Finding rank and nullity of a linear map.

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The discussion focuses on determining the rank and nullity of the linear map T defined by T(x) = (ax)a, where a is the vector [2,3,1] in R3. Participants express confusion about how to find the rank and nullity without using matrix row reduction methods. It is clarified that the image of T is one-dimensional, consisting of multiples of the vector a, and that the zero vector is included in the image. The distinction between the kernel and the image is debated, with suggestions to refer to definitions in textbooks for clarity. Understanding these concepts is essential for accurately calculating the rank and nullity of the transformation.
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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