Matrices+ 03 - Eigenvalues in 3D: UNIZOR.COM - Math+ & Problems - Matrices

Notes to a video lecture on http://www.unizor.com

Matrices+ 02
Eigenvalues in 3D

Problem A

Find all eigenvalues and eigenvectors of this 3⨯3 matrix, if it's known that one of the eigenvalues is 10.

4	8	−4
4	−2	−4
−8	−4	−10

Note A
We specify one of the eigenvalues because general calculation of all eigenvalues in 3D leads to a polynomial equation of the 3rd degree.
Since we want to avoid the necessity to solve it, we specify one eigenvalue, which leads to finding the other two by solving a quadratic equation, which should not present any problem.

Solution A

If matrix A transforms vector v to a collinear one with the magnitude of the original one multiplied by a factor λ, the following matrix equation must hold
A·v = λ·v
or in coordinate form for 3⨯3 matrix

a1,1 a1,2 a1,3 a2,1 a2,2 a2,3 a3,1 a3,2 a3,3

·

v1 v2 v3

=

λ

·

v1 v2 v3

which is equivalent to
(a_1,1−λ)·v₁+a_1,2·v₂+a_1,3·v₃ = 0
a_2,1·v₁+(a_2,2−λ)·v₂+a_2,3·v₃ = 0
a_3,1·v₁+a_3,2·v₂+(a_3,3−λ)·v₃ = 0

This is a system of three linear equations with four unknowns λ, v₁, v₂ and v₃.

One trivial solution would be v₁=0, v₂=0 and v₃=0, in which case λ can take any value.
This is not a case worthy of analyzing.

If the matrix of coefficients of this system has a non-zero determinant, this trivial solution would be the only one.
Therefore, if we are looking for a non-trivial solution, the matrix's determinant must be zero, which gives a specific condition on the value of λ.

Therefore, a necessary condition for existence of eigenvectors v other than null-vector is
det(A) =
= (a_1,1−λ)·(a_2,2−λ)·(a_3,3−λ)+
+a_2,1·a_3,2·a_1,3+a_1,2·a_2,3·a_3,1−
−a_1,3·(a_2,2−λ)·a_3,1−
−a_1,2·a_2,1·(a_3,3−λ)−
−(a_1,1−λ)·a_2,3·a_3,2 = 0
Using values of a matrix of this problem, this equation is
(4−λ)·(−2−λ)·(−10−λ)+
+8·(−4)·(−8)+4·(−4)·(−4)−
−(−4)·(−2−λ)·(−8)−
−8·4·(−10−λ)−
−(4−λ)·(−4)·(−4) = 0
Simplifying this equation, we get
−λ³−8·λ²+108·λ+720=0

First of all, we can check if the eigenvalue 10 given in the problem is the root of this equation.
Indeed, the following is true.
−10³−8·10²+108·10+720=0

Since we know one root λ₁=10 of cubic equation, we can represent the left side of this equation as a product of (λ−10) and a quadratic polynomial with easily calculated coefficients, getting equation
(λ−10)·(−λ²−18·λ−72) = 0

To get all the roots of the original cubic equation, we have to solve a quadratic equation
−λ²−18·λ−72 = 0
or, using a canonical form,
λ²+18·λ+72 = 0
Its roots are
λ_2,3 = −9±√81−72 = −9±3

So, we have three eigenvalues for our matrix: −12, −6 and 10.

Consider now that we have determined eigenvalue λ and would like to find eigenvector v=||v₁,v₂,v₃|| transformed into a collinear one by matrix A with this exact factor of change in magnitude.

If some vector v=||v₁,v₂,v₃|| that is transformed into a collinear one with a factor λ exists, vector s·v, where s is any real non-zero number, would have exactly the same quality because of associativity and commutativity of multiplication by a scalar.
A·(s·v) = (A·s)·v = (s·A)·v =
= s·(A·v) = s·(λ·v)= λ·(s·v)

Therefore, we don't need to determine exact values v₁, v₂ and v₃, we just need to determine only the direction of vector v=||v₁,v₂,v₃||.

Let's start with the first eigenvalue λ₁=−12.
Using it in the equation A·v=λ·v, we have the following system of equations
4·v₁+8·v₂−4·v₃=−12·v₁
4·v₁−2·v₂−4·v₃=−12·v₂
−8·v₁−4·v₂−10·v₃=−12·v₃
Bringing everything to the left side, get this system
16·v₁+8·v₂−4·v₃=0
4·v₁+10·v₂−4·v₃=0
−8·v₁−4·v₂+2·v₃=0

As expected, the determinant of the coefficients of this system is zero and this system of equations is linearly dependent (third equation multiplied by −2 gives the first), as it should be, because otherwise the only solution would be a null-vector.
So, we drop the first equation and resolve the remaining two equations for v₂ and v₃ in terms of v₁.
Here is what remains
4·v₁+10·v₂−4·v₃=0
−8·v₁−4·v₂+2·v₃=0
Divide by 2 all of them to simplify and find v₃ from the second equation
2·v₁+5·v₂−2·v₃=0
v₃ = 4·v₁+2·v₂
Substituting v₃ into the first equation:
2·v₁+5·v₂−2·(4·v₁+2·v₂)=0
or
−6·v₁+v₂=0
Therefore,
v₂=6·v₁ and v₃ = 16·v₁

Regardless of the value of v₁, vector ||v₁,6·v₁,16·v₁|| is an eigenvector.
Set v₁=1 for simplicity, and vector ||1,6,16|| should be an eigenvector.

Let's check it out.

4 8 −4 4 −2 −4 −8 −4 −10

·

1 6 16

=

=

4+48−64 4−12−64 −8−24−160

=

−12 −72 −192

=

=

−12·

1 6 16

Which confirms that vector ||1,6,16|| (and any collinear to it) is an eigenvector with −12 as its eigenvalue.

Let's do the same calculations with the second eigenvalue λ₂=−6.
Using it in the equation A·v=λ·v, we have the following system of equations
4·v₁+8·v₂−4·v₃=−6·v₁
4·v₁−2·v₂−4·v₃=−6·v₂
−8·v₁−4·v₂−10·v₃=−6·v₃
Bringing everything to the left side, get this system
10·v₁+8·v₂−4·v₃=0
4·v₁+4·v₂−4·v₃=0
−8·v₁−4·v₂−4·v₃=0
Dividing the first equation by 2, the second - by 4 and the third - by −4, all these equations yield a simpler system
5·v₁+4·v₂−2·v₃=0
v₁+v₂−v₃=0
2·v₁+v₂+v₃=0

As expected, the determinant of the coefficients of this system is zero and the equations are linearly dependent (the third equation equals the first minus triple the second one), as it should be, because otherwise the only solution would be a null-vector.
So, we drop the first equation and resolve the remaining two equations for v₂ and v₃ in terms of v₁.
Here is what remains
v₁+v₂−v₃=0
2·v₁+v₂+v₃=0

Find v₃ from the first equation
v₃ = v₁+v₂
Substituting v₃ into the second equation:
2·v₁+v₂+(v₁+v₂)=0
or
3·v₁+2·v₂=0

Therefore,
v₂=−(3/2)·v₁ and v₃ = −(1/2)·v₁

Regardless of the value of v₁, vector ||v₁,−(3/2)·v₁,−(1/2)·v₁|| is an eigenvector.
Set v₁=2 for simplicity, and vector ||2,−3,−1|| should be an eigenvector.

Let's check it out.

4 8 −4 4 −2 −4 −8 −4 −10

·

2 −3 −1

=

=

8−24+4 8+6+4 −16+12+10

=

−12 18 6

=

=

−6·

2 −3 −1

Which confirms that vector ||2,−3,−1|| (and any collinear to it) is an eigenvector with −6 as its eigenvalue.

Finally, let's do the same for the third eigenvalue 10.
Using it in the equation A·v=λ·v, we have the following system of equations
4·v₁+8·v₂−4·v₃=10·v₁
4·v₁−2·v₂−4·v₃=10·v₂
−8·v₁−4·v₂−10·v₃=10·v₃
Bringing everything to the left side, get this system
−6·v₁+8·v₂−4·v₃=0
4·v₁−12·v₂−4·v₃=0
−8·v₁−4·v₂−20·v₃=0
Dividing the first equation by −2, the second - by 4 and the third - by −4, all these equations yield a simpler system
3·v₁−4·v₂+2·v₃=0
v₁−3·v₂−v₃=0
2·v₁+v₂+5·v₃=0

As expected, the determinant of the coefficients of this system is zero and the equations are linearly dependent (the first equation multiplied by 7 equals the first multiplied by 11 plus the third multiplied by 5), as it should be, because otherwise the only solution would be a null-vector.
So, we drop the first equation and resolve the remaining two equations for v₂ and v₃ in terms of v₁.
Here is what remains
v₁−3·v₂−v₃=0
2·v₁+v₂+5·v₃=0

Find v₃ from the first equation
v₃ = v₁−3·v₂
Substituting v₃ into the second equation:
2·v₁+v₂+5·(v₁−3·v₂)=0
or
7·v₁−14·v₂=0
or
v₁−2·v₂=0

Therefore,
v₂=(1/2)·v₁ and v₃ = −(1/2)·v₁

Regardless of the value of v₁, vector ||v₁,(1/2)·v₁,−(1/2)·v₁|| is an eigenvector.
Set v₁=2 for simplicity, and vector ||2,1,−1|| should be an eigenvector.

Let's check it out.

4 8 −4 4 −2 −4 −8 −4 −10

·

2 1 −1

=

=

8+8+4 8−2+4 −16−4+10

=

20 10 −10

=

=

10·

2 1 −1

Which confirms that vector ||2,1,−1|| (and any collinear to it) is an eigenvector with 10 as its eigenvalue.

Answer A

Matrix

4 8 −4 4 −2 −4 −8 −4 −10

has three eigenvalues:
−12, −6 and 10.
Their corresponding eigenvectors are:
||1,6,16||, ||2,−3,−1|| and ||2,1,−1||.
Of cause, any vector collinear to a particular eigenvector would also be an eigenvector with the same eigenvalue.