Professor Dave again, let’s get diagonal.
Now that we have learned about eigenvalues and eigenvectors, it’s time to learn about
their applications, so let’s start by learning
about something called diagonalization.
Diagonalization involves taking a matrix and
writing it as a product of matrices, one of
which only has values along the diagonal.
For example, if we have matrix A, we will
write it as X D X-1 where D is a diagonal
matrix. If we move the X matrices to the other
side, we see that X-1 A X = D. The X matrix
is said to diagonalize A, because multiplying
A in this way gives us the diagonal matrix
D. This process only works for matrices that
have unique eigenvalues, or if there are duplicate
eigenvalues then there must exist linearly
independent eigenvectors for the duplicate
values. The reason for this is that the diagonal
matrix D is simply made up of the eigenvalues
of A, while the matrix X is made up of the eigenvectors.
So fortunately, once we know the eigenvalues
and eigenvectors of a matrix, the diagonalization
process is easy. Let’s assume an n by n
matrix A, which has eigenvalues λ1, λ2,
to λn. Then the diagonal matrix D is just
those eigenvalues placed along the diagonal,
with zeros for all the off-diagonal elements.
Now if x1, x2, to xn are the eigenvectors
for those eigenvalues, the matrix X will have
columns made up of those vectors, in the same
order as the eigenvalues, going down the diagonal,
(x1, x2, to xn). It’s important to note
that while we can place the eigenvalues and
eigenvectors in any order we want once we
find them, we must remain consistent in how
we order them in these matrices. The first
column of the X matrix must be the eigenvector
associated with the eigenvalue in the first
term for the diagonal of D, same with the
second column of X and the second term for
the diagonal of D, and so on.
Let’s look at an example. Consider the matrix
A, with entries -3, -4; 5, 6. We must first
find the eigenvalues, which means we must
solve for the values of lambda that satisfy
this expression, where the determinant of
A - λI = 0. Taking this determinant, we get
(-3 - λ)(6 - λ) - (-4)(5) = 0. Foiling these
binomials we get -18 + 3λ - 6λ + λ2 + 20
= 0, which when simplified gives us λ2 - 3λ
+ 2 = 0. Now by factoring this we see that
(λ - 1)(λ - 2) = 0, which means our solutions
and eigenvalues are λ = 1 and λ = 2. Now
to find the eigenvectors, we will plug in
these values of λ into the equation (A - λI)x
= 0, and solve for a possible vector x. Starting
with λ = 1, our matrix A - λI becomes -4,
-4; 5, 5. The top row ends up telling us that
-4x1 - 4x2 = 0. Dividing by -4 we get x1 +
x2 = 0, or simply x1 = -x2. We don’t have
to worry about the second row here because
it ends up giving us the same thing. We will
choose x2 = 1, which makes x1 equal to -1,
so our eigenvector for λ = 1 is (-1, 1).
Now we will plug in λ = 2. Our matrix A - λI
becomes -5, -4; 5, 4. Both rows will again
give the same result, but looking at the top
row we get -5x1 - 4x2 = 0. Solving for x1
we get x1 = (-4/5)x2. Choosing x2 = 1, we
get x1 = -4/5, making our eigenvector for
λ = 2 equal to (-4/5, 1). Now that we have
all of our eigenvalues and eigenvectors, we
can write out the diagonal matrix D and the
matrix X. D will simply be the eigenvalues
along the diagonal: D = 1, 0; 0, 2. X will
have the eigenvectors as columns, but now
that we have chosen an order for the eigenvalues
we must keep that order in mind. The eigenvector
for λ = 1 must be our first column, while
the eigenvector for λ = 2 will be our second,
making our matrix X = -1, -4/5; 1, 1. We have
obtained the main matrices for diagonalizing
the matrix A. The only remaining piece is
the inverse of X. Since X is a 2 by 2 matrix,
we simply swap the top left and bottom right
elements, multiply the off-diagonal elements
by -1, and divide the whole matrix by the
determinant. The determinant will be (-1 times
1) - (-4/5 times 1) which is -1 + 4/5, which
equals -1/5. The inverse will therefore be
the matrix 1, 4/5; -1, -1, divided by the
determinant -1/5, which means we just multiply
by -5, giving us -5, -4; 5, 5.
We have found all the components to diagonalize
A, those being X, D, and the inverse of X.
But in order to show that setting up the matrices
in this way does in fact give back A, following
the equation A = XDX-1, let’s do the multiplication
to verify. First take D times X-1 and we get
-5, -4; 10, 10. Next we will multiply X by
this new matrix, giving us 5 minus (4/5 times
10), 4 minus (4/5 times 10); -5 + 10, -4 +
10. This becomes 5 - 8, 4 - 8; 5, 6. We end
up with -3, -4; 5, 6, which was indeed the
matrix A we started with.
Diagonalization helps make computations with
matrices much easier and computer friendly,
making it a valuable process to remember,
so let’s check comprehension.
