Finding the basis of a null space

In summary, the matrix given has a null space with dimensions 2, and to find the basis for the null space, the matrix must be row reduced. The resulting matrix should have a basis of (-3, 0), (-5, 2), (1, 0), and (0, 1). However, the computer keeps saying the answer is wrong because the row space vectors are dependent on each other. Another row vector that is not a scalar multiple of the others must be chosen to obtain a correct answer.
  • #1
angiep410
39
0

Homework Statement


The matrix is:
-2 -2 -4 4
-1 1 2 -2
-1 0 -3 0
-4 1 -7 -2

I know the dimensions for the null space are 2

Homework Equations


I know that to find the basis for a null space Ax=0, so I row reduced it and I got
1 0 3 0
0 1 5 -2
0 0 0 0
0 0 0 0

The Attempt at a Solution


Since I got this matrix, I thought the basis for the null space would be
-3 and 0
-5 2
1 0
0 1

but my computer keeps saying it is wrong. Can someone tell me what I'm doing wrong?
 
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  • #2
Hi angiep410! :smile:

Did you try to multiply your original matrix with your solutions?
Does the product come out as the zero vector?
 
  • #3
No it doesn't come out to zero, but I don't know how to get it to come out to zero. Do you have any suggestions?
 
  • #4
angiep410 said:

Homework Statement


The matrix is:
-2 -2 -4 4
-1 1 2 -2
-1 0 -3 0
-4 1 -7 -2

I know the dimensions for the null space are 2

Homework Equations


I know that to find the basis for a null space Ax=0, so I row reduced it and I got
1 0 3 0
0 1 5 -2
0 0 0 0
0 0 0 0
This is not the row reduction I get.

The Attempt at a Solution


Since I got this matrix, I thought the basis for the null space would be
-3 and 0
-5 2
1 0
0 1

but my computer keeps saying it is wrong. Can someone tell me what I'm doing wrong?
 
  • #5
What is the row reduction you get?
 
  • #6
angiep410 said:
No it doesn't come out to zero, but I don't know how to get it to come out to zero. Do you have any suggestions?

What HoI said!

Redo your row reduction.
If you get the same row reduction, perhaps you can show a couple of the steps you did?
 
  • #7
Row Operation 1:
multiply the 1st row by 1/2
1 -1 -2 2
-1 1 2 -2
-1 0 -3 0
-4 1 -7 -2
Row Operation 2:
add 1 times the 1st row to the 2nd row
1 -1 -2 2
0 0 0 0
-1 0 -3 0
-4 1 -7 -2
Row Operation 3:
add 1 times the 1st row to the 3rd row
1 -1 -2 2
0 0 0 0
0 -1 -5 2
-4 1 -7 -2
Row Operation 4:
add 4 times the 1st row to the 4th row
1 -1 -2 2
0 0 0 0
0 -1 -5 2
0 -3 -15 6
Row Operation 5:
interchange the 2nd row and the 3rd row
1 -1 -2 2
0 -1 -5 2
0 0 0 0
0 -3 -15 6
Row Operation 6:
multiply the 2nd row by -1
1 -1 -2 2
0 1 5 -2
0 0 0 0
0 -3 -15 6
Row Operation 7:
add 3 times the 2nd row to the 4th row
1 -1 -2 2
0 1 5 -2
0 0 0 0
0 0 0 0
Row Operation 8:
add 1 times the 2nd row to the 1st row
1 0 3 0
0 1 5 -2
0 0 0 0
0 0 0 0
 
  • #8
Right at the first step you seem to have lost a minus sign.

The rest looks just fine, except that.
 
  • #9
I like Serena said:
Right at the first step you seem to have lost a minus sign.

The rest looks just fine, except that.

where? I don't see it
 
  • #10
sorry. i see what you're talking about. i typed the matrix in wrong on this forum. the actual matrix is:
2 -2 -4 4
-1 1 2 -2
-1 0 -3 0
-4 1 -7 -2
 
  • #11
angiep410 said:
sorry. i see what you're talking about. i typed the matrix in wrong on this forum. the actual matrix is:
2 -2 -4 4
-1 1 2 -2
-1 0 -3 0
-4 1 -7 -2

In that case you have a correct basis that spans the null space.
Your matrix multiplied by each of your 2 solutions yields the zero vector, and there is no other independent vector with that property.

Why would you say that your computer keeps saying that your answer is wrong?
 
  • #12
I am asked to find the basis for the row space, which I said was:
(2,-2,-4,4), (-1,1,2,-2)

the column space, which I said was:
2 -2
-1 1
-1 0
-4 1

and the null space, which I said was:
-3 0
-5 -2
1 0
0 1

and my computer says that at least 1 thing is wrong, but it doesn't tell me what it is.
 
  • #13
Your row space vectors are dependent on each other.
 
  • #14
I like Serena said:
Your row space vectors are dependent on each other.

so what does that mean in terms of my answer?
 
  • #15
angiep410 said:
so what does that mean in terms of my answer?

You need to pick another row vector.
One that is not a scalar multiple of the other.
 
  • #16
but how do i pick another row vector?
 
  • #17
Keep one row and replace the other one by any other row you like.

Actually, you should check that if you left-multiply the row with the matrix, that it is not zero.
 
  • #18
I tried keeping (2,-2,-4,4) and putting in the 2 other rows, but my computer still says it's wrong
 
  • #19
In that case, I believe you're having trouble with entering your data in a format that your computer accepts.

Did you type in your vectors the way you showed them in your post?
 
  • #20
yes, I entered it exactly as shown. The program gives me spaces to put the numbers in.
 
  • #21
It seems weird to me that you would enter columns vectors next to each other the way you showed them in your post.

Other than that, I can't help you without more information on what your computer asks and shows exactly.
 
  • #22
My computer asks exactly what I wrote before and it shows:

Determine a basis for the row space:
(___ ___ ___ ___), (___ ___ ___ ___)

Determine a basis for the column space:
___ ____
___ ____
___ ____
___ ____

determine a basis for the null space:
___ ____
___ ____
___ ____
___ ____
 
  • #23
Then I'm out of ideas, assuming you picked another row properly and did not make any typos.
 
  • #24
For the row space, try entering the two non-zero rows from the reduced matrix. That's probably what the computer is expecting.

For the null space, your original answer was fine. The answer you gave in post #12 has a sign error. Which answer are you using now?
 
  • #25
I was using the one I originally gave and you were right! I needed the two non-zero rows from the reduced matrix! thanks!
 
  • #26
vela said:
For the row space, try entering the two non-zero rows from the reduced matrix. That's probably what the computer is expecting.

How did you come up with that?
 
  • #27
Row reduction just results in linear combinations of the original rows, so at the end, the non-zero rows, being independent, form a basis for the row space.
 
  • #28
vela said:
Row reduction just results in linear combinations of the original rows, so at the end, the non-zero rows, being independent, form a basis for the row space.

True.
But any 2 rows of the matrix also form a basis (except the first 2 rows).
So why wouldn't those qualify?
 
  • #29
The computer said so. :wink: It's probably just a limitation of the software.
 
  • #30
vela said:
The computer said so. :wink: It's probably just a limitation of the software.

More likely a limitation of the person formulating the questions and entering the acceptable answers. :tongue2:
 

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 the zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the matrix.

2. Why is finding the basis of a null space important?

Finding the basis of a null space is important because it allows us to understand the structure of the solutions to a system of linear equations. It also helps us to determine if a system is consistent or inconsistent, and to find special solutions such as the trivial solution or the unique solution.

3. How do you find the basis of a null space?

To find the basis of a null space, we first need to put the matrix A into reduced row echelon form. Then, we can identify the pivot columns and the free columns. The basis of the null space will be the columns of the original matrix that correspond to the free columns in the reduced row echelon form. These columns form a linearly independent set of vectors that span the null space.

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

Yes, the basis of a null space can be empty. This occurs when the matrix A has no free columns in its reduced row echelon form, which means that all columns are pivot columns. In this case, the null space only contains the zero vector, and there is no basis for the null space.

5. How does the dimension of the null space relate to the rank of the matrix?

The dimension of the null space, also known as the nullity, is related to the rank of the matrix by the rank-nullity theorem. This theorem states that the rank of a matrix plus its nullity is equal to the number of columns in the matrix. In other words, the rank and nullity are complementary, and as one increases, the other decreases.

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