Linear Transformations,Find basis of kernel and range

[nehap.2491]

suppose that vectors in R3 are denoted by 1*3 matrices, and define T:R4 to R3 by T9x,y,z,t)=(x-y+z+t,2x-2y+3z+4t,3x-3y+4z+5t).Find basis of kernel and range.

 

[micromass]

First you'll need to calculate what the kernel and image is of T.

For the kernel, you'll need to solve the following system

$$\left\{ \begin{array}{c}<br /> x-y+z+t = 0\\<br /> 2x-2y+3z+4t = 0\\<br /> 3x-3y+4z+5t = 0<br /> \end{array}\right.$$

For the image, you'll have to solve

$$\left\{ \begin{array}{c}<br /> x-y+z+t = \alpha\\<br /> 2x-2y+3z+4t = \beta\\<br /> 3x-3y+4z+5t = \gamma<br /> \end{array}\right.$$

where alpha, beta, gamma are parameters. You'll then have to see for which alpha, beta, gamma this system has a solution.

 

[micromass]

Ow, for the image you won't need to do all that stuff, I'm sorry.

You'll first have to find a basis of R4, call this {e1,e2,e3,e4}. Then {T(e1),T(e2),T(e3),T(e4)} is a set which spans the image. If this set is linear independent, then it's a basis. If not, then remove some vectors until it is linear independent...

 

[Outlined]

If you are open to it, the equation below (in case f : V -> W) might even help:

dim(ker(f)) + dim(Im(f)) = dim(V)

 

[nehap.2491]

Ow, for the image you won't need to do all that stuff, I'm sorry.

You'll first have to find a basis of R4, call this {e1,e2,e3,e4}. Then {T(e1),T(e2),T(e3),T(e4)} is a set which spans the image. If this set is linear independent, then it's a basis. If not, then remove some vectors until it is linear independent...

Thank you!

 

[nehap.2491]

If you are open to it, the equation below (in case f : V -> W) might even help:

dim(ker(f)) + dim(Im(f)) = dim(V)

Thank you!