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

**x3bnm** · November 16th 2011, 11:33 AM

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

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

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

Part of Solution:
The problem is solved like this:

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

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

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

So the solutions are:

$x_1 = 0, x_2 = 0, x_3 = 0$

My question:

How do you find the basis for $ker(T)$ now because $x_1 = x_2 = x_3 = 0$ ?

**TheEmptySet** · November 16th 2011, 11:36 AM

Originally Posted by x3bnm

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

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

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

Part of Solution:
The problem is solved like this:

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

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

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

So the solutions are:

$x_1 = 0, x_2 = 0, x_3 = 0$

My question:

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

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

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

**HallsofIvy** · November 16th 2011, 11:52 AM

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.

**x3bnm** · November 16th 2011, 11:53 AM

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.

**x3bnm** · November 16th 2011, 12:12 PM

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

**x3bnm** · November 16th 2011, 12:42 PM

Originally Posted by x3bnm

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

I found the answer.

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

HallsofIvy Thanks again.

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