How do I find the orthonormal basis for the intersection of subspaces U and V?

In summary, the conversation revolves around finding the orthonormal basis for two subspaces U and V, with the initial attempt leading to the wrong answer. After discussing the approach and steps taken, it is suggested that the two null spaces of U and V must be combined into a 4x2 matrix, which is then row reduced to obtain the basis. Finally, the obtained basis must be orthonormalized to get the correct answer.
  • #1
Intothephy7
3
0

Homework Statement



Hi, i am trying to do the question on the image, Can some one help me out with the steps.

[PLAIN]http://img121.imageshack.us/img121/6818/algebra0.jpg

Solution in the image is right but my answer is so off from the current one.

Homework Equations





The Attempt at a Solution



This is the steps i took

1. found basic solution for U, 3 by 4, matrix to find U-perp
1. found basic solution for V, 3 by 4, matrix to find V-perp
3. Then i put V and U - perpendicular vectors to gether and got basic solutions with dim = 2 and 2 basic solutions.

than i did gram schmitt to find othogonal basic of those two

F1 = X1
F2 = X2 - {(F1 dot X2) / ||F1||2}F1

where did i went wrong
 
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  • #2
Found the answer, I actually have to find orthonormal vectors which is last step i missed

Q1 = (1/||F1||)F1
Q2 = (1/||F2||)F2
 
  • #3
Old thread, I know, but same question.

I tried doing the things OP said he did but ended up with the wrong answer. Here's what I did.

First I took the vectors that spanned U and made them the rows of a 3x4 matrix A. Then I got the RREF form of this matrix and found the 4x1 matrix X where AX = 0. This matrix was in the form (transposed) t[-1.4 3 -2.4 1]. Just used random numbers for this example.

Then I did the same thing for the other vectors and also got another 4x1 matrix.

Then I used the Gram-Schmidt algorithm to find the orthogonal basis of those two column matrices which gave me two 4x1 matrices which I then orthonormalized (if that's the word haha).

Can someone please let me know where I went wrong? Thanks.
 
  • #4
Mod Note: Moving this thread into the Calculus & Beyond section.
 
  • #5
BluePhone said:
Old thread, I know, but same question.

I tried doing the things OP said he did but ended up with the wrong answer. Here's what I did.

First I took the vectors that spanned U and made them the rows of a 3x4 matrix A. Then I got the RREF form of this matrix and found the 4x1 matrix X where AX = 0. This matrix was in the form (transposed) t[-1.4 3 -2.4 1]. Just used random numbers for this example.

Instead of showing us random numbers, how about showing the actual result you got.



BluePhone said:
Then I did the same thing for the other vectors and also got another 4x1 matrix.

Same thing here; show your result.


BluePhone said:
Then I used the Gram-Schmidt algorithm to find the orthogonal basis of those two column matrices which gave me two 4x1 matrices which I then orthonormalized (if that's the word haha).

Can someone please let me know where I went wrong? Thanks.
 
  • #6
The vectors I got were different but the question was the same; new numbers are generated every submission and I didn't write the matrix I got down, just scratchwork. But here's how I would solve the one in the original post:

First I'd rewrite the vectors as matrices:

2 0 -6 6
6 6 -2 -1
5 1 3 4

and

21 3 -9 30
6 2 18 -4
0 -2 3 -1

then I'd find the RREF of each

1 0 0 1.4674
0 1 0 -1.8043
0 0 1 -0.5109

and

1 0 0 1.2308
0 1 0 -0.3846
0 0 1 -0.5897

so I'd take the 4th variable as the parameter and make these two matrices which are the solutions to AX = 0

consider this X1

-1.4674
1.8043
0.5109
1

and

this X2

-1.2308
1.2308
0.5897
1

then I found F1 and F2

F1 = X1
F2 = X2 - {(F1 dot X2) / ||F1||^2}F1

and finally Q1 and Q2

Q1 = (1/||F1||)F1
Q2 = (1/||F2||)F2

for example, doing so would give me Q1

-0.56818
0.6986
...
...

which is incorrect
 
  • #7
The subspace U is a subspace because 3 vectors of 4 elements each cannot fully span R(4); similarly for the subspace V. This implies that there is a non-empty null space associated with U and V. The vectors X1 and X2 you found each span the associated null space.

The intersection of U and V is the complement (perpendicular subspace) of the union of the two null spaces. What you found so far is a pair of vectors that span the union of the two null spaces.

You must find a space that is perpendicular to the union of the null spaces. What you need to do is combine X1 and X2 into a 4x2 matrix (working in integers):

Code:
[ -135/92  -16/13  ]
[  83/46      5/13   ]
[  47/92      23/39 ]
[     1         1    ]

transpose it and row reduce:

Code:
[ 1   0   -11/21  -6/7 ]
[ 0   1   -1/7     -1/7 ]

Kill the first two columns and make the 3rd and 4th variables parameters:

Code:
[ 11/21  6/7 ]
[  1/7    1/7 ]
[   1       0  ]
[   0       1  ]

There's your basis; just orthonormalize and you're done. You might even get the same answer as shown in the first post.
 
  • #8
Excellent! Thank you very much!
 

1. What is an orthonormal basis?

An orthonormal basis is a set of vectors in a vector space that are mutually perpendicular and have a length of 1. This means that the dot product of any two vectors in the basis is 0, and the norm (or length) of each vector is 1.

2. Why is finding an orthonormal basis important?

Finding an orthonormal basis is important because it can simplify calculations and make it easier to understand and solve problems in linear algebra. It is also a fundamental concept in areas such as signal processing, quantum mechanics, and machine learning.

3. What does it mean to find an orthonormal basis for the intersection of two subspaces?

When finding an orthonormal basis for the intersection of two subspaces, we are looking for a set of vectors that are both contained in both subspaces and are mutually perpendicular with a length of 1. This basis will help us represent the intersection of the two subspaces in a simpler and more organized way.

4. How do you find an orthonormal basis for the intersection of two subspaces?

To find an orthonormal basis for the intersection of two subspaces, we can use the Gram-Schmidt process. This involves taking a set of linearly independent vectors from the two subspaces and using orthogonal projections to find a set of mutually perpendicular vectors with a length of 1.

5. Can there be more than one orthonormal basis for the intersection of two subspaces?

Yes, there can be more than one orthonormal basis for the intersection of two subspaces. This is because there are multiple ways to choose a set of linearly independent vectors from the two subspaces, and the Gram-Schmidt process can produce different results depending on the chosen vectors.

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