Finding linear mappings $f, g$ with given properties.

Question

I'm having trouble finding two linear maps $f : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ and $g : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that the image of $f$ is equal to the kernel which equals $g$ is equal an axis.

Thanks.

Answer 1

Note that the $x$ axis forms a subspace of the real plane (whose equation is $y = 0$ as a line), and along with the $y$ axis provides a coordinate system for $\mathbb R^2$, in that every point can be represented uniquely as a linear combination of the unit vector pointing in the $x$ and the $y$ axis directions.

The idea, therefore, is to take a point in $\mathbb R^2$ and consider only its $x$ value, setting the $y$ value to zero : this automatically ensures the image to be a subset of the $x$ axis. More rigorously, define $F(x,y) = (x,0)$. This is a linear transformation(check this) whose image is the $x$ axis, as I explained earlier. You may try to write the matrix(in standard basis) for this if you like.

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Now, what about $g$? The point is again, that if the kernel is to be exactly the $x$ axis, then every point of non-zero $y$ coordinate goes to some non-zero number pair.

The answer should now be in front of you : $g(x,y) = (0,y)$. Now, $g$ will send a point to zero only if the $y$ coordinate is zero i.e. the point is on the $x$ axis. You may again see the matrix of $g$ in standard basis if you like.

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Attempting to find matrices satisfying constraints before working out the geometric aspects can sometimes become a blind search for many variables satisfying some constraints, so sometimes it is better to see if you can get a good picture of a desirable linear transformation by seeing what it does to a point $(x_1,...,x_n)$, rather than as a matrix.

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The usual "computation of matrices given linear transformation" with standard basis works as follows :order the basis : in our case, it is $\{(1,0),(0,1)\}$. Now, $F(1,0) = (1,0)$ and $F(0,1) = (0,0)$. Therefore, writing these image vectors as columns, $F = \begin{bmatrix}1 \quad 0 \\ 0 \quad 0\end{bmatrix}$.

Similarly, $G(1,0) = (0,0)$ and $G(0,1) = (0,1)$, so this matrix $G = \begin{bmatrix} 0 \quad 0 \\ 0 \quad 1\end{bmatrix}$. Check that $FG = 0$ (Since $\mbox{im} F = \ker G$, obviously $GF = 0$). However, below I give an example where $FG \neq 0$.