When i have a basis for the image of a three dimensional transformation, how do i check if a specific vector y is part of that image?

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$$ T(X) = \left( \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right) = \left( \begin{array}{c} -x_{1}-3x_{2}+x_{3} \\ 3x_{1}+4x_{2}+2x_{3} \\ x_{2}-x_{3} \end{array} \right) for \text{ } x = \left( \begin{array}{c} x_{1}\\ x_{2}\\ x_{3} \end{array} \right) \in \mathbb{R}^{3} $$

I had to find basis of the image of this transformation, so i brought it to row echelon form $$ \begin{bmatrix} -1 & -3 & 1 \\ 0 & -5 & 5 \\ 0 & 0 & 0 \end{bmatrix} $$

and got the basis:

$$ \left( \begin{array}{cc} -1 & -3 \\ 3 & 4 \\ 0 & 1 \end{array} \right) $$

Now i have to find a vector $ Y \not \in \text{ the image} $.

Then i tried just a random vector $ r = \left( \begin{array}{cc} 0 \\ 0 \\ 1 \end{array} \right) $

But i think what i did was to check if it was part of the basis, and i did a calculation of that - but that's not really my assignment. I am not sure how to check if this vector is in the image??

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Consider the system given by the augmented matrix

$\left[\begin{array}{ccc|c} -1 & -3 & 1 & a \\ 3 & 4 & 2 & b \\ 0 & 1 & -1 & c\end{array}\right]$

We can row reduce this, just to the point where the bottom of the coefficient matrix is all $0$'s, and then it will be clear what conditions on $a$, $b$ and $c$ would make the system inconsistent. Thus:

$\begin{align} \left[\begin{array}{ccc|c} -1 & -3 & 1 & a \\ 3 & 4 & 2 & b \\ 0 & 1 & -1 & c\end{array}\right] &\sim \left[\begin{array}{ccc|c} -1 & -3 & 1 & a \\ 0 & -5 & 5 & 3a+b \\ 0 & 1 & -1 & c\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c} -1 & -3 & 1 & a \\ 0 & -5 & 5 & 3a+b \\ 0 & 5 & -5 & 5c\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c} -1 & -3 & 1 & a \\ 0 & -5 & 5 & 3a+b \\ 0 & 0 & 0 & 3a+b+5c\end{array}\right] \end{align}$

From this last matrix, it is clear that any vector $\left[\begin{matrix}a \\ b \\ c\end{matrix}\right]$ satisfying $3a+b+5c\neq 0$ will not be in the image of the transformation.


Another method, which works because we're in $\mathbb{R}^3$, would be to take the cross-product of your two basis vectors of the range. That would give you something pointing out of the plane, i.e., not in the image.


To check whether a certain vector $v$ is in the image, if you're sitting on a basis for the image, $\{b_1,b_2\}$, you can just set up the system:

$$b_1x+b_2y = v,$$

and check for consistency. With your basis, and your vector $\left[\begin{matrix}0\\0\\1\end{matrix}\right]$, that would look like:

$$\left[\begin{array}{cc|c} -1 & -3 & 0 \\ 3 & 4 & 0 \\ 0 & 1 & 1 \end{array}\right]$$

It only takes a couple of row operations to check that this is inconsistent.