hi I'm Daniel Chan from UNSW Australia
in this linear algebra video we'll be
looking at the concept of
diagonalisability so we'll look at it in
terms of this question here we ask is
this 2x2 matrix a 2 1 0 2 diagonalizable
so diagonalisability is a question that
involves the theory of eigenvalues and
eigenvectors and the way we'll approach
this question is to ask ourselves can we
find a basis of eigenvectors for A so to
do that we'll actually find the
eigenvalues and find the eigenvectors so
let's look at the eigenvalues first so
as usual to find the eigenvalues it's
what you need to do is to work out the
roots of the characteristic polynomial
so you solve 0 equals determinant A minus lambda I 
this determinant of this 2 by 2 matrix
is found by subtracting lambdas from the
diagonals of this matrix here, so you have 2 minus lambda, 2 minus lambda on the
diagonals and these off diagonal entries
stay the same so to compute this
determinant this is actually upper
triangular so you just have to multiply
the diagonal entries so that's just 2 minus lambda squared and of course for this to equal 0 each of the
factors has to equal 0 and since there's
a repeated root here we'll find there's
only one eigenvalue and that eigenvalue is
2 so lambda equals 2 is the only eigenvalue
so remember for a 2x2 matrix if you get
two distinct eigenvalues you know
straight away that it's diagonalisable
but here you have a multiple root
they're not distinct so you still can't
tell we have to go one step further and
actually compute the eigenvectors so
let's compute the eigen vectors now
eigen vectors for lambda equals 2 so remember how do you get the eigenvectors they're the non zero
vectors inside the kernel of a minus two I
so we need to solve a minus 2 I x equals
zero and let's write this out as a
augmented matrix so for the A minus 2 I and
the coefficient matrix we subtract two
from the diagonals so we get zero and
off diagnonal entries remain the same
and this we want equal to zero it's a
homogeneous equation and so we see
straight away that for this system of
linear equations to hold the second
coordinate of x has to be zero so
solution
is all vectors of form x1 0 and this is for
any x1 in R
so of course this is a subspace and which
subspace is it one way to say this is of
course it's just all scalar multiples
from 1 0 so this is just a span of 1 0 so if
we want a basis of eigenvectors what do
we need we need to have two linearly
independent vectors from this one
dimensional subspace which is not
possible so since we don't have a basis
of eigenvectors we see that A is
actually not diagonalisable
A cannot have a basis of eigenvectors so A is not diagonalizable
and that completes this question
