Does there exist a $\mathbb{R}$-linear function $f : \mathbb{R}^3 → \mathbb{R}^2$ with $f(1; 0; 3) = (1;1)$ and $f(−2; 0; −6) = (2; 1)$?

Question

Hey I have this problem where I want to see if my solutions are right.

Does there exist a $\mathbb{R}$-linear function $f : \mathbb{R}^3 → \mathbb{R}^2$ with $f(1; 0; 3) = (1;1)$ and $f(−2; 0; −6) = (2; 1)$?

To check whether there exists an $\mathbb{R}$-linear function $f : \mathbb{R}^3 → \mathbb{R}^2$ with $f(1; 0; 3) = (1;1)$ and $f(-2; 0; -6) = (2; 1)$, we check whether the two given vectors $(1, 1)$ and $(2, 1)$ are in the image of the linear transformation. If both vectors are in the image, then such a linear transformation exists; otherwise, it does not.

To see whether $(1, 1)$ and $(2, 1)$ are in the image of the linear transformation, we can form the matrix $M$ whose columns are the two input vectors, i.e.,

$$M = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$$

We can then check whether $M$ has rank $2$, since the rank of $M$ is the same as the dimension of the image of the linear transformation. If $M$ has rank $2$, then the columns of $M$ span $\mathbb{R}^2$, and hence both $(1, 1)$ and $(2, 1)$ are in the image of the linear transformation. If $M$ has rank less than $2$, then the columns of $M$ do not span $\mathbb{R}^2$, and hence at least one of $(1, 1)$ and $(2, 1)$ is not in the image of the linear transformation.

Calculating the determinant of $M$, we get $\det(M) = -1$, which is nonzero, so $M$ has rank $2$. Therefore, there exists an $\mathbb{R}$-linear function $f : \mathbb{R}^3 → \mathbb{R}^2$ with $f(1; 0; 3) = (1;1)$ and $f(-2; 0; -6) = (2; 1)$.

Am I right?

Answer 1

What you showed is that the image vectors of $f$ span $\mathbb R^2$.

But $f$ is not linear.

For $f$ to be linear, we would need $f(u+v)=f(u)+f(v)$ and $f(cv)=cf(v),$

for all $u,v\in \mathbb R^3$ and $c\in\mathbb R$.

But $f(-2(1;0;3))\ne-2f(1;0;3)$.

Answer 2

There can be no linear map with those properties.

The points $(1,0,3)$ and $(-2,0,-6)$ lie on the line described by $$z=3x\text{ and }y = 0$$

These points would have to get mapped to $(1,1)$ and $(2,1)$, respectively, which lie on the line given by $y=1$. Since a line is completely defined by two points, this means that the line in $\mathbb R^3$ is mapped to the line $y=1$ in $\mathbb R^2$. But the line in $\mathbb R^3$ contains the origin which would have to get mapped to a point in $\mathbb R^2$ with a $y$ coordinate equal to $1$.

Therefore, the map cannot be linear.