Finding the basis of its kernel of $T([x, y, z]^T)=x-2y-3z$

Question

Consider the linear map$:T:\mathbb{R}^3 → \mathbb{R}$ with $$T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right)=x-2y-3z$$ Find the basis of its kernel.

My try

Since the plane is the nullspace of the matrix $$A=\begin{bmatrix} 1 & -2 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

But I am stuck here. Can anyone explain this furthur

Answer 1

The "kernel" of a linear transformation, $A$, is defined as the subspace, $\{v\}$, of its domain such that $Av= 0$. Here that is $A(x, y, z)= x- y+ 3z= 0$. We can solve for $x$: for any $y, z$, $x= y- 3z$. A vector in the kernel is of the form $(x, y, z)=(y- 3z, y, z)= (y, y, 0)+ (-3z, 0, z)= y(1, 1, 0)+ z(-3, 0, 1)$. A basis for the kernel is ${(1, 1, 0), (-3, 0, 1)}$.

Answer 2

Set $x=1$ and solve $0=x-2y-3z$ for $y$ with $z=0$ and vice versa. Once you have found $y',z'$ satisfying $0=1-2y'$ and $0=1-3z'$, then you know that $f(1,y',0)=f(1,0,z')=0$ and so you have two vectors in the kernel of $f$. Verify that they are linearly independent. Finally, as $\dim(\Bbb R^3)=\dim(Im(f))+\dim(\ker(f))$, we have $\dim(\ker(f))=2$ and so the vectors form a basis.

Answer 3

The Kernel of $T$ is a two dimensional space.

You need to find two linearly independent vectors satisfying $$ x-2y-3z=0 $$

Finding such vectors is achieved by choosing $x$ and $y$ values and solving for $z$

I have found $(3,0,1)$ and $(0,3,-2)$ to serve as a basis for the kernel of T.