For this discussion, we will be going over problem Exercise EE.C20 in Section E.
Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so.
\[ \begin{bmatrix} -1 & 2 \\ -6 & 6 \end{bmatrix} \] I really like this breakdown of eigenvalues and vectors. It into detail regarding the actual applications of these concepts in addition to the thory http://math.mit.edu/~gs/linearalgebra/ila0601.pdf
Before we solve this problem, we must examine what we are being asked. Lets break it up into small steps. We are first tasked with finding what is an eigenvalue.
An eigen value is a scalar coming from a set associated with a linear system of equations. They are found by computing the root of the characteristic polynomial. You can think of the eigen value as a way to measure the shrink or stretch on some vector x. Think back to pre-calculus where the concept of shrinking or stretching on a function is introduced.
The eigen vector is a vector that only changes according to some scalar when a specific linear transform is applied to it. The eigen vector is also sometimes known as the characteristic vector.
This is a pretty good blurb on what an eigenspace is http://mathworld.wolfram.com/Eigenspace.html but long story short, the zero vector union with the set of all eigen vectors is a subspace or R known otherwise as a eigenspace of the eigen value (lambda).
With all these concepts in mind, lets solve our problem.
We need the following formulas in order to proceed: \[ Ax=\lambda x\\det(A-I\lambda)=0 \]
The first step is to find our charateristic polynomial and solve for the roots by subtracting some unknown from the diagonal terms of A.
\[ \begin{bmatrix} -1-x & 2 \\ -6 & 6-x \end{bmatrix} \]
We then proceed to compute the determinant https://math.dartmouth.edu/archive/m8s00/public_html/handouts/matrices3/node7.html
\[ \begin{bmatrix} -1-x & 2 \\ -6 & 6-x \end{bmatrix}=(-1-x)(6-x)-(2)(-6) \]
Use Algebra to simplify. Don’t forget to watch the signs when doing the distribution. It is always a good idea to arrange your polynomial in order of degree once you have carried out all distributions. Its more of a common convention in algebra. \[ (-1-x)(6-x)-(2)(-6)\\-6+x-6x+x^{2}+12\\x^{2}-5x+6 \]
Solve for the roots by setting the equation equal to zero. Here is a good sourcefor aquick reviewon factoring http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx \[ x^{2}-5x+6\\(x-3)(x-2)\\x=2, x=3 \]
We found the roots to be 2 and 3. The roots of our polynomial are the eigenvalues of this system. \[ \lambda=2\\\lambda=3 \]
The first case is to find vectors x which satisft the expression when lambda is 2 \[ (A-\lambda I)x=0\\(A-2 \lambda)\\\begin{bmatrix} -1-2 & 2 \\ -6 & 6-2 \end{bmatrix}\ \\\begin{bmatrix} -3 & 2 \\ -6 & 4 \end{bmatrix} \]
We now have a singular matrix, hence why lambda is an eigen value. We need to solve Ax=0 as follows: \[ \begin{bmatrix} -3 & 2 \\ -6 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} \] We need to perform row operations in order to reduce this system using Gauss-Jordan Elimination (note that column one is for x and column 2 is for y) \[ \begin{bmatrix} -3 & 2 & 0 \\ -6 & 4 & 0 \end{bmatrix} \]
Divide row 1 by -3 to get \[ \begin{bmatrix} 1 & -2/3 & 0 \\ -6 & 4 & 0 \end{bmatrix} \]
Add -6 times the 1st row to the second row to get \[ \begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
We are then able to pull the following equations \[ x-\frac { 2 }{ 3 } y=0 \]
Which can be re-written as \[ x=\frac { 2 }{ 3 } y \] Here we need to do some trial and error by picking values to make the above equation true. If we let x=1, then y=3/2 therefore the eigen vector is [1, 3/2] or we can re-write as [2,3]
We need to repeat the same process for lambda=3 \[ (A-\lambda I)x=0\\(A-3 \lambda)\\\begin{bmatrix} -1-3 & 2 \\ -6 & 6-3 \end{bmatrix}\ \\\begin{bmatrix} -4 & 2 \\ -6 & 3 \end{bmatrix} \]
Re-Write the system as follows: \[ \begin{bmatrix} -4 & 2 \\ -6 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}\\ \begin{bmatrix} -4 & 2 & 0 \\ -6 & 3 & 0 \end{bmatrix} \]
We perform row reduction in order to end up with the following system \[ \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
Which yields the equation \[ x-\frac { 1 }{ 2 } y=0 \] re-write as \[ x=\frac { 1 }{ 2 } y \]
We pick values for x and y, lets let x=1, therfore y is 2, thus our eigen vector for lambda=2 is [1,2]
Verify Eigen Values and vectors
A <- matrix(c(-1,2,-6,6), 2, 2, byrow=TRUE)
A## [,1] [,2]
## [1,] -1 2
## [2,] -6 6Compute the eigen values
e <- eigen(A)
values<-e$values
values## [1] 3 2We conclude that each eigenspace does indeed contain a single basis vector. We also know the multiplicity for each eigen value is 1. Hence we have found the eigen values, eigen vectors, and eigenspaces.
This is a good tutorial if you want to do it by hand https://rpubs.com/aaronsc32/eigenvalues-eigenvectors-r