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

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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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Found the answer, I actually have to find orthonormal vectors which is last step i missed

Q1 = (1/||F1||)F1
Q2 = (1/||F2||)F2
 
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.
 
Mod Note: Moving this thread into the Calculus & Beyond section.
 
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.
 
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
 
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.
 
Excellent! Thank you very much!
 

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