>>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.

Best video on this topic! Thank you so much!

Very nice explanation

if 0-i?