Basis of kernel(T) where T is a linear transformation

• Nov 16th 2011, 11:33 AM
x3bnm
Basis of kernel(T) where T is a linear transformation
Problem:
I can't find answer to a problem. The problem is like this:

Find the basis for $\displaystyle \text{kernel of}(T)$ where $\displaystyle T$ is a linear transformation.

Suppose $\displaystyle T:R^3 \to R^3,\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x)$

Part of Solution:
The problem is solved like this:

$\displaystyle A = \left[{\begin{array}{ccc} 1 & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1 \end{array}}\right]$

The reduced row echelon form of augmented matrix $\displaystyle [A\vdots \mathbf{0}]$ is:

$\displaystyle \left[{\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}}\right]$

So the solutions are:

$\displaystyle x_1 = 0, x_2 = 0, x_3 = 0$

My question:

How do you find the basis for $\displaystyle ker(T)$ now because $\displaystyle x_1 = x_2 = x_3 = 0$?
• Nov 16th 2011, 11:36 AM
TheEmptySet
Re: Basis of kernel(T) where T is a linear transformation
Quote:

Originally Posted by x3bnm
Problem:
I can't find answer to a problem. The problem is like this:

Find the basis for $\displaystyle \text{kernel of}(T)$ where $\displaystyle T$ is a linear transformation.

Suppose $\displaystyle T:R^3 \to R^3,\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x)$

Part of Solution:
The problem is solved like this:

$\displaystyle A = \left[{\begin{array}{ccc} 1 & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1 \end{array}}\right]$

The reduced row echelon form of augmented matrix $\displaystyle [A\vdots \mathbf{0}]$ is:

$\displaystyle \left[{\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}}\right]$

So the solutions are:

$\displaystyle x_1 = 0, x_2 = 0, x_3 = 0$

My question:

How do you find the basis for $\displaystyle ker(T)$ now?

You need to make your vectors the columns and not the rows!!

Your proceedure is correct but you have used the wrong matrix.
• Nov 16th 2011, 11:52 AM
HallsofIvy
Re: Basis of kernel(T) where T is a linear transformation
But you are still going to get the same thing: T(x, y, z)= (x+ 2y, y+ 2z, z+ 2x)= (0, 0, 0). That is, we must have x+ 2y= 0, y+ 2z= 0, and z+ 2x= 0. From the first equation, x= -2y. From the second, y= -2z so x= -2(-2z)= 4z. But from the third equation z= -2x so that x= 4(-2x)= -8x which is only satified for x= 0. Then z= -2x= 0 and y= -2z= 0. The entire kernel is the single 0 vector. That does not have a basis.
• Nov 16th 2011, 11:53 AM
x3bnm
Re: Basis of kernel(T) where T is a linear transformation
Quote:

Originally Posted by TheEmptySet
You need to make your vectors the columns and not the rows!!

Your proceedure is correct but you have used the wrong matrix.

Sorry I didn't understand what you said. Can you kindly explain a little? Sorry about that.
• Nov 16th 2011, 12:12 PM
x3bnm
Re: Basis of kernel(T) where T is a linear transformation
Quote:

Originally Posted by HallsofIvy
But you are still going to get the same thing: T(x, y, z)= (x+ 2y, y+ 2z, z+ 2x)= (0, 0, 0). That is, we must have x+ 2y= 0, y+ 2z= 0, and z+ 2x= 0. From the first equation, x= -2y. From the second, y= -2z so x= -2(-2z)= 4z. But from the third equation z= -2x so that x= 4(-2x)= -8x which is only satified for x= 0. Then z= -2x= 0 and y= -2z= 0. The entire kernel is the single 0 vector. That does not have a basis.

Thanks HallsofIvy. That makes sense. One quick question.

Can $\displaystyle \left( \left[{\begin{array}{ccc} 0\\ 0\\ 0 \end{array}}\right]\right)$ ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.
• Nov 16th 2011, 12:42 PM
x3bnm
Re: Basis of kernel(T) where T is a linear transformation
Quote:

Originally Posted by x3bnm
Can $\displaystyle \left( \left[{\begin{array}{ccc} 0\\ 0\\ 0 \end{array}}\right]\right)$ ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.