Intro to Linear Algebra - Nullspace of Rank 1 Matrix

[fractalizard]

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.

 

[pasmith]

An $(n-1)$ dimensional subspace in $\mathbb{R}^n$ is analagous to a plane in $\mathbb{R}^3$: they are given by an equation of the form $\mathbf{x} \cdot \mathbf{n} = 0$.

 

[FactChecker]

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.

Why isn't the nullspace an n-1 dimensional space within R^n?

It is. You can verify the properties of a subspace.

 

[Mark44]

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 $\mathbb R^4$.

 

[fractalizard]

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:

And from the solution manual:

The highlighted line is the piece I am having trouble understanding, the rest of the solution makes sense to me.

 

[pasmith]

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 $(n - 1)$ dimensional subspace is $$\{ \mathbf{x} \in \mathbb{R}^n: \mathbf{x} \cdot \mathbf{a} = x_1a_1 + \cdots + x_na_n = 0\}$$ for some $\mathbf{a} \neq 0$. This is analagous to a plane in $\mathbb{R}^3$. In higher dimensions one might more properly call it a hyperplane.

 

[Steve4Physics]

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!)

 

[Office_Shredder]

As I stated above, every $(n - 1)$ dimensional subspace is $$\{ \mathbf{x} \in \mathbb{R}^n: \mathbf{x} \cdot \mathbf{a} = x_1a_1 + \cdots + x_na_n = 0\}$$ for some $\mathbf{a} \neq 0$. This is analagous to a plane in $\mathbb{R}^3$. 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.

 

[Mark44]

I would think of a plane with no prefix as being 2 dimensions.

This...