We're given the 3-by-3 matrix A
with all entries equal to 1,
and asked to determine the eigenvalues,
determine what are the defects of the eigenvalues,
and finally, find the general solution
of x prime equals A times x.
To find the eigenvalues,
we set up the equation the determinant
of the difference of A and lambda I equals 0,
and then solve for lambda,
so using our matrix A, here's our initial setup.
Simplifying in the parentheses,
we now need to evaluate the 3-by-3 determinant,
set it equal to 0, and solve for lambda.
Using row 1, the determinant
is equal to the entry in row 1, column 1,
which is 1 minus lambda
times the determinant of the matrix
after deleting row 1 and column 1,
and then, minus the entry in row 1, column 2, which is 1,
times the determinant of the matrix
after deleting row 1 and column 2,
and then, finally, plus the entry in row 1, column 3,
which is 1 times the determinant of the matrix
after deleting row 1 and column 3,
and now we evaluate the 2-by-2 determinants.
Multiplying all of this out and simplifying,
we end up with negative lambda cubed
plus 3 lambda squared equals 0.
Factoring out negative lambda squared,
we have negative lambda squared
times the quantity lambda minus 3 equals 0.
Notice the eigenvalues are 0 and 3,
but because of the lambda squared,
the eigenvalue of 0 has an algebraic multiplicity of 2,
so we can say the eigenvalues
are lambda sub 1 and lambda sub 2 equals 0,
and lambda sub 3 equals 0,
or we can say the eigenvalues are 0 and 3.
The eigenvalue of 0 has an algebraic multiplicity of 2.
For part b,
we're asked to determine the defects of the eigenvalues.
The only eigenvalue that might have a defect
would be the eigenvalue of 0
because it has an algebraic multiplicity of more than 1,
in this case 2,
so if we're able to determine
two linearly independent eigenvectors
for the eigenvalue of 0,
the eigenvalue of 0 won't have a defect,
but, for example, if we could only find one
because the algebraic multiplicity is 2,
and we could only find one eigenvector,
2 minus 1 is equal to 1,
which would mean the eigenvalue of 0 has a defect of 1,
so now let's see if we can find
two linearly independent eigenvectors for lambda equals 0.
To do this, we set up the equation.
The difference of A and lambda I
times the vector v equals a 0 vector
and then determine a vector v,
so again, here's our initial setup.
Because lambda's 0, simplifying in the parentheses
this results in matrix A times vector v equals a 0 vector.
Notice all three equations are the same,
which are v1 plus v2 plus v3 equals 0.
Because of this, we will be able to determine
two linearly independent eigenvectors
for the eigenvalue of 0.
For example, for the first eigenvector v1,
we can let v1 equal 1, v2 equal 0, and v3 equal -1,
and for a second linearly independent eigenvector,
we can let v1 equal 0, v2 equal 1, and v3 equal -1,
which gives us the eigenvector v2,
so because the eigenvalue of 0,
which has an algebraic multiplicity of 2,
has two linearly independent eigenvectors,
meaning it has a geometric multiplicity of 2 as well,
the eigenvalue of 0 does not have any defects,
so again, we found two linearly independent eigenvectors
for the eigenvalue of 0 with multiplicity 2.
Therefore, there are no defects of the eigenvalue of 0.
There's also no defects for the eigenvalue of 3
because it has multiplicity of 1.
We also know, for part c,
when we go to find the general solution
to x prime equals A times x,
these eigenvectors
give us two linearly independent solutions,
which are x1 and x2
where x1 is equal to the eigenvector v1
times e to the power of 0t,
and x2 is equal to the eigenvector v2
times e to the power of 0t.
Of course, e to the power of 0t simplifies to 1.
We'll simplify this later,
and now, to find the general solution
to x prime equals A times x,
we still have to determine an eigenvector
corresponding to the eigenvalue of 3.
Let's do this on the next slide.
So again, for lambda sub 3 equals 3,
here's our initial setup.
Simplifying, we now need to solve the system
shown here on the right.
Let's use an augmented matrix,
and render it in reduced row echelon form.
Notice v3 is a free variable.
Row 1 indicates that v1 equals v3.
Row 2 indicates v2 equals v3,
and again, v3's a free variable,
so if we let v3 equal 1,
we have the eigenvector v3
corresponding to lambda sub 3 equals 3
of the vector 1, 1, 1,
which means the third independent solution
to x prime equals A times x
is x3 equals the vector 1, 1, 1
times e to the power of 3t,
and now we have all the information we need
to write the general solution
to the system given by x prime equals A times x.
Using the eigenvalues and the corresponding eigenvectors,
we have the general solution is x of t equals c1
times the eigenvector v1
times e to the power of 0t plus c2
times the eigenvector v2
times e to the power of 0t plus c3
times the eigenvector v3
times e the power of 3t.
Simplifying e to the power of 0t,
we have our general solution.
I hope you found this helpful.