Question

Find all linear maps whose kernel is spanned by a plane

Answer 1

Call the transformation
[math]T.[/math] Its domain is
[math]\mathbf R^4,[/math] and its kernel is dimension 2, so its image is dimension 2, so let's look for a transformation
[math]T:\mathbf R^4\to\mathbf R^2.[/math]
Note that since
[math](1,2,3,4)[/math] and
[math](0,1,1,1)[/math] generate the kernel, [math](1,0,1,2)[/math] is also in the kernel. It's a nicer vector than
[math](1,2,3,4)[/math] so let's use it instead.
Let [math]T[/math] be represented by the matrix [math]A[/math]
[math]A=\begin{bmatrix}a&b&c&d\\e&f&g&h\end{bmatrix}[/math]
We'll figure out some good values for the entries of this matrix.
Since [math]T(1,0,1,2)=(0,0)[/math] and [math]T(0,1,1,1)=(0,0)[/math] we need
[math]A= \begin{bmatrix}a&b&c&d\\e&f&g&h\end{bmatrix} \begin{bmatrix}1&0\\0&1\\1&1\\2&1\end{bmatrix}= \begin{bmatrix}0&0\\0&0\end{bmatrix}[/math]
We can do that if we make
[math]a=-c-2d, e=-g-2h,b=-c-d,[/math] and [math]f=-g-h.[/math]
But we also want the image to be all of [math]\mathbf R^2.[/math] We have to choose [math]c,d,g,[/math] and [math]h[/math] to make that so. The easiest way to do that is to let [math]c=1,d=0,g=0,[/math] and
[math]h=1.[/math] That gives us this as the matrix [math]A[/math]
[math]A=\begin{bmatrix}-1&-1&1&0