Zero Dimensional Null Space (What's the meaning of this?)

  • #1
kosovo dave
Gold Member
35
0

Main Question or Discussion Point

So a question on my linear algebra homework asks for the dimensions of Nul(A) and Col(A).
Let A =
\begin{pmatrix}
-4 & -3\\
-1 &4\\
-3& -7
\end{pmatrix}

I row reduced the above matrix to
\begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}

Now, the T.A. for my section told us that to find the dimension of Nul(A) you look at the number of free variables in Nul(A). There are no free variables, so the dimension of Nul(A) is 0? What does this mean? I think I may be a little confused on what it means to find the dimension of a space. Why should the number of free variables in the null space tell you anything about the dimension of the null space?
 

Answers and Replies

  • #2
33,087
4,793
So a question on my linear algebra homework asks for the dimensions of Nul(A) and Col(A).
Let A =
\begin{pmatrix}
-4 & -3\\
-1 &4\\
-3& -7
\end{pmatrix}

I row reduced the above matrix to
\begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}
You should have another zero row in your matrix.
kosovo dave said:
Now, the T.A. for my section told us that to find the dimension of Nul(A) you look at the number of free variables in Nul(A). There are no free variables, so the dimension of Nul(A) is 0? What does this mean?
Yes, dim(Nul(A)) is 0. It means that the nullspace is just the zero vector. The null space will always contain the zero vector, but could have other vectors as well.
kosovo dave said:
I think I may be a little confused on what it means to find the dimension of a space. Why should the number of free variables in the null space tell you anything about the dimension of the null space?
Your matrix represents a transformation from ##\mathbb{R}^2## to ##\mathbb{R}^3##. In finding the nullspace, the matrix you ended with says that x = 0 and y = 0. There are only two variables, so there are no free variables that can take on arbitrary values. If you had ended up with a free variable, it would mean that the nullspace is a line (dimension 1) through the origin. If you had ended up with two free variables, the nullspace would be a plane (dimension 2) through the origin, and so on.
 
  • #3
kosovo dave
Gold Member
35
0
Since I only have 2 variables, can't I throw away that row of 0's in the RREF matrix?

You should have another zero row in your matrix.
Yes, dim(Nul(A)) is 0. It means that the nullspace is just the zero vector. The null space will always contain the zero vector, but could have other vectors as well.


Your matrix represents a transformation from ##\mathbb{R}^2## to ##\mathbb{R}^3##. In finding the nullspace, the matrix you ended with says that x = 0 and y = 0. There are only two variables, so there are no free variables that can take on arbitrary values. If you had ended up with a free variable, it would mean that the nullspace is a line (dimension 1) through the origin. If you had ended up with two free variables, the nullspace would be a plane (dimension 2) through the origin, and so on.
I still think I'm missing something here. I don't quite see why a free variable makes the null space a line, plane, etc. Maybe I'm approaching what the null space is in the wrong way. I'm thinking about it as the set of vectors you can put into x so that Ax=0. Is there a better way to think about it?
 
  • #4
Stephen Tashi
Science Advisor
6,928
1,184
I'm thinking about it as the set of vectors you can put into x so that Ax=0. Is there a better way to think about it?
It's best to keep in mind that the set of such vectors is itself a vector space. For example [itex] Ax = 0 [/itex] and [itex] Ay = 0 [/itex] implies [itex] A(x + y)\ = \ Ax + Ay \ = \ 0 + 0 = 0 [/itex] . If you are thinking of vectors as directed line segments originating at the origin of Euclidean 3-D space then the possible proper subspaces are lines, planes, and the zero dimensional space.
 
  • #5
33,087
4,793
Since I only have 2 variables, can't I throw away that row of 0's in the RREF matrix?
When you row reduce a matrix, you get another matrix that is equivalent to the one you start with. A 3 x 2 matrix can't be equivalent to a 2 x 2 matrix.
 

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