Finding the null space, matrix fun wee

In summary: No, the determinant of the matrix is not 0. It is just the product of all the entries in the matrix, and so it is not zero if any row or column is 0.
  • #1
mr_coffee
1,629
1
Hello everyone I'm confused on finding the null space on this problem:
the matrix is:
0 2 0 -5
0 1 4 0
0 0 1 0
0 0 0 1

null(A) =
2b - 5d = 0
b + 4c = 0;
c = 0
d = 0;
b = 0;
a = ?
You don't know what a is, so I'm quite confused. Any help?
 
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  • #2
Just a thought..your matrix row reduces to:

0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

Somebody else may have to check this.. but it suggests to me that the only values for b c and d are for the trivial solution, and a could be anything.
 
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  • #3
Ok, so here is your matrix:
[tex] A = \left( \begin{array}{cccc} 0 & 2 & 0 & -5 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) [/tex]
You want to know what the nullspace is of the matrix [itex] A [/itex]. So this means you have an equation of the form.
[tex] A\vec x = \vec 0 [/tex]
which will look like (sorry LaTeX wasn't playing nice for me here, I'll just write it out):
[0 2 0 -5][x1] = [0]
[0 1 4 0][x2] = [0]
[0 0 1 0][x3] = [0]
[0 0 0 1][x4] = [0]
So to solve this you could:
Use gaussian-jordon elimination and reduce it to:
[tex] \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) [/tex]
as Hammie pointed out.

So this leaves you with:
[tex] x_1 = \alpha [/tex]
[tex] x_2 = 0 [/tex]
[tex] x_3 = 0 [/tex]
[tex] x_4 = 0 [/tex]

Which is a matrix of the form:
[tex] \left( \alpha,0,0,0 \right) ^T [/tex]

which typically you want to factor out the variables to yield:
[tex] \vec x = \alpha \left( 1,0,0,0 \right) ^T [/tex]Now thinking of this in the algebraic sense, the solution to the system is:
(going back to the [itex] a, b, c, d [/itex] notation:
[tex] a = \alpha [/tex]
[tex] b = 0 [/tex]
[tex] c = 0 [/tex]
[tex] d = 0 [/tex]

Which, just as Hammie said. b,c,d are for the trivial solution while a could be anything.

I think an easy general method for solving for the null space is to:
1) reduce it as much as possible:
ex) (1 0 0, 0 1 0, 0 0 0)^T type of matrix

2) any row that is all 0's then set it equal to some variable.
3) write down a column vector describing the solutions, and factor out the variables.
 
Last edited:
  • #4
Wow excellent explanation! thank you very much, both of you! :)
 
  • #5
mr_coffee said:
Hello everyone I'm confused on finding the null space on this problem:
the matrix is:
0 2 0 -5
0 1 4 0
0 0 1 0
0 0 0 1
null(A) =
2b - 5d = 0
b + 4c = 0;
c = 0
d = 0;
b = 0;
a = ?
You don't know what a is, so I'm quite confused. Any help?

Yes, that's precisely correct. You can't solve for a because a doesn't appear in any of the equations. In particular if b= c= d= 0, No matter what a is, the linear transformation is will take (a, 0, 0, 0) to (0, 0, 0, 0).
What does that tell you about one basis vector for the null space?

You also have b+ 4c= 0 so that c= -(1/4)b as well as 2b- 5d= 0 so that
d= (2/5)b. That seems to say that if you take b to be anything, c= -(1/4)b, d= (2/5)b and a to be anything, all the equations will be satisfied. What does that tell you about the dimension of the kernel? What is a basis for the kerne?
 
  • #6
Quick question, I'm now trying to find the imagine space of this matrix, but I took the determinant of the matrix and it is equal to 0, doesn't that mean there is no image space? Because there is a row of 0's it can't span R^4, so there can't be a unique solution for all varaibles right?
 

1. What is the null space of a matrix?

The null space of a matrix is the set of all vectors that, when multiplied by the matrix, result in a zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the matrix and x is a vector.

2. How do you find the null space of a matrix?

To find the null space of a matrix, you can use the techniques of row reduction or Gaussian elimination. These methods involve performing elementary row operations on the matrix until it is in reduced row echelon form, where the null space can be easily identified.

3. Why is the null space important in linear algebra?

The null space is important in linear algebra because it provides information about the solutions to a system of linear equations. It can also be used to determine the rank and dimension of a matrix, which are important properties in various applications of linear algebra.

4. Can the null space of a matrix be empty?

Yes, it is possible for the null space of a matrix to be empty. This occurs when there are no solutions to the equation Ax = 0, which can happen if the matrix is invertible or if there are no free variables after row reduction.

5. How does the null space relate to the concept of linear independence?

The null space is closely related to the concept of linear independence. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. This is equivalent to saying that the null space of the matrix formed by these vectors is empty.

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