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 where is a linear transformation.

Suppose

**Part of Solution:**

The problem is solved like this:

The reduced row echelon form of augmented matrix is:

So the solutions are:

**My question:**

How do you find the basis for now because ?

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

where

is a linear transformation.

Suppose

**Part of Solution:**
The problem is solved like this:

The reduced row echelon form of augmented matrix

is:

So the solutions are:

**My question:**
How do you find the basis for

now?

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

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

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.

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.

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 ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.

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

Quote:

Originally Posted by

**x3bnm** Can

ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.

I found the answer.

Physics Forums - View Single Post - a question about (0,0,0) vector

HallsofIvy Thanks again.