Finding rank and nullity of a linear map.

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Homework Help Overview

The discussion revolves around determining the rank and nullity of a linear map defined by T(x) = (ax)a, where a is a specific vector in R3. Participants are exploring the implications of this definition in the context of linear transformations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to relate the concepts of rank and nullity to the given linear map, questioning how to find the dimension of the image and kernel without traditional matrix methods. There is also a discussion about whether the image includes the zero vector and the definitions of the image and kernel.

Discussion Status

The discussion is active, with participants questioning their understanding of the image and kernel of the linear transformation. Some have suggested looking up definitions to clarify these concepts. There is no explicit consensus yet, as different interpretations of the definitions are being explored.

Contextual Notes

Participants express uncertainty about the definitions of image and kernel, particularly regarding the inclusion of the zero vector and the relationship between these two concepts. There is a suggestion to refer to textbook definitions for clarity.

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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