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Linear Algebra Decoded
 

Linear Algebra Decoded Example

Linear Algebra Decoded
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This example has been taken directly from the solution given by Linear Algebra Decoded to the formulated problem.
Find the sum of the subspaces E and F.

E =  {(x, y, z, w)∈ R4 | -y+w = 0; x-2y-2z+w = 0 }

F =  {(x, y, z, w)∈ R4 | x-2z-w = 0; x+4y+4z+w = 0; 2y+3z+w = 0 }

How to solve this problem?
The sum of two subspaces E and F, written E + F, consists of all sums u + v, where u belongs to E and v belongs to F. It is the smallest of all the subspaces containing both subspaces. In practice, to determine the sum subspace, just find the subspace spanned by the union of two sets of vectors, one that spans E and other that spans F. In this case, first it must be determined two sets of vectors that span E and F respectively, specifically two bases, one for the subspace E and another one for the subspace F.

Step 1: Find a basis for the subspace E.

- Implicit equations of the subspace E.
-y+w = 0
x-2y-2z+w = 0

- Represent the system of linear equations composed by the implicit equations of the subspace E in matrix form.
┌                   ┐
│ 0  -1   0  1 |  0 │
│ 1  -2  -2  1 |  0 │
└                   ┘

- Transform the augmented matrix to row echelon form.
r1 <---> r2
┌                   ┐
│ 1  -2  -2  1 |  0 │
│ 0  -1   0  1 |  0 │
└                   ┘

- Translate the row echelon form matrix to the associated system of linear equations, eliminating the null equations.
x-2y-2z+w = 0
-y+w = 0

- Select the free variables.
{z, w }

- Back substitution.

-y+w = 0
y = w

x-2y-2z+w = 0
x = 2y+2z-w
x = 2•(w)+2z-w
x = 2z+w

- Parametric representation.
E =  {(2z+w ; w ; z ; w) | z, w ∈ R }

- Basis for the subspace E. Let's call it A.
A = {(1  1  0  1); (2  0  1  0)}

Step 2: Find a basis for the subspace F.

- Implicit equations of the subspace F.
x-2z-w = 0
x+4y+4z+w = 0
2y+3z+w = 0

- Represent the system of linear equations composed by the implicit equations of the subspace F in matrix form.
┌                   ┐
│ 1  0  -2  -1 |  0 │
│ 1  4   4   1 |  0 │
│ 0  2   3   1 |  0 │
└                   ┘

- Transform the augmented matrix to row echelon form.
r2 <---> r2 - r1
┌                   ┐
│ 1  0  -2  -1 |  0 │
│ 0  4   6   2 |  0 │
│ 0  2   3   1 |  0 │
└                   ┘
r2 <---> r3
┌                   ┐
│ 1  0  -2  -1 |  0 │
│ 0  2   3   1 |  0 │
│ 0  4   6   2 |  0 │
└                   ┘
r3 <---> r3 - 2•r2
┌                   ┐
│ 1  0  -2  -1 |  0 │
│ 0  2   3   1 |  0 │
│ 0  0   0   0 |  0 │
└                   ┘

- Translate the row echelon form matrix to the associated system of linear equations, eliminating the null equations.
x-2z-w = 0
2y+3z+w = 0

- Select the free variables.
{z, w }

- Back substitution.

2y+3z+w = 0
y = (-3z-w)/2

x-2z-w = 0
x = 2z+w

- Parametric representation.
F =  {(2z+w ; (-3z-w)/2 ; z ; w) | z, w ∈ R }

- Basis for the subspace F. Let's call it B.
B = {(2  -1  0  2); (4  -3  2  0)}

Step 3: Find the subspace spanned by the vectors of both bases: A and B.

- Write the associated augmented matrix.
┌                  ┐
│ 1  2   2   4 | x │
│ 1  0  -1  -3 | y │
│ 0  1   0   2 | z │
│ 1  0   2   0 | w │
└                  ┘

- Transform the augmented matrix to row echelon form taking into account the last column composed of variables.
r2 <---> r2 - r1
r4 <---> r4 - r1
┌                      ┐
│ 1   2   2   4 | x    │
│ 0  -2  -3  -7 | -x+y │
│ 0   1   0   2 | z    │
│ 0  -2   0  -4 | -x+w │
└                      ┘
r2 <---> r3
┌                      ┐
│ 1   2   2   4 | x    │
│ 0   1   0   2 | z    │
│ 0  -2  -3  -7 | -x+y │
│ 0  -2   0  -4 | -x+w │
└                      ┘
r3 <---> r3 + 2•r2
r4 <---> r4 + 2•r2
┌                        ┐
│ 1  2   2   4 | x       │
│ 0  1   0   2 | z       │
│ 0  0  -3  -3 | -x+y+2z │
│ 0  0   0   0 | -x+2z+w │
└                        ┘

Step 4: Subspace E + F.

E + F =  {(x, y, z, w)∈ R4 | -x+2z+w = 0 }

Parametric representation of the subspace.

Free variables.
{y, z, w }

Back substitution.

-x+2z+w = 0
x = 2z+w

Parametric representation.
E + F =  {(2z+w ; y ; z ; w) | y, z, w ∈ R }

  
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