Find the linear transformation when given the kernel.

Question

Write down a linear transformation $T:\mathbb R^3 \to\mathbb R$ whose kernel is $\{(x,y,z)\in\mathbb R^3:x + 3y - z = 0\}$.

Any tips, it is the only question on the paper that I cannot tackle. I just don't know what to do for the first part, I know how to calculate nullity.

Answer 1

Recall the definition of the kernel of a linear transformation: it’s the set of vectors that are mapped to zero. In this case, you have a description of that set as the vectors that satisfy the equation $x+3y-z=0$, but the l.h.s. of this equation describes a linear transformation, so...

Answer 2

The map $f:\mathbb R^3\to\mathbb R$ defined by $$f(x,y,z)=x+3y-z$$ seems to do the trick.

Answer 3

Define T((x,y,z))=x+3y-z. It is easy to see that this is a linear transformation. So Ker(T)={x+3y-z=0 | x,y,z are real numbers}