Intro to Linear Algebra - Nullspace of Rank 1 Matrix

In summary: Is the problem statement correct? In summary, the nullspace is a plane in R^n, but the nullspace isn't an n-1 dimensional space within R^n.
  • #1
fractalizard
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Homework Statement
Given A is a mxn matrix with rank 1, the nullspace is a _____ in R^n.
Relevant Equations
N(A) = Linear combination of "special solutions" to A
The published solutions indicate that the nullspace is a plane in R^n. Why isn't the nullspace an n-1 dimensional space within R^n? For example, if I understand things correctly, the 1x2 matrix [1 2] would have a nullspace represented by any linear combination of the vector (-2,1), which would be a line.
 
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  • #2
An [itex](n-1)[/itex] dimensional subspace in [itex]\mathbb{R}^n[/itex] is analagous to a plane in [itex]\mathbb{R}^3[/itex]: they are given by an equation of the form [itex]\mathbf{x} \cdot \mathbf{n} = 0[/itex].
 
  • #3
fractalizard said:
Homework Statement: Given A is a mxn matrix with rank 1, the nullspace is a _____ in R^n.
Relevant Equations: N(A) = Linear combination of "special solutions" to A

The published solutions indicate that the nullspace is a plane in R^n.
That sounds wrong. Are you sure that you copied the problem statement exactly? The problem statement or the book answer might have a typo. Or the book might just need some more proofreading. It's very hard to eliminate all errors from a book.
fractalizard said:
Why isn't the nullspace an n-1 dimensional space within R^n?
It is. You can verify the properties of a subspace.
 
  • #4
fractalizard said:
Homework Statement: Given A is a mxn matrix with rank 1, the nullspace is a _____ in R^n.
Relevant Equations: N(A) = Linear combination of "special solutions" to A

The published solutions indicate that the nullspace is a plane in R^n. Why isn't the nullspace an n-1 dimensional space within R^n? For example, if I understand things correctly, the 1x2 matrix [1 2] would have a nullspace represented by any linear combination of the vector (-2,1), which would be a line.
Perhaps the authors of the solution meant that the nullspace was a hyperplane; an object of dimension one less than that of ##\mathbb R^n##.

For a 2x2 matrix of rank 1, the domain is ##\mathbb R^2## and the nullspace is of dimension 1, so the nullspace is a line in ##\mathbb R^2##.

For a 3x4 matrix of rank 1, the domain is ##\mathbb R^4## and the nullspace is of dimension 3, so the nullspace is a hyperplane in ##\mathbb R^4##.
 
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  • #5
Wow, this place is great - thanks for the quick replies. I suspected that this might be an error in the solution. FYI, my posted problem statement was only the piece of the original problem that I was having trouble with, here is the full problem statement:

40 only.png
And from the solution manual:

40 answer.png


The highlighted line is the piece I am having trouble understanding, the rest of the solution makes sense to me.
 
  • #6
fractalizard said:
The highlighted line is the piece I am having trouble understanding, the rest of the solution makes sense to me.

As I stated above, every [itex](n - 1)[/itex] dimensional subspace is [tex]\{ \mathbf{x} \in \mathbb{R}^n: \mathbf{x} \cdot \mathbf{a} = x_1a_1 + \cdots + x_na_n = 0\}[/tex] for some [itex]\mathbf{a} \neq 0[/itex]. This is analagous to a plane in [itex]\mathbb{R}^3[/itex]. In higher dimensions one might more properly call it a hyperplane.
 
  • #7
fractalizard said:
And from the solution manual:

View attachment 326075
The highlighted line is the piece I am having trouble understanding, the rest of the solution makes sense to me.
Hi @fractalizard. Welcome to PF.

In addition to the other excellent replies I’d like to add a (non-mathematician's) example which might help.

Say ##A## is a ##4 \times 4## matrix. If the rank = 1 then any row is a scalar multiple of any other row, For example:

##A = \begin {bmatrix}
a&b&c&d \\
3a&3b&3c&3d\\
-2a&-2b&-2c&-2d\\
4a&4b&4c&4d
\end{bmatrix}##

So in this example we have rows: ##R_2=3R_1, ~R_3= -2R_1## and ##R_4 =4R_1##.

Take the vector ##\textbf {x}=\begin {bmatrix} x_1\\x_2\\x_3\\x_4 \end {bmatrix}##

##A \textbf {x}=\begin {bmatrix}
R_1\cdot \textbf {x}\\
3R_1\cdot \textbf {x}\\
-2R_1\cdot \textbf {x}\\
4R_1\cdot \textbf {x}
\end {bmatrix}##

To make ##A \textbf {x}=0 ## (i.e. for ##\textbf {x}## to be in A’s null space) we require only that ##R_1\cdot \textbf {x} = 0##.

This gives the required ‘single equation’ (mentioned in the soution manual): ##ax_1 + bx_2 + cx_3 + dx_4= 0##

This equation defines a ‘plane’ in ##\mathbb R^n##. Though, as already has been said, for n>3 the term ‘hyperplane’ could be used to avoid confusion with 2D planes. (Also, note that for n = 2, the so-called ‘plane’ would a line!)
 
  • #8
pasmith said:
As I stated above, every [itex](n - 1)[/itex] dimensional subspace is [tex]\{ \mathbf{x} \in \mathbb{R}^n: \mathbf{x} \cdot \mathbf{a} = x_1a_1 + \cdots + x_na_n = 0\}[/tex] for some [itex]\mathbf{a} \neq 0[/itex]. This is analagous to a plane in [itex]\mathbb{R}^3[/itex]. In higher dimensions one might more properly call it a hyperplane.

The same way a line is always 1 dimension, I would think of a plane with no prefix as being 2 dimensions.
 
  • #9
Office_Shredder said:
I would think of a plane with no prefix as being 2 dimensions.
This...
 

FAQ: Intro to Linear Algebra - Nullspace of Rank 1 Matrix

What is the nullspace of a matrix?

The nullspace of a matrix A, also known as the kernel, is the set of all vectors x such that A*x = 0. It essentially represents all the solutions to the homogeneous equation A*x = 0.

How do you find the nullspace of a rank 1 matrix?

To find the nullspace of a rank 1 matrix, you need to solve the equation A*x = 0. Since a rank 1 matrix has only one linearly independent row, you can usually express the nullspace as a span of vectors that satisfy this equation. Typically, you will end up with a vector that spans the nullspace.

What is the dimension of the nullspace for a rank 1 matrix?

The dimension of the nullspace for a rank 1 matrix is n - 1, where n is the number of columns in the matrix. This is because the rank of the matrix plus the dimension of the nullspace must equal the number of columns (Rank-Nullity Theorem).

Why is the nullspace important in linear algebra?

The nullspace is important because it provides insight into the solutions of linear systems. It helps in understanding the structure of the matrix and the linear transformations it represents. In practical applications, the nullspace can indicate dependencies between variables and can be used in optimization problems, differential equations, and more.

Can the nullspace of a rank 1 matrix be empty?

No, the nullspace of a rank 1 matrix cannot be empty. The nullspace will always contain at least the zero vector. For a rank 1 matrix, the nullspace will actually be a subspace of dimension n - 1, meaning it will contain infinitely many vectors in addition to the zero vector.

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