4 \end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2} \end{pmatrix}=0 $$ $$ \begin{cases} x_{1}+2x_{2}=0\\ 3x_{1}+4x=0 \end{cases} $$
This gives the idea of finding $A$ s.t $$ ImT=Ker(A) $$
Now, say we find a basis for $Im(T)$, we can complete this basis for $Im(T)$ to a basis of $\mathbb{R}^{4}$.
So, say for example $\{v_{1},v_{2}\}$ is a basis for $Im(T)$ and $\{v_{1},v_{2},v_{3},v_{4}\}$ is a basis for $\mathbb{R}^{4}$.
If we define a mapping $L:\mathbb{R}^{4}\to\mathbb{R}^{4}$ s.t $$ L(v_{3}),L(v_{4}) $$
are linearly independent, and $$ L(v_{1})=L(v_{2})=0 $$
then $$ Ker(L)=sp\{v_{1},v_{2}\}=Im(T) $$
let $A$ be the representative matrix of $L$, so that $$ Ker(A)=Ker(L)=Im(T) $$
This was the idea - I leave it up to you to plug in the details of your case
First find the matrix $A$ of the linear transformation.
Now you want a system of equations that says for which $\overline b$ is the system $A\overline x = \overline b$ consistent (cause this happens when $\overline b$ is in the image). Well, let $$\overline b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$$ and then follow the algorithm for solving $A\overline x = \overline b$. By that I mean write down the augmented matrix $[A \ | \ \overline b]$ and start row reducing.
Eventually you'll get to a point where the remaining unreduced rows only have stuff in the very last column. You won't know whether you can divide to make a $1$ in that column because the "stuff" will have variables $b_i$ in it, so you won't know if it's zero or nonzero.
When you get to that point stop. Remember that $A \overline x = \overline b$ is consistent if and only if you don't get a leading $1$ in the last column when reducing the augmented matrix. So the system of equations you're looking for is given by taking those rows that only have stuff in the last column, and setting that stuff equal to zero.
For example, if my augmented matrix was $$\begin{bmatrix} 1 & 2 & b_1 \\ 2 & 4 & b_2 \\ 1 & 2 & b_3\end{bmatrix}$$ then I would do the row operations $R_2 - 2R_1 \to R_2$ and $R_3 - R_1 \to R_3$ to get $$\begin{bmatrix} 1 & 2 & b_1 \\ 0 & 0 & b_2 - 2b_1 \\ 0 & 0 & b_3 - b_1 \end{bmatrix}$$ and then I would stop and say that my system of equations was $b_2 - 2b_1 = 0$ and $b_3 - b_1 = 0$.
Once you have the system of equations that define the image, write them in matrix form $A'\overline b = \overline 0$. The nullspace of the matrix $A'$ is the solution space to that system of equations, which in turn is the image of $T$.
