Problem of the week - Kernel of a linear transformation

Linear transformations are one of the key concepts of Linear Algebra, and they are considered the most useful part of this branch of mathematics. In the post "Linear algebra and digital image processing. Part III. Affine transformations", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity.

There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. As linear transformations can be represented by a matrix, most of the problems related to linear transformations can be solved using the same procedures we use to find the solution to many of the basic problems of matrices and system of linear equations.

This problem of the week will deal with the kernel (the set of vectors in the starting vector space which are transformed to the zero vector) and nullity of a linear transformation, and its solution only requires to know how to work with matrices and make elementary row operations.

Let f(x, y, z) = (-3x + 2y + 3z; x - αy + z; -4x + 4y + αz) a linear transformation from R³ to R³.

Find α so the nullity (dimension of the kernel) of f is equal to 1.

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.

For the dimension of the kernel to be equal to 1, the degrees of freedom for the system of linear equations resulting from making to zero all the components of the linear transformation formula, must be equal to 1, which is equivalent to the matrix of the system have rank equal to 2.

The system of linear equations is:

-3x + 2y + 3z = 0
  x - αy +  z = 0
-4x + 4y + αz = 0

And its matrix form is:

┌                ┐
│ -3   2  3 |  0 │
│  1  -α  1 |  0 │
│ -4   4  α |  0 │
└                ┘

Transform the augmented matrix to row echelon form.

r2 <———> 3•r2
r3 <———> -3•r3
┌                   ┐
│ -3    2    3 |  0 │
│  3  -3α    3 |  0 │
│ 12  -12  -3α |  0 │
└                   ┘
r2 <———> r2 + r1
r3 <———> r3 + 4•r1
┌                      ┐
│ -3     2      3 |  0 │
│  0  2-3α      6 |  0 │
│  0    -4  12-3α |  0 │
└                      ┘
r2 <———> r3
┌                      ┐
│ -3     2      3 |  0 │
│  0    -4  12-3α |  0 │
│  0  2-3α      6 |  0 │
└                      ┘
r3 <———> 4•r3
┌                       ┐
│ -3      2      3 |  0 │
│  0     -4  12-3α |  0 │
│  0  8-12α     24 |  0 │
└                       ┘
r3 <———> r3 + (2-3α)•r2
┌                                 ┐
│ -3   2                   3 |  0 │
│  0  -4               12-3α |  0 │
│  0   0  24 + (12-3α)(2-3α) |  0 │
└                                 ┘

For the matrix to have rank equal to 2, the term 24 + (12-3α)(2-3α) must be equal to 0. So let's find the values of α that make the equation equal to 0.

24 + (12-3α)(2-3α) = 0
24 + 24 - 36α - 6α + 9α2 = 0
9α2 - 42α + 48 = 0

dividing the equation by 3

3α2 - 14α + 16 = 0
(3α - 8)(α - 2) = 0

3α - 8 = 0
α = 8/3

α - 2 = 0
α = 2

The answer is: α = 8/3 or α = 2