chapter 5: linear algebra applications

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Chapter 5: Linear Algebra Applications

I. Homogeneous Linear EquationsII. Non-homogeneous equationIII. Eigen value problem

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5.1 Hom

ogeneous linear Equations

N=3; The problem is to solve : There is an obvious solution:

But is it the only one?Consider the determinant By using the rules of determinants that we proved in chapter 4, we compute the following:

But the left column is zero by assumption of our linear system of equations, and thus. And thus if our starting homogeneous equations are to be satisfied we need: and likewise we could just as easily have gotten: and .

These equations are satisfied if x=y=z=0 (our obvious solution) but can also have a solution for non-zero (x,y,z)’s if det[a]=0Conclusion: Non-zero solutions to homogeneous linear equations are possible only if the matrix of the coefficients of the equations has a zero determinant.

5.1. Solution to homogeneous linear equations.

{𝑎1𝑥+𝑏1𝑦+𝑐1 𝑧=0𝑎2𝑥+𝑏2𝑦+𝑐2𝑧=0𝑎3𝑥+𝑏3 𝑦+𝑐3𝑧=0

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5.1 Hom

ogeneous linear Equations

5.1 Solution to homogeneous linear equations cont’d.

E5.1-1 Consider the system of equations

What are the possible solutions? If applicable, find the equation of the set of solutions.

E5.1-2 Consider the system of equations What are the possible solutions? If applicable, find the equation of the set of solutions.

{𝑥− 𝑦=0𝑥+𝑦=0

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5.2 Inhomogeneous Equations:Again for n=3 (but the generalization to any n is obvious) the problem is:

In matrix form we have: where: and

Assuming a non-zero determinant, we can use the inverse to solve for:

and thus we get the solution for immediately by:

In case the determinant of [a] is zero, the inverse of [a] does not exist. Since the determinant is zero, it means that the left hand sides of the equations are linearly dependent. For instance in our n=3 case above we could have: (1st row = * 2nd row)Then we have 2 possible cases:

• in which case the 2 proportional equations are consistent. and the system has an infinite number of solutions points. Since we have 3 unknowns (x,y,z) and only 2 independent equations the space of solution is a line (the intersection of 2 planes).

• Note; we could also have further dependence: say, the 3rd row is proportional to the second one. In that case there would be one independent equation (i.e. the 3 equations are proportional to each other) then the space of solutions would be would be a plane of equation, say, if equations 2 and 3 are consistent or, again, we have the possibility of no solutions if equations 2 and 3 are NOT consistent: for instance but

{𝑎1 𝑥+𝑏1 𝑦+𝑐1𝑧=d 1𝑎2𝑥+𝑏2 𝑦+𝑐2𝑧=d 2𝑎3𝑥+𝑏3 𝑦+𝑐3𝑧=d 3

E 5.2-2 : Consider the matrix Find and discuss solution(S) to


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5.2 Inhomogeneous Equations:


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5.3 Eigenvalue PROBLEM

PROBLEM STATEMENT: Given a matrix A, the problem is to find out if there exist vectors V such that: [5.3-1]Such vectors are called eigenvectors associated with the eigen value

Necessary Condition For The Existences Eigenvectors: If V exist then: [5.3-2] and thus: [5.3-3]5.3-1 prove this using known matrix operation: add. and mult. of matrices and mult. of matrix by numberFrom what we learned in 5.2, we know that in order to get a non-zero (i.e. =0) solution of [5.3-3], the determinant of matrix must be zero.

Requiring the determinant to be zero we get: det

This is called the characteristic equation. In n=3 this is a cubic equation in that will have at most three roots called the eigenvalues of the matrix. (for nxn determinants the char. Equation will have at most n eigenvalues.)For each eigenvalue we can find a corresponding eigenvector satisfying [5.3-1]. Note however that if is a eigenvector of the eigenvalue then so is the vector since . Thus if a vector is an eigenvector, then so will be all vectors proportional to it.

NOTE: Some eigenvalues can be of higher multiplicity. For instance, if the characteristic equation has a factor of the form (…..) then appears as a an eigenvalue of multiplicity 2.

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EXAMPLE: consider the matrix Lets compute e-values and e-vectors.

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SOME PROPERTIES of EIGENVECTORS and EIGENVALUES:

• Theorem: Eigenvectors associated with distinct eigenvalues form a linearly independent set.Proof (by contradiction): Let’s do it for n=3 For simplicity. Assume the set is linearly dependent. Thus we can write: [5.3-4] Now let’s apply the matrix a on the equation: since the vi’s are e-vectors:

Which implies that: and finally comparing this to [5.3-4] we conclude that and which contradicts our assumption that the e-values are different. And thus the vi’s cannot be linearly dependent.

• For eigenvalues of multiplicity s more than one (also called DEGENERATE eigenvalues), one can find at most s linearly independent eigenvectors.

• For matrices over complex numbers, the characteristic equations always n roots (although some could be degenerate.

• Notice that if =0 is an eigenvalue of a, then a is not invertible.Proof: since is an e-value then det and since =0, thus det (a) =0 as well; proving that a is not invertible.

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5.3 Eigenvalue PROBLEM: Diagonalization of matrices

An nxn matrix A is said to be diagonalizable if there exist a matrix P such that A=PDP-1 (or equivalently P-1AP=D) where D is a diagonal matrix. Theorem: P exists if and only if there exist n independent eigenvectors of A; Consider the n independent eigenvectors of A: in that case P is made up of the eigenvectors of A, entered as column coefficients. D is in addition found to be made up of the corresponding eigenvalues of AProof for n=3 (general case identical)

Consider the 3 independent eigenvectors of A: Let’s compute AP (notice that the elements of P are the e-vectors components in COLUMN form):

11 12 13 11 21 31 11 11 12 12 13 13 11 21 12 22 13 23 11 31 12 32 13 33

21 22 23 12 22 32 21 11 22 12 23 13 21 21 22 22 23 23 21 31 22 32 23 33

31 32 33 13 23 33 31

a a a v v v a v a v a v a v a v a v a v a v a v

a a a v v v a v a v a v a v a v a v a v a v a v

a a a v v v a v

11 32 12 33 13 31 21 32 22 33 23 31 31 32 32 33 33a v a v a v a v a v a v a v a v

Since in an eigenvector, we know that: and thus if we look, for instance, at the x component we get: and likewise for the other components of v1 as well as the other vectors; for instance: This means that we can rewrite the above product of matrices as:

1 11 2 21 3 31

1 12 2 22 3 32

1 13 2 23 3 33

v v v

v v v

v v v

31 21 32 22 33 2 2 2 22 23 3 (3rd row of )Ava v a a vv v v ��

E5.3-2 In the above is equal to which element of the matrix AP?

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Now let’s compute PD where D is the diagonal matrix of the eigenvalues arranged in the same order as their corresponding eigenvectors in the matrix P. We get:

11 21 31 1 1 11 2 21 3 31

12 22 32 2 1 12 2 22 3 32

13 23 33 3 1 13 2 23 3 33

0 0

0 0

0 0

v v v v v v

v v v v v v

v v v v v v

But that’s exactly the same answer as we got on the previous page computing AP!!!So we can conclude that: AP=PD and thus, if P is invertible (i.e. its det is non-zero and thus the eigenvector are linearly independent!), we finally obtain:

A=PDP-1

where P is the matrix of the eigenvectors of A in column placement and D is the diagonal matrix of the eigenvalues of A is in the same position as the eigenvectors in P:

11 21 31 1

12 22 32 2

13 23 33 3

0 0

0 0

0 0

v v v

P v v v D

v v v

If some of the eigenvalues are degenerate one can carry out the same procedure as long as one can find n linearly independent eigenvectors.In the next chapter we’ll look at a few examples of this procedure in concrete physical cases.

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Example of diagonalization of A=1 3 3

3 5 3

3 3 1

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E5.3-3 Consider the matrix: Find the characteristic equations, eigenvalues and

eigenvectors (normalized). Explain whether or not the matrix can be diagonalized, considering the eigen vectors. If it can, continue and find matrices P and D and verify that P does indeed diagonalize A.

E5.3-4 Repeat above steps for and then check with mathematica eigenvalues and eigenvectors.

E5.3-5 Consider the matrix of a 30 degree counterclockwise rotation in 2 dimensions. Can you find real eigenvalues/ eigenvectors? Comment on your result as it pertains to the diagonalization of a rotation matrix.

E5.3-6 Take an 4x4 matrix of your choice. Compute its eigenvectors and eigenvalues using mathematica. Still using mathematica compute P-1AP and verify that you indeed get the diagonal matrix of its corresponding eigenvalues. If the matrix yu found is not diagonalizable choose a different one and repeat calculation

2 1 0

1 2 1

0 1 2

1 1 1

0 2 1

0 0 3

chapter 5: linear algebra applications

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