Find all linear mappings $\phi : \mathbb{R}^2 \to \mathbb{R}^2$, that map the straight line $y = x$ into the straight line $y = 3x$

Question

I'm a bit lost and would like your help

If I understand correctly, the straight line $y=x$ can be represented by the matrix/vector

$\begin{bmatrix} x \\ x \end{bmatrix}$

Similarly, $y=3x$ can be represented by

$\begin{bmatrix} x \\ 3x \end{bmatrix}$

Using matrix multiplication, we would need to multiply the first matrix by the matrix

$\begin{bmatrix} 1 & 3 \end{bmatrix}$ $\begin{bmatrix} x \\ x \end{bmatrix}$

which gives $\begin{bmatrix} x & 3x \end{bmatrix}$

Then we would need to take the transpose of this matrix to get $\begin{bmatrix} x \\ 3x \end{bmatrix}$

Is this correct?

Answer 1

A linear map is not defined by a vector, it is defined by a matrix. Here we want a $2×2$ matrix, and if we can map $(1,1)$ onto any nonzero point on the line we are done by properties of linear maps. $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}k\\3k\end{bmatrix}$$ $$\implies3(a+b)=c+d$$ We should also have $a+b\ne0$, otherwise the map's range is not the full line. Hence $3(a+b)=c+d\ne0$ is the condition for a linear map to have the question property.

Answer 2

Notice that $y=x$ and $y=3x$ are simply 1-dimensional linear subspaces of $\mathbb{R}^2$. To map the former onto the latter, it is equivalent to map some basis of the former onto some basis of the latter. Suppose under standard basis, $$\phi(x,y)=\begin{bmatrix}a & b \\ c & d\end{bmatrix}\cdot(x,y)$$ It suffices that we consider $(1,1)$ as the basis of $y=x$, we also know any basis of the latter is of the form $(\lambda,3\lambda)$. We hence want $\phi(1,1)=(\lambda,3\lambda)$ for $\lambda\neq 0$. What can you conclude?