# Finding rank and nullity of a linear map.

1. Oct 21, 2012

### sg001

1. The problem statement, all variables and given/known data

let a be the vector [2,3,1] in R3 and let T:R3-->R3 be the map given by T(x) =(ax)a

State with reasons, the rank and nullity of T

2. Relevant equations

3. The attempt at a solution

Im having trouble understanding this... I know how to do this with a matrix ie row reduce and no. of leading cols = rank ,, and then no. of non leading cols = nullity.

But im stuck on how to go about it with this eqn.?

And that the rank = dim(image)

but how would I finad that...

or alternatively the dim(kernal)?

Thanks.

2. Oct 21, 2012

### LCKurtz

Isn't the image 1 dimensional, being multiples of a?

3. Oct 21, 2012

### sg001

but does the image include zero values?

4. Oct 21, 2012

### LCKurtz

Do you mean the zero vector? If $x = \theta$, the zero vector, what is $T(\theta)$ by your formula?

5. Oct 22, 2012

### sg001

still the zero vector

6. Oct 22, 2012

### LCKurtz

7. Oct 23, 2012

### sg001

yes, so the image cointains the zero value... but I thought the image was the set of all function values except 0, I thought that was the kernal

or is it that the kernal is a proper subset of the image?

8. Oct 23, 2012

### LCKurtz

At this point, I suggest you look up the definition of the kernel of a linear transformation in your text.