Finding rank and nullity of a linear map.

sg001
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Homework Statement

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

Homework Equations


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 I am 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.
 
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sg001 said:

Homework Statement




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

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

but does the image include zero values?
 
sg001 said:
but does the image include zero values?

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

still the zero vector
 
sg001 said:
but does the image include zero values?

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

sg001 said:
still the zero vector

Does that answer your question?
 
LCKurtz said:
Does that answer your question?

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?
 
sg001 said:
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?

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

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