in this problem we're given the matrix a
6 -2 6 and -1 and were asked to find
the eigenvalues and the eigenvectors of
that matrix
now the eigenvalues and eigenvectors are
scalars lambda the eigenvalues and
vectors v the eigenvectors so that A
times v is equal to lambda times v and
we'd like to find those so I'm going to
so we're going to start by finding the
eigen values lambda and we do that as
follows we compute the characteristic
polynomial of that matrix and we find
the zeros of the characteristic
polynomial so characteristic polynomial
p of lambda is defined as the
determinant of the matrix A minus lambda
times the identity matrix so let's
compute that so we get the determinant
of the matrix 6 minus lambda -2 6
and -1 minus lambda we'd like to compute
the determinant of this so we get
so we get this characteristic polynomial
lambda squared -5 lambda plus 6 and
we want to find the zeros of that and so
we see that these factors as lambda
minus 2 so this matrix has eigenvalues
two and three
so now let's compute the eigenvectors
for the eigenvalue lambda 1 equal to 2
so we have to find all the nonzero
vectors in the kernel of a minus 2 times
the identity matrix where 2 here is the
eigenvalue so this is the kernel of the
matrix 4 -2 6 and -3 so how do
I compute the kernel of a matrix
well I solve the linear system 4 -2
6 -3 and 0 0 because the kernel
means exactly those vectors were
mapped to 0 0 by this matrix so let's do
some row operation first I'm going to
divide the first row by 2 so this is 2
-1 0 6 -3 and 0 and then let's
subtract 3 times the first row from the
second row so 2 -1 0 and 0 0 0 ok
so by looking at this matrix we see we
can easily see one vector that's in the
kernel for example 1 & 2 so the kernel
of A minus 2I is the span of for example
the vector 1 2 and now we know that the
eigen vectors corresponding to the eigen
value lambda 1 are all the nonzero
multiples of this vector so
eigenvector for lambda 2are all the t
times 1 2 where t is not 0 now we'd like
to find the eigenvectors for the
eigenvalue lambda 2 equal to 3 and we
could do it exactly in the same way that
we did it for the first eigenvalue but
we're going to do something a bit faster
and we're going to use the fact that we
know if all the roots of the
characteristic polynomial are distinct
then all the kernels are one-dimensional
so that's exactly the case we're in
right here we've got two distinct
eigenvalues and therefore we know that
the corresponding kernels are
one-dimensional so here for lambda 2
equal to 3 we'd want to look at the
kernel of a minus 3 times the identity
matrix so this is the kernel of 3 -2 6 and -4 and it's actually
it's very easy to see to find a vector
in that kernel just by looking at the
first row for example we see that 2 & 3
is a vector so we know that this will be
the span for example of 2 & 3 and
therefore we know that our eigenvectors
for lambda 2 will be all the multiples t
times 2 3 where t is not 0
