[Linear Algebra] Nullspace equals Column space

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Homework Help Overview

The discussion revolves around the properties of nullspace and column space in the context of a 3 by 3 matrix in linear algebra. The original poster questions why a 3 by 3 matrix cannot have a nullspace that equals its column space.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between nullspace and column space, providing a specific matrix example. Some participants clarify definitions and dimensions of these spaces, while others explore the implications of the rank-nullity theorem.

Discussion Status

Participants are actively engaging with the concepts, with some providing clarifications about the definitions and relationships between the nullspace and column space. There is an exploration of the rank-nullity theorem and its implications for the dimensions of these spaces.

Contextual Notes

There is a focus on the dimensions of the nullspace and column space, particularly in relation to the rank of the matrix. The original poster expresses confusion regarding the equation n - r = r and its implications, indicating a need for further clarification on the terms used.

Dafe
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Homework Statement



Why does no 3 by 3 matrix have a nullspace that equals its column space?

Homework Equations



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The Attempt at a Solution



[tex] A =<br /> \begin{bmatrix}<br /> 0 & 0 & 1 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0<br /> \end{bmatrix}<br /> \][/tex]

[tex] C(A) =<br /> \begin{bmatrix}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br /> \][/tex]

Does not then N(A) = C(A)?
I think I am missing something here.

Thank you for your time.
 
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I don't know what you are missing because I don't know what you are thinking! The "column space" of a matrix is the space spanned by its columns thought of as vectors. The column space of your matrix is the one dimensional space spanned by <0, 0, 1>= [itex]\vec{k}[/itex]. The null space of a matrix, A, is the set of all vectors, [itex]\vec{v}[/itex] such that [itex]A\vec{v}= \vec{0}[/itex]. For this matrix that is the space spanned by <1, 0, 0>= [itex]\vec{i}[/itex] and <0, 1, 0>= [itex]\vec{j}[/itex]. They are not at all the same. In fact the two are orthogonal complements.

It is true for any n by n matrix, with n odd, that the null space cannot be the same as the column space because, for any n by n matrix, the sum of the dimension of the column space and the dimension of the null space must equal n. If the two dimensions are the same, their sum is an even number.
 
Hi,

I forgot the "space" in nullspace.

The book writes:

n - r = r

n = 3

3 = 2r is impossible.

The n - r = r is confusing me. Is r meant to be the number of pivot columns?

Thanks!
 
Dafe said:
Hi,

I forgot the "space" in nullspace.

The book writes:

n - r = r

n = 3

3 = 2r is impossible.

The n - r = r is confusing me. Is r meant to be the number of pivot columns?

Thanks!


You have probably learned a theorem like "rank + nullity = number of columns" or "rank + dimension of null space = number of columns." Yes, rank = number of pivot entries in rre form.

Also "rank= row rank = col rank = dim of col sp."

r = dim of col sp
n-r = dim of null sp

Set n-r equal to r; is the resulting equation possible?
 
Ah yes, n - r is the number of special solutions, number of free variables and the dimension of the nullspace.
Thank you very much!
 

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