# Finding a Linear Transformation given its Kernel

• Feb 2nd 2011, 08:15 AM
Lord Voldemort
Finding a Linear Transformation given its Kernel
Question: Give an example of a linear transformation whose kernel is the plane x + 2y + 3z = 0 in |R3.

I'm lost here. I found the vectors [1, 1, -1] and [-5, 4, -1] span the Kernel, but I really have no idea where to go from here... would it be the product of the matrices that have kernels in the two lines in the direction of the two vectors?
• Feb 2nd 2011, 08:35 AM
tonio
Quote:

Originally Posted by Lord Voldemort
Question: Give an example of a linear transformation whose kernel is the plane x + 2y + 3x = 0 in |R3.

I'm lost here. I found the vectors [1, 1, -1] and [-5, 4, -1] span the Kernel, but I really have no idea where to go from here... would it be the product of the matrices that have kernels in the two lines in the direction of the two vectors?

You're in the right track! Now just define a lin. trans. T s.t. $\displaystyle T(1,1,-1)=T(-5,4,-1)=(0,0,0)$ . For this, complete these two vectors to a basis of $\displaystyle \mathbb{R}^3$ and then define the l.t. on the third vector to be anything but the zero vector, and extend by linearity.

Tonio

Pd. I'm assuming you meant the plane $\displaystyle x +2y +3z = 0$ ...(Thinking)
• Feb 2nd 2011, 08:36 AM
FernandoRevilla
Choose, for example, the basis of $\displaystyle \mathbb{R}^3$:

$\displaystyle B=\{(1,1,-1),(-5,4,-1),(0,0,1)\}$

Then,

$\displaystyle T \equiv\begin{Bmatrix}T(1,1,-1)=(0,0,0)\\T(-5,4,-1)=(0,0,0)\\ T(0,0,1)=(a,b,c)\end{matrix}$

such that $\displaystyle (a,b,c)\neq (0,0,0)$ determines a family of linear transformations with the given kernel.

Fernando Revilla

Edited: Sorry, I didn`t see tonio's post.
• Feb 2nd 2011, 08:49 AM
Lord Voldemort
Sorry, we haven't covered the basis of a matrix/space, so completing a vector to a basis of |R3 doesn't mean anything yet.
• Feb 2nd 2011, 09:02 AM
Lord Voldemort
Ok maybe not? I really don't know what you mean by completing something to a basis. Also as you said, "define the l.t. on the third vector to be anything but the zero vector", I don't understand how defining a new LT will do anything.
• Feb 2nd 2011, 10:49 AM
FernandoRevilla
Quote:

Originally Posted by Lord Voldemort
Ok maybe not? I really don't know what you mean by completing something to a basis. Also as you said, "define the l.t. on the third vector to be anything but the zero vector", I don't understand how defining a new LT will do anything.

If you haven't covered the concept of matrix of a linear transformation with respect to determined basis, it is rather difficult to solve the above doubt.

There are alternatives to solve your problem, but we need to know what tools we can use.

Fernando Revilla
• Feb 2nd 2011, 06:45 PM
Tinyboss
• Feb 2nd 2011, 10:06 PM
FernandoRevilla
Quote:

Originally Posted by Tinyboss
$\displaystyle f_{ (\alpha,\beta,\gamma)}(x,y,z)=(\alpha(x+2y+3z),\be ta(x+2y+3z),\gamma(x+2y+3z))$
with $\displaystyle (\alpha,\beta,\gamma)\neq (0,0,0)$ .