Finding eigenvalues using the characteristic equation

Finding eigenvalues using the characteristic equation


>>Let’s talk about the
characteristic equation. So recall, the scaler lambda
is called an eigenvalue of a square matrix a if and only if a minus lambda times the
identity has a nontrivial null space. We also learned that a square
matrix has a nontrivial null space if and only
if the determinant of that matrix is zero. So we can combine these
two statements to come up with a method for computing
eigenvalues of a matrix a. So we’re — the idea is just
to treat lambda as an unknown and look at the determinant of a
minus lambda times the identity, when is that equal to zero? And so this equation, determinant a minus lambda times
the identity equals zero is called a characteristic
equation, right. So it’s an equation that
has this equals sign in it. We’re going to solve the
characteristic equation for — solve for lambda to
find the eigenvalues. The left side of this
equation is polynomial, determinant a minus lambda i, is called the characteristic
polynomial. So let’s take a look
at an example and try to find the real
eigenvalues of this matrix a which is minus 4, minus 1, 6, 1. We compute a minus
lambda times the identity, and this just subtracts
off lambda from the main diagonal
of the matrix. And then we compute determinant
using the formula that we saw for two-by-two matrices. We multiply everything out
until we get the characteristic polynomial, lambda squared
plus 3 lambda plus 2, and the characteristic equation, lambda squared plus 3
lambda plus 2 equals zero. We’re going to solve this
characteristic equation for lambda to see what
the eigenvalues of a are. I can factor this quadratic
and see that the eigenvalues, the roots of the
characteristic polynomial which give me the
eigenvalues of the matrix, these are just minus
2 and minus 1. Now that we have the
eigenvalues, if we wanted to go and find the eigenvectors,
we would just take each of those values for lambda
and find the null space of a minus lambda i. And so for
two-by-two we can sometimes do that by inspection, or we
can go back all the way to the beginning and just
do row reduction to solve. Let’s take a look at a
triangular example — and it’s slightly larger — suppose we want to
find the eigenvalues of this four-by-four matrix a,
given by 5, 5, zero, 2; zero, 2, minus 3, 6; zero, zero, 3,
minus 2; zero, zero, zero, 5. And so again we’re going to compute a minus
lambda times the identity. This subtracts lambda from the
main diagonal of the matrix. We still get a triangular
matrix out, notice. And so that means the
determinant is going to be easy to compute. It’s just the product
of the diagonal entries. The product of the
entries that are on the main diagonal
of this matrix. We get 5 minus lambda
shows up twice, a 2 minus lambda,
and a 3 minus lambda. So notice I’m not
going to multiply this out because the goal is
to find out what the roots of this characteristic
polynomial are. And so I’m going to
leave it in factored form so that I can read
off the roots. So we see the eigenvalues
are 5, 2, and 3. And notice 5 minus lambda
term has this exponent of 2. And so we say that the
algebraic multiplicity of the eigenvalue 5 is 2. And likewise, the
multiplicity of 2 is 1. And the multiplicity of 3 is 1. And so this trick that we just
did — or that we just noticed, works in general for
any triangular matrix. So if a is an n-by-n
triangular matrix, then a minus lambda times
the identity is also going to be triangular
because all this is going to do is subtract lambda from
the main diagonal entries. And then the determinant of a
minus lambda times the identity, since a minus lambda identity
is triangular, the determinant of that matrix is
just the product of the entries on
the main diagonal. So if the matrix
is upper-triangular or lower triangular like this, when I subtract off
lambda times the identity, it subtracts off
the main diagonal. Computing the determinant, I just multiply all the
entries on the main diagonal. It gives me the characteristic
polynomial that — already in factored form. And so I can read off that the
eigenvalues are just the entries on the diagonal,
a1, a2, up to an.

3 thoughts on “Finding eigenvalues using the characteristic equation

  • January 8, 2019 at 1:43 pm
    Permalink

    Best video on this topic! Thank you so much!

    Reply
  • March 14, 2019 at 2:28 am
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    Very nice explanation

    Reply

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