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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 a basis and the parametric representation of the kernel of the linear transformation f.

f(x, y, z, w) = (y+2z-w; 2x+8y+2z-6w; 2x+7y-5w)

How to solve this problem?
The kernel or null-space of a linear transformation is the set of all the vectors of the input space that are mapped under the linear transformation to the null vector of the output space. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find the vector subspace where the implicit equations are the homogeneous equations obtained when the components of the linear transformation formula are equalled to zero.

Step 1: System of linear equations associated to the implicit equations of the kernel, resulting from equalling to zero the components of the linear transformation formula.

y+2z-w = 0
2x+8y+2z-6w = 0
2x+7y-5w = 0

Step 2: Represent the system of linear equations in matrix form.

┌                  ┐
│ 0  1  2  -1 |  0 │
│ 2  8  2  -6 |  0 │
│ 2  7  0  -5 |  0 │
└                  ┘

Step 3: Transform the augmented matrix to row echelon form.

r1 <---> r3
┌                  ┐
│ 2  7  0  -5 |  0 │
│ 2  8  2  -6 |  0 │
│ 0  1  2  -1 |  0 │
└                  ┘
r2 <---> r2 - r1
┌                  ┐
│ 2  7  0  -5 |  0 │
│ 0  1  2  -1 |  0 │
│ 0  1  2  -1 |  0 │
└                  ┘
r3 <---> r3 - r2
┌                  ┐
│ 2  7  0  -5 |  0 │
│ 0  1  2  -1 |  0 │
│ 0  0  0   0 |  0 │
└                  ┘

Step 4: Translate the row echelon form matrix to the associated system of linear equations, eliminating the null equations.

2x+7y-5w = 0
y+2z-w = 0

Step 5: Solution.

The implicit equations of the kernel are the equations obtained in the previous step.
Ker(f) =  {(x, y, z, w)∈ R4 | 2x+7y-5w = 0; y+2z-w = 0 }

Step 6: Find the parametric representation of the kernel, and a basis.

- Select the free variables.
{z, w }

- Back substitution.

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

2x+7y-5w = 0
x = (-7y+5w)/2
x = (-7•(-2z+w)+5w)/2
x = 7z-w

-  Parametric representation.
Ker(f) =  {(7z-w ; -2z+w ; z ; w) | z, w ∈ R }

- For each free variable, give the value 1 to that variable and value 0 to the others, obtaining a vector of the kernel. The set of vectors obtained is a basis for the kernel.
Basis(Ker(f)) = {(-1  1  0  1); (7  -2  1  0)}

  
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