We're given matrix a,
and asked to find the eigenvector
corresponding to the
eigenvalue three plus six i
in the form where where the
x component is positive one,
and the y component is a plus b i.
Looking at our notes below,
the eigenvectors of a matrix A
corresponding to a particular lamba
are the nonzero solutions
to this equation here.
Where in our case, we'd
have this equation here,
where here's lambda times the
two by two identity matrix,
minus the given matrix
A, times the eigenvector
equals the zero vector.
So, subtracting these matrices,
we have lambda minus
seven, two, negative 26,
and lambda plus one.
Now we'll substitute three
plus six i for lambda,
and solve the system of equations.
So if our lambda equals three plus six i,
we'd have three plus six i minus seven,
well three minus seven is negative four,
so we have negative four plus six i,
two, negative 26,
and then here we'd have
three plus six i plus one.
Of course three plus one is four,
so we have four plus six i.
Now, let's go ahead and
write the augmented matrix
from this matrix equation,
where the first row would
come from the equation
negative four plus six i times x sub one,
plus two times x sub two equals zero,
so the first row would be
negative four plus six i, two, and zero.
The second row would
come from the equation
negative 26 x sub one
plus four plus six i times
x sub two equals zero.
So we have negative 26,
four plus six i, and zero.
Now we'll write this in
reduced row echelon form,
so for our first step,
let's get a real number
in this position here
by multiplying the first
real by the conjugate
of negative four plus six i.
So we'll replace row one with
negative four minus six i times r sub one.
Well, negative four plus six i
times negative four minus
six i ends up being 16,
and then minus 36 i squared,
which is 16 minus 36 times
negative one, which equals 52,
so the first row is 52,
and then two times
negative four minus six i
is negative eight minus twelve i,
and then we have zero.
Second row stays the same.
Let's continue on the next slide.
Notice two times negative
26 plus 52 would be zero.
Let's replace row two with two
times row two plus row one.
Let's also get a one in this position
by replacing row one with
one over 52 times row one.
One over 52 times 52 is one,
and then one over 52 times
negative eight minus twelve i
would be negative eight
minus twelve i over 52,
but all these share a
common factor of four.
We can write this as four times
negative two minus three i,
all over four times 13.
The fours simplify out,
so we have
the quantity negative two
minus three i divided by 13,
the third element is zero,
and then for two times
row two plus row one,
two times negative 26 plus 52 is zero,
two times four plus six i
would be eight plus 12 i,
and if we add negative eight
minus 12 i, that would be zero.
Then we have zero.
So as expected, we do have a row of zeros,
which means there are an
infinite number of eigenvectors
corresponding to the given eigenvalue,
but this first row tells us that x sub one
plus the quantity
negative two minus three i
divided by 13 times x
sub 2 must equal zero,
so if we solve this for x sub 1,
we can say that x sub one
equals positive two plus
three i divided by 13
times x sub 2.
If we were looking for
all the eigenvectors
corresponding to the given eigenvalue,
we could let x sub 2 be equal to t,
and therefore x sub
one would be equal to--
let's write this as 2/13
plus 3/13 i
times t.
Remember, the eigenvector
can't be the zero vector,
so we can say t can't equal zero.
Again, in general, we can see that
the eigenvector x would
have an x component
of 2/13 plus 3/13
i times t,
times t
and a y component of t,
again, given t doesn't equal zero,
but remember if we go
back to the first slide,
we want the eigenvector,
where the x component is positive one,
and the y component is in the
form of a plus or minus b i.
So we want to solve this for x sub 2.
Multiplying both sides by the reciprocal,
we'd have x sub two equals 13,
over the complex number two plus three i
times x sub one.
Again, the eigenvector
that we're looking for,
we know has an x component of one,
which means x sub one is one,
so now we need to find
x sub two of one,
which would just be 13
over the complex number two plus three i,
but we need this in the
form of a plus or minus b i,
so we'll go ahead an multiply
this by two minus three i
over two minus three i.
In the numerator we have
13 times two minus three i.
Denominator, we're going to have four,
and then minus six i
plus six i, that's zero,
minus nine i squared,
which becomes plus nine,
so we have 13 times two
minus three i divided by 13.
13s simplify out, so x sub two
equals two minus three i,
when the x component is positive one.
So we have two minus three i,
so this is the particular
eigenvector we're looking for.
Going back to our first slide,
we only only enter the values of a and b.
Notice how we have plus sign here,
so a is equal to positive two,
and b would be negative three.
I hope you found this helpful.
