Finding the basis of the nullspace, can you see if i'm doing this right?

In summary, the author found out that the null space of their 5 by 3 matrix is not in R3, but in R5. They thanks their user for the help and explanation.
  • #1
mr_coffee
1,629
1
Hello everyone, I think i did this right, but i want to make sure this is what they want. The questions says:
In each case find a basis for and caclulate the dimension of nullA:
Here is my work and problem!
http://img207.imageshack.us/img207/2948/lastscan6ib.jpg [Broken]

I just found out its wrong...but why? the book got:
{[-2 4 1 0 0 ]^T, [7 -11 0 2 1]^T}
dim(nullA) = 2.
Those kind of resemble my columns, i got
[2 -4 0 0 0] and [-7 11 2 0 0], hm...
 
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  • #2
First of all, do you understand where the null space is?

If A: U->V is a linear transformation from U to V, then the null space of L is the subspace of U such that if x is in the null space, then Ax= 0.

Your A is a 5 by 3 matrix: from R5 to R3, but you give {[1,0,0], [0,1,0], [0,0,1]} as the basis for the null space. That's impossible! The nullspace of A a subpace of R5, not R3.

Those kind of resemble my columns, i got
[2 -4 0 0 0] and [-7 11 2 0 0], hm..."
Now, that's really mystiftying! In your image, your columns contain only 3 numbers, your rows, 5!

I haven't checked all of your arithmetic (you shift rows around in a way I find confusing) but assuming the last, row-reduced, matrix is correct, the equation AX= 0 becomes (letting X= [x, y, z, u, v]),
x+ 2z- 7v= 0, y- 4z+ 11v= 0, and u- 2v= 0. That is 3 equations in 5 unkowns so we can solve for 3 of them in terms of the other 2: in this row-reduced form, x= -2z+ 7v, y= 4z- 11v., u= 2v.
Taking z= 1, v= 0, that gives x= -2, y= 4, u= 0. One vector in the nullspace is [-2, 4, 1, 1, 0].
Taking z= 0, v= 1 gives x= 7, y= -11, u= 2. Another vector in the nullspace is [7, -11, 0, 2, 1].
Those 2 vectors constitute a basis for the nullspace.

The nullspace is 2 dimensional. That's no surprise. In order to "squeeze" a 5 dimensional space into a 3 dimensional space, at least 2 dimensions have to go to 0. Assuming no other dependencies, that would be exactly 2.
 
  • #3
ohh wow, i did totally screw that one up, thank you so much for the detailed explanation as always :biggrin: !
 

1. What is the nullspace and why is it important in scientific research?

The nullspace is the set of all vectors that produce a result of zero when multiplied by a given matrix. It is important in scientific research because it helps us identify linearly dependent variables and find solutions to systems of equations.

2. How do you find the basis of the nullspace?

To find the basis of the nullspace, we can use the row reduction technique or the Gaussian elimination method to reduce the given matrix to its reduced row echelon form. The basis of the nullspace will be the set of non-pivotal columns in the reduced matrix.

3. Can you explain the concept of linear independence in relation to the nullspace?

Linear independence refers to a set of vectors where none of them can be written as a linear combination of the others. In the context of the nullspace, if a set of vectors are linearly independent, then they will not contribute to the nullspace and will be considered as the basis for the nullspace.

4. How can we check if we have correctly found the basis of the nullspace?

We can check if we have correctly found the basis of the nullspace by multiplying the basis vectors by the original matrix and checking if the result is a zero vector. If the result is zero, then we have correctly identified the basis of the nullspace.

5. Are there any applications of the nullspace in real-world problems?

Yes, the nullspace has several applications in real-world problems such as image compression, data analysis, and network analysis. It helps in reducing the dimensionality of data and identifying relationships between variables, making it a useful tool in various scientific fields.

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