Hello, welcome to Maths with Jay
I'm going to answer each part of this
question in a separate screencast so
let's start with Part A so we're looking
to find the eigenvalues of this matrix
first of all. Let's just remind ourselves
what this is all about. We've got a
matrix A and the eigenvector x so
multiplying matrix A by the
eigenvector x will give us the same
result as multiplying the vector x by
the eigenvalue lambda, so in the first
part of the question we're looking to
find the eigenvalues, so we're expecting
to find up to three so let's just
rearrange this: we're going to subtract
(lambda x) from both sides, and because
this is a matrix equation we're going to
have A minus lambda times I, so I being
the identity matrix and that's
multiplying x, and that gives us the zero
vector
so the eigenvector x is not going to be
the zero vector so we can see that the
determinant of A minus (lambda I) is equal
to zero and that gives us our
characteristic equation and that's what
we start off by writing down, so
basically what we're doing is
subtracting lambda from each of our
terms on the main diagonal so we're
looking at the matrix A and writing down
the characteristic equation
so let's put in the main diagonal first
of all, we're subtracting lambda from each
of the numbers there, which all happen
to be the same, and then the rest of the
numbers stay as they were in A, so we've
got ones here and zero there so that's
the determinant, and the characteristic
equation is the determinant is equal to zero.
And then the simplest thing to do to
expand that is to use - well, you could use
the first row, the last row, the first
column, the last column - just one of the
ones that's got zero in it - so I'm going
to go with the first row, so we've got (2
minus lambda) multiplying the determinant
(2 minus lambda) 1, 1, (2 minus lambda) and
then we're going to subtract 1 times the
determinant 1, 0, 1, (2 minus lambda) and
that's equal to 0, so expanding all that
lot, so we've got (2 minus lambda)
multiplied by itself, subtract 1, and then
the second one will give us 1 times (2
minus lambda) minus 0, so that's just
going to be (2 minus lambda) so we can see
that we can take a (2 minus lambda)
outside both of these terms and we could
multiply out the lambda ... (2 minus lambda)
all squared is going to be lambda
squared minus 4 lambda plus 4, and then
subtracting 1, and then, so that's the
first, so that's the terms in the square
brackets - and then let's take away one
for the other (two minus lambda) and then
that's equal to 0, so we've got (2 minus
lambda) times lambda squared minus 4
lambda plus 2
is zero, so we've either got the( 2 minus
lambda) is 0, so lambda is 2, or - now
unfortunately this quadratic (lambda
squared minus 4 lambda plus 2) doesn't
factorize does it? - so we're going to have
to use the quadratic formula, so not too
bad, the numbers aren't too bad are they(?)
so minus negative 4, so that's going to
be 4 plus or minus the square root of 4
squared, so 16 minus 4ac, so minus 4 times
2, all over 2 times 1, so that will be 4
plus or minus 8 over 2, and that will be
2 plus or minus root 2, so we would leave
that in an exact form, so now we can
write down our different values for
lambda; it's a good idea to use
subscripts here, so our first value or a
value for lambda, our first eigenvalue
is 2, another eigenvalue is 2 minus root
2, so it's lambda2 and our third eigenvalue
lambda3 is 2 plus root 2, so it doesn't
matter what order you write those down
in, so long as you've got those 3 as your
eigenvalues, and it would be nice to do
a check if you've got time: do you know a
check? It involves the determinant of a
matrix A ,and the product of the
eigenvalues, those two things should be
the same as each other, so let's check
that out: so the determinant of A is
going to be equal to 2 times...again I'm
using the first row of A...it
makes life easier doesn't it? so 2 times
the determinant of 2 1 1 2 and minus the
determinant 1 1 0 2 so that will be
two times four minus one, minus two,
that will be zero, and so that's going to
be two times three minus two, so that's
going to be four, and then the product of
our eigenvalues is equal to two times
(2 minus root 2) times( 2 plus root 2) so
that's 2 times 2 squared minus root 2
squared, which is 2 so that's 2 times 4
oh no it's NOT is it?  It's 2 times 2 so
that's 2 times 2 which is also 4, so that
tells us that there is a good chance that
we've got the right answer, of course it
doesn't prove that it actually is the
right answer, but we know that if we have
got the the right values for the the
eigenvalues then their product
would be the same as the determinant, so
next time we'll go on and use these
eigenvalues to find the eigenvectors for
part b of the question.
