For us to find the eigenvectors
of the given two by two matrix,
the eigenvectors corresponding to Lambda sub one
equals negative three
and Lambdas up two equals negative five,
can be written as vector v sub one
equals the vector of 1 a
and vector v sub two equals the vector 1 b.
So we want to find the eigenvectors
in the form of vector v sub one and vector v sub two.
The eigenvectors of matrix A corresponding to Lambda
are the non-zero solutions to the equation
where we have the difference of matrix A
and the product of Lambda and the identity matrix
times vector x the eigenvector equals the zero vector.
So let's begin by determining the eigenvector v sub one
that corresponds to Lambda equals negative three.
So we have difference of matrix A the given matrix
and Lambda, which is negative three
times a two by two identity matrix
times eigenvector, which is the vector v sub one,
which is the vector 1 a
equals a zero vector.
Let's write minus negative three
times the two-by-two identity matrix
as plus the matrix three zero, zero three.
Let's also write the zero vector as the vector zeros zero.
And now to sum the two matrices,
45 plus three is -42.
30 plus zero of course is -30,
56 plus zero is 56 and 37 plus three is 40
times the vector 1 a,
which is the eigenvector that we're looking for,
equals the zero vector.
Notice for this example, there's only one variable.
So for the next step,
let's write the corresponding system of equations.
The first equation is negative 42 times one
or negative 42 plus negative 30 times a
or minus 30a equals zero.
Also we could stop here and solve for a
because we have one equation with one variable.
Let's go ahead and write the system
where the second equation is 56 times one
and 56 plus 40 times a plus 40a and equals zero.
So again, we could also just solve
the second equation for a, to determine the value of a,
to find the components of the eigenvector,
the vector v sub one.
Let's just go ahead and show both.
If we want to solve this first equation for a,
let's add 38 of both sides of the equation
that would give us negative 42 equals 30a
dividing both sides by 30 to solve for a
and simplifying, we get a equals negative seven fifths.
Let's just show we get the same result
from the second equation.
To solve the second equation for a,
we would subtract 56 on both sides,
which gives us 40a equals negative 56,
dividing both sides by 40
and simplifying, we also get a equals negative seven fifths.
So now we know the eigenvector,
the vector v sub one that corresponds to Lambda
equals negative three,
where the x component is one
is the vector one negative seven fifths.
And now we'll find the vector v sub two
the eigenvector that corresponds to Lambda
equals negative five.
So setting up the same equation,
we have the given matrix a minus Lambda
where Lambda now is negative five
times the two by two identity matrix
times the eigenvector, which is now the vector v sub two,
which is the vector 1 b equals the zero vector.
And again, let's write minus negative five
times the two by two identity matrix
as plus the matrix five zero zero five.
And let's write the zero vector as the vector zero zero.
Next we sum the two matrices
negative 45 plus five is negative 40,
negative 30 plus zero is negative 30,
56 plus zero is 56 and 37 plus five is 42.
Times the vector 1 b equals the vector zero zero.
Again, we only have one variable here.
Let's write the corresponding system of equations
where the first equation is negative 40 times one
or negative 40 plus negative 30 times b
or minus 30b equals zero.
And the second equation is 56 times one or 56
plus 42 times b equals zero.
And again, we can solve either equation to find b.
Again, let's show both.
Solving the first equation for b,
let's add 30b to both sides,
which would give us negative 40 equals 30b,
dividing both sides by 30,
simplifying, we have b equals negative four thirds.
So again we could stop here,
let's go ahead and show we get the same result
by solving the second equation for b.
To solve for the second equation for b
let's subtract 56 on both sides,
which gives us 42b equals negative 56,
dividing both sides by 42 and simplifying,
we do get the same result of negative four thirds.
So now we know the specific eigenvector,
the eigenvector v sub two
that corresponds to the eigenvalue of negative five
is the vector 1 b,
which is the vector one and negative four thirds.
I hope you found this helpful.