Find a basis for the null space

  • #1

Homework Statement



You're given two matrices (A and B). You want to find a basis for the space {x|x = Ay where By =0}.



Homework Equations





The Attempt at a Solution



You're looking for all vectors x=Ay such that y is in the null space of B. So you're looking for a basis for only a part of R(A). My initial though was to find a basis for the null space of B, multiply A by that basis, and then find a basis for that new matrix. But the more I think about it, the less sense that makes. Any suggestions?
 
Last edited:

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


Homework Statement



You're given two matrices (A and B). You want to find a basis for the space {x|x = Ay where By =0}.



Homework Equations





The Attempt at a Solution



You're looking for all vectors x=Ay such that y is in the null space of B. So you're looking for a basis for only a part of R(A). My initial though was to find a basis for the null space of B, multiply A by that basis, and then find a basis for that new matrix. But the more I think about it, the less sense that makes. Any suggestions?

How about multiplying the basis vectors for the null space of B by the matrix A? You will want to check for any vectors that might be in the null space of A.
 
  • #3


How about multiplying the basis vectors for the null space of B by the matrix A? You will want to check for any vectors that might be in the null space of A.

Matrix A is a 4x4 matrix of full rank, so the zero vector is the only vector in its null space. B is a 2x4 matrix of rank 2, so it's null space is of dimension 2. So if I mulitply the basis vector of B by matrix A, is that already a basis for x?
 
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


Matrix A is a 4x4 matrix of full rank, so the zero vector is the only vector in its null space. B is a 2x4 matrix of rank 2, so it's null space is of dimension 2. So if I mulitply the basis vector of B by matrix A, is that already a basis for x?

You could try to prove that it's a basis. It's kind of easy since if rank(A)=n, then A is invertible. You can assume a basis {u_i} for X and compare it to a basis {v_i} for null(B).
 
  • #5


A*(basis vectors for null(B)) is a 4x2 matrix of full rank, so it's columns form a basis for the the range of that matrix. I'm just a bit confused as to why that's a basis for x.
 
  • #6
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


A*(basis vectors for null(B)) is a 4x2 matrix of full rank

You're probably having trouble because you're trying to think of a basis as a matrix. It's more useful to use the fact that any vector in X can be written as a linear combination of basis vectors. Use that to relate vectors in X to vectors in null(B) and you should be able to make a connection between the bases as well.
 
  • #7


You're probably having trouble because you're trying to think of a basis as a matrix. It's more useful to use the fact that any vector in X can be written as a linear combination of basis vectors. Use that to relate vectors in X to vectors in null(B) and you should be able to make a connection between the bases as well.

So I should think of it as A times the first basis vector of null(B) and A times the second basis vector of null(B)?

So if y could be any vector (i.e., x is the range of A), then the basis vectors of y would be (e1,e2,e3,e4)? Is that a good way of looking at it?
 
  • #8
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


So I should think of it as A times the first basis vector of null(B) and A times the second basis vector of null(B)?

That's what you get if you start with the basis for null(B). If you want to turn this into a proof, you might try to assume that there is a vector in X that cannot be written as a linear combination of those vectors. You would then attempt to use the information given in the problem to show a contradiction.

So if y could be any vector (i.e., x is the range of A), then the basis vectors of y would be (e1,e2,e3,e4)? Is that a good way of looking at it?

I wouldn't say that's a good way of looking at it, since you really haven't defined what you mean by (e1,e2,e3,e4). Is that a collection of vectors or a vector with components given by e1, e2, etc? What is true is that

[tex] \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}[/tex]

is a basis for range(A). However that's not really useful for solving this problem, since X is not range(A).
 
  • #9


I'm sorry for all the questions, but how can you state that that's a basis for R(A) when you don't even know A?

I was just trying to look at it as if we were looking for the range, meaning that there were no restrictions on y. So in that case y could be expressed as a linear combination of (e1,e2,e3,e4).
 
Last edited:
  • #10
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


I'm sorry for all the questions, but how can you state that that's a basis for R(A) when you don't even know A?

True. But if rank(A) =4 and the elements of A are real, then [tex]\text{range}(A) = \mathbb{R}^4[/tex], so that's a basis.

I was just trying to look at it as if we were looking for the range, meaning that there were no restrictions on y. So in that case y could be expressed as a linear combination of (e1,e2,e3,e4).

Yes, if you define {e1,e2,e3,e4} (note the use of brackets since this is a set of vectors) to be a basis for range(A), then any vector in X can be expressed as a linear combination of them. The added condition will place relations among the coefficients of the expansion, reducing the dimension of the space.
 
  • #11


What if A were rank deficient? You couldn't automatically say that A*{b1,b2} (where b1 and b2 are the basis vectors of null(B)) is a basis for x? Wouldn't you have to check first?
 
  • #12


I guess what I'm trying to say is that isn't it possible for different matrices A and B that A*(basis vectors of null(B)) could just be a spanning set and not a basis for x?
 
Last edited:
  • #13
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


What if A were rank deficient? You couldn't automatically say that A*{b1,b2} (where b1 and b2 are the basis vectors of null(B)) is a basis for x? Wouldn't you have to check first?

If null(A) is not empty, then some linear combinations of the basis vectors of null(B) will be in null(A). Whatever linear combinations map to range(A) will span X.

I guess what I'm trying to say is that isn't it possible for different matrices A and B that A*(basis vectors of null(B)) could just be a spanning set and not a basis for x?

You should be able to use the ranks of A and B to find a relationship between dim X and dim null(B). If some linear combinations of the basis vectors get mapped to zero, you can use those equations to reduce the spanning set to a basis.
 
  • #14


Thanks for answering all my questions.

What am I going to do next may seem absurd, but I just wanted to make sure I understand the concept.

If there were no restrictions on y, then y could be any vector in R^4. So a basis of y would be {e1,e2,e3,e4}. Then the vector space in question would be spanned by A*({e1,e2,e3,e4}). Now whether or not that also forms a basis for the space would depend upon the spanning vectors being linearly independent. In this particular problem they are since A is nonsingular.

Now with the restriction that y must be in the 2-dimensional null space of the 2x4 matrix B, the vector space in question is spanned by A*{b1,b2} (where {b1,b2} is a basis for the null space of B). And whether or not they form a basis is dependent upon no linear combination of {b1.b2} being mapped into the null space of A. So you're just checking that the spanning vectors are linearly independent, right?
 
Last edited:
  • #15
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


Now with the restriction that y must be in the 2-dimensional null space of the 2x4 matrix B, the vector space in question is spanned by A*{b1,b2} (where {b1,b2} is a basis for the null space of B). And whether or not they form a basis is dependent upon no linear combination of {b1.b2} being mapped into the null space of A. So you're just checking that the spanning vectors are linearly independent, right?

Yes, that's all correct. But to put all that into a proof would really mean finding the relationship between dimensions of the spaces and then actually showing that a vector in X can be written as a linear combination of the basis vectors. You have all the pieces, you just want to put them into a formal order.
 
  • #16


Can you informally say that if A is a 4x4 matrix whose column space spans all of R^4, a basis for the null space of B (a 2x4 matrix with null space of dimension 2) will also form a basis for space X?
 
  • #17
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


Can you informally say that if A is a 4x4 matrix whose column space spans all of R^4, a basis for the null space of B (a 2x4 matrix with null space of dimension 2) will also form a basis for space X?

The basis for null(B) maps to the basis for X. The basis vectors for X might not be linear combinations of the basis vectors for null(B). You can see this if you take

[tex] A = \begin{pmatrix} 1 & 1 & 1& 1 \\ 0 & 1 & 1 &1 \\ 0& 0 & 1 & 1 \\ 0& 0& 0&1 \end{pmatrix}, [/tex]

and suppose that

[tex]\begin{pmatrix} 1\\ -1 \\ 0 \\ 0 \end{pmatrix},~\begin{pmatrix} 0\\ 0 \\ 1 \\ -1 \end{pmatrix}[/tex]

are basis vectors for some null(B).

As I've been saying, it's fine to say things in words, but at some point you will want to prove statements using some equations.
 

Related Threads on Find a basis for the null space

  • Last Post
2
Replies
29
Views
4K
Replies
6
Views
2K
Replies
9
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
4K
Replies
1
Views
9K
Replies
2
Views
3K
Replies
5
Views
17K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
Top