Let W be the subspace of R5 spanned by the vectors w1, w2, w3, w4, w5, where
w1 =
2 −1 1 2 0
,
w2 =
1 2 0 1 −2
,
w3 =
4 3 1 4 −4
,
w4 =
3 1 2 −1 1
,
w5 =
2 −1 2 −2 3
.
Find a basis for W ⊥.
Why did the vector go to therapy?
Because it had some linear dependencies it needed to sort out! Now let's find a basis for W ⊥, shall we?
To find the basis for the orthogonal complement, we can use the method of finding the null space of the transpose of the matrix formed by the vectors w1, w2, w3, w4, w5.
So, let's calculate the null space of the transpose of the matrix:
⎡
⎢ 2 1 4 3 2 ⎥
⎢ ⎥
⎢-1 2 3 1 -1 ⎥
⎢ ⎥
⎢ 1 0 1 2 2 ⎥
⎢ ⎥
⎢ 2 1 4 -1 -2⎥
⎢ ⎥
⎣ 0 -2 -4 1 3 ⎦
Now solve for the null space to find a basis for W ⊥.
Hmm, it seems like my mathematical skills are a bit rusty. My jokes are better than my linear algebra! Let's say the basis for W ⊥ is B = {v1, v2, v3}, where vi ∈ W ⊥.
To find a basis for the subspace W⊥, we need to find vectors that are orthogonal (perpendicular) to all the vectors in W.
Let's find the orthogonal complement of W step by step.
Step 1: Write down the augmented matrix [A|0], where A represents the matrix with the given vectors w1, w2, w3, w4, w5 as its columns, and 0 represents the zero vector.
A = [w1, w2, w3, w4, w5]
=
[2 1 4 3 2;
-1 2 3 1 -1;
1 0 1 2 2;
2 1 4 -1 -2;
0 -2 -4 1 3]
Step 2: Perform row reduction on [A|0] to reduced row echelon form using Gaussian elimination.
Rref(A|0) =
[1 0 1 0 -1 | 0;
0 1 2 0 3 | 0;
0 0 0 1 2 | 0;
0 0 0 0 0 | 0;
0 0 0 0 0 | 0]
Step 3: Write down the equations corresponding to the reduced row echelon form.
From the reduced row echelon form, we obtain the following equations:
x1 + x3 - x5 = 0
x2 + 2x3 + 3x5 = 0
x4 + 2x5 = 0
Step 4: Solve the system of equations.
From the equations, we can express the variables in terms of the free variables:
x1 = -x3 + x5
x2 = -2x3 - 3x5
x4 = -2x5
Step 5: Express the solutions in terms of vectors.
Using the values obtained in step 4, we can express the solutions as vectors. Let's denote the free variables as t1 and t2.
The solutions can be written as:
[x1; x2; x3; x4; x5] = [-t1 + t2; -2t1 - 3t2; t1; -2t2; t2]
= t1[-1; -2; 1; 0; 0] + t2[1; -3; 0; -2; 1]
Step 6: Write down a basis for W⊥.
The vectors [-1; -2; 1; 0; 0] and [1; -3; 0; -2; 1] form a basis for W⊥.
Therefore, a basis for W⊥ is:
B = {[-1; -2; 1; 0; 0], [1; -3; 0; -2; 1]}
To find a basis for the subspace W⊥, first, we need to find the orthogonal complement of W.
The orthogonal complement of W, denoted as W⊥, consists of all vectors v such that v is perpendicular to every vector in W.
To find v, we can set up an augmented matrix by stacking the vectors w1, w2, w3, w4, w5 as columns of a matrix:
W = [w1 | w2 | w3 | w4 | w5]
Finding the basis for W⊥ is equivalent to finding the null space of the matrix W.
1. Write the matrix W as a row reduced echelon form (RREF):
To start, perform row operations on the matrix W to reduce it into an upper triangular form.
2. Set up the augmented matrix with the columns of W and perform row operations to reduce it to RREF:
[W | 0] = [w1 | w2 | w3 | w4 | w5 | 0]
Apply row operations such as multiplying rows by scalars, adding rows, or swapping rows to transform the matrix into row reduced echelon form.
3. Identify pivot columns:
The pivot columns are columns that have leading non-zero entries in the row reduced echelon form. In this case, pivot columns correspond to independent variables.
4. Express remaining variables in terms of pivot variables:
Express all the variables in the non-pivot columns (free variables) in terms of the variables in the pivot columns. This will give you a parametric representation of the null space.
5. Write the basis vectors for W⊥:
The basis vectors for W⊥ are obtained by setting the free variables equal to the values in the parametric representation and solving for the pivot variables.
Once you have the parametric representation, you can determine the basis vectors for W⊥ by assigning specific values to the free variables and solving for the pivot variables. The resulting vectors will form a basis for W⊥.
To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors.
Define projection of v on u as
p(u,v)=u*(u.v)/(u.u)
we need to proceed and determine u1...u5 as:
u1=w1
u2=w2-p(u1,w2)
u3=w3-p(u1,w3)-p(u2,w3)
u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4)
u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5)
so that u1...u5 will be the new basis of an orthogonal set of inner space.
However, the given set of vectors is not independent, since
w1+w2=w3,
therefore an orthogonal basis cannot be found.
For further reading and examples, see for example
http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process