Finding the basis of a null space

[angiep410]

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?

 

[I like Serena]

Hi angiep410!

Did you try to multiply your original matrix with your solutions?
Does the product come out as the zero vector?

 

[angiep410]

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?

 

[HallsofIvy]

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?

 

[angiep410]

What is the row reduction you get?

 

[I like Serena]

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?

 

[angiep410]

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

 

[I like Serena]

Right at the first step you seem to have lost a minus sign.

The rest looks just fine, except that.

 

[angiep410]

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

 

[angiep410]

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

 

[I like Serena]

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?

 

[angiep410]

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.

 

[I like Serena]

Your row space vectors are dependent on each other.

 

[angiep410]

Your row space vectors are dependent on each other.

so what does that mean in terms of my answer?

 

[I like Serena]

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.

 

[angiep410]

but how do i pick another row vector?

 

[I like Serena]

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.

 

[angiep410]

I tried keeping (2,-2,-4,4) and putting in the 2 other rows, but my computer still says it's wrong

 

[I like Serena]

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?

 

[angiep410]

yes, I entered it exactly as shown. The program gives me spaces to put the numbers in.

 

[I like Serena]

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.

 

[angiep410]

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:
___ ____
___ ____
___ ____
___ ____

 

[I like Serena]

Then I'm out of ideas, assuming you picked another row properly and did not make any typos.

 

[vela]

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?

 

[angiep410]

I was using the one I originally gave and you were right! I needed the two non-zero rows from the reduced matrix! thanks!

 

[I like Serena]

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?

 

[vela]

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.

 

[I like Serena]

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?

 

[vela]

The computer said so. It's probably just a limitation of the software.

 

[I like Serena]

The computer said so. 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: