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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 vector subspace E spanned by the set of vectors V.

V = {(-2  -4  2  -4); (-1  2  0  1); (1  6  -2  5)}

How to solve this problem?
The span of a set of vectors V is the set of all possible linear combinations of the vectors of V. It will be use the notation [V] to denote the span of V. In practice, the problem of determining the implicit equations of the subspace spanned by V, is equivalent to determine when the system of linear equations where the coefficient matrix is composed by the vectors of V as columns, and a generic vector of the space specified by means of variables as the vector used to compose the augmented matrix, has solution. To solve this problem just transform the augmented matrix to row echelon form. The rank of the coefficient matrix must be equal to the rank of the augmented matrix, therefore the entries in the most right column (which are equations) associated to null rows in the coefficient matrix must be zero.

Step 1: Write the augmented matrix of the system of linear equations where the coefficient matrix is composed by the vectors of V as columns, and a generic vector of the space specified by means of variables as the additional column used to compose the augmented matrix.

┌                ┐
│ -2  -1   1 | x │
│ -4   2   6 | y │
│  2   0  -2 | z │
│ -4   1   5 | w │
└                ┘

Step 2: Transform the augmented matrix to row echelon form taking into account the last column composed of expressions depending on variables.

r2 <---> r2 - 2•r1
r3 <---> r3 + r1
r4 <---> r4 - 2•r1
┌                    ┐
│ -2  -1   1 | x     │
│  0   4   4 | -2x+y │
│  0  -1  -1 | x+z   │
│  0   3   3 | -2x+w │
└                    ┘
r2 <---> r3
┌                    ┐
│ -2  -1   1 | x     │
│  0  -1  -1 | x+z   │
│  0   4   4 | -2x+y │
│  0   3   3 | -2x+w │
└                    ┘
r3 <---> r3 + 4•r2
r4 <---> r4 + 3•r2
┌                      ┐
│ -2  -1   1 | x       │
│  0  -1  -1 | x+z     │
│  0   0   0 | 2x+y+4z │
│  0   0   0 | x+3z+w  │
└                      ┘

Step 3: For the system to have solution is necessary that the entries in the last column, corresponding to null rows in the coefficient matrix be zero (equal ranks). The equations defined by those expressions, are the implicit equations of the vector subspace spanning for the set of vectors.

E = [V] =  {(x, y, z, w)∈ R4 | 2x+y+4z = 0; x+3z+w = 0 }

Parametric representation of the subspace.

Find the equivalent system of implicit equations transforming the associated matrix to row echelon form.
┌                 ┐
│ 2  1  4  0 |  0 │
│ 1  0  3  1 |  0 │
└                 ┘
r1 <---> r2
┌                 ┐
│ 1  0  3  1 |  0 │
│ 2  1  4  0 |  0 │
└                 ┘
r2 <---> r2 - 2•r1
┌                   ┐
│ 1  0   3   1 |  0 │
│ 0  1  -2  -2 |  0 │
└                   ┘

Equivalent system of implicit equations.
x+3z+w = 0
y-2z-2w = 0

Free variables.
{z, w }

Back substitution.

y-2z-2w = 0
y = 2z+2w

x+3z+w = 0
x = -3z-w

Parametric representation.
E = [V] =  {(-3z-w ; 2z+2w ; z ; w) | z, w ∈ R }

  
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