### Video Transcript

Calculate the eigenvalues of the
matrix π΄ equals negative one, zero, zero, negative one.

Well, first when we take a look at
our matrix, we can see that itβs a two-by-two matrix. Well, we know that an π by π
matrix will always have π eigenvalues. So therefore, we know that with our
matrix being a two-by-two matrix, weβre gonna have two eigenvalues. Okay, great, so we know thatβs what
weβre looking for. But what can we do to work out the
eigenvalues? So to help us solve this problem
and find our eigenvalues, we have a relationship. And that one is that the
determinant of the matrix π΄ minus ππΌ, where πΌ is the identity matrix β β well,
this is equal to zero. And Itβs the π in our relationship
thatβs gonna help us calculate our eigenvalue.

So therefore, using our
relationship, we can say that the determinant of the matrix π΄, which is negative
one, zero, zero, negative one, minus π multiplied by the identity matrix, which is
the matrix one, zero, zero, one, is just gonna be equal to zero. And as we said, the identity matrix
is one thatβs in this form. But we have these terms as one. So the diagonal running from the
top left to bottom right is one. So therefore, if we multiplied by
the scaler, which was our π , weβre gonna get negative one, zero, zero, negative
one minus π, zero, zero, π. And we want to find the determinant
of this. So once we carried out the
subtraction of our matrices, weβre gonna get the determinant of the matrix. Negative one minus π, zero, zero,
negative one minus π is equal to zero.

So when we look at a two-by-two
matrix and we want to find the determinant of that, if you remind ourselves what we
do, well, we multiply diagonally and then subtract. So we have ππ minus ππ. So if we have the matrix π, π,
π, π the determinant of that is ππ minus ππ. So therefore, if we do that, weβre
gonna get negative one minus π multiplied by negative one minus π minus zero
equals zero. So then, if we distribute the first
parentheses over the second parentheses, weβre gonna get β well, first of all, weβre
gonna get one plus π. Thatβs cause negative one
multiplied by negative one is one and negative one multiplied by negative π is just
π. So itβs one plus π. Then weβve got plus another π and
then plus π squared. Then this is equal to zero. So then if we tidy this up, we get
π squared plus two π plus one equals zero.

So now what we need to do is solve
to find π cause itβs gonna give us our eigenvalues. Well now to solve this, what we can
do is factor because we can factor our expression because weβve got π squared plus
two π plus one. Well, we need two values. Theyβre gonna multiply together to
give us one and add together to give us two. So therefore, weβre gonna get π
plus one multiplied by π plus one is equal to zero. And thatβs because if we have one
multiplied by one is one; one add one is two. So they satisfy this. So now what we need to do is solve
this to find our π value. Well, what weβre gonna do is set
one of our parentheses equal to zero. And thatβs because to get a result
of zero, we need to have zero multiplied by something to get zero. So Iβm gonna have π plus one
equals zero.

So therefore, π is gonna be equal
to negative one. And actually, as you can see in
this one, weβve got π plus one multiplied by π plus one. This is same as π plus one
squared. So weβve got repeated roots. However, we were expecting two
eigenvalues cause we showed that earlier. There should be two eigenvalues for
two-by-two matrix. But weβve got a repeated
eigenvalue, so repeated root. So therefore, what we say is that
our eigenvalue, or eigenvalues, is negative one. But negative one is degenerate. And itβs degenerate because both of
our eigenvalues are identical because weβve got, as we said, repeated roots.