finding eigenvalues, char poly

Announcements

Ï Quiz 4 will be on Thurs Feb 18 on sec 3.3, 5.1 and 5.2.

Ï "Curved" quiz grades (3 point curve as I mentioned in theemail) will be posted by today evening.

Ï I have updated the homework set (again!!). I decided to skip5.5 and start chap 6 after 5.3.

Last Week

1. De�ned eigenvalues and eigenvectors of a square matrix

2. How to test whether a given vector is an eigenvector of a givenmatrix (multiply and check whether we get a scalar multiple ofthe original vector)

3. How to test whether a given number λ is an eigenvalue (checkwhether the matrix A−λI has linearly dependent columns)

4. How to �nd the eigenvectors of a given eigenvalue (row reduceA−λI and solve for basic in terms of free variables).

5. Eigenvalues of any triangular matrix are the entries of themain diagonal

6. Zero is an eigenvalue of A if and only if A is not invertible.

7. We did not see how to �nd eigenvalues of a general squarematrix.

Last Week

How to �nd eigenvalues of any square matrix A?

Idea: An eigenvalue λ is a scalar such that the equation(A−λI )x= 0 has free variables. This means

1. The matrix A−λI is not invertible or

2. The determinant of the matrix A−λI is zero3. Solve the equation det(A−λI )= 0 for λ

4. For a 2×2 matrix it is easy, we get a quadratic equation, sofactorize and �nd λ.

5. For a 3×3 matrix, it is not always easy to solve for λ (exceptin some carefully constructed matrices)

6. For a 4×4 and larger matrices, use ofcalculator/software/approximate numerical schemes areprefered.

1. The matrix A−λI is not invertible or2. The determinant of the matrix A−λI is zero

3. Solve the equation det(A−λI )= 0 for λ

1. The matrix A−λI is not invertible or2. The determinant of the matrix A−λI is zero3. Solve the equation det(A−λI )= 0 for λ

Example 2, section 5.2

Find the eigenvalues of the matrix[5 33 5

Solution: We have to look at the determinant of the matrix[5 33 5

]−λ

[1 00 1

[5 33 5

[λ 00 λ

[5−λ 33 5−λ

In other words, form a matrix where you subtract λ from thediagonal elements (no change to the o� diagonal elements). This isthe case with any square matrix of any size.

]−λ

[1 00 1

[5 33 5

[λ 00 λ

[5−λ 33 5−λ

]−λ

[1 00 1

[5 33 5

[λ 00 λ

[5−λ 33 5−λ

Example 2, section 5.2Let us look at the determinant of the new matrix∣∣∣∣ 5−λ 3

3 5−λ∣∣∣∣= (5−λ)2−9.

Simplify this quantity. Some high school algebra will be handy.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

This quantity must be equal to zero. In other words,

λ2−10λ+16= 0.

Factorize this and we get,

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

are the 2 eigenvalues.

3 5−λ∣∣∣∣= (5−λ)2−9.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

λ2−10λ+16= 0.

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

3 5−λ∣∣∣∣= (5−λ)2−9.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

λ2−10λ+16= 0.

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

3 5−λ∣∣∣∣= (5−λ)2−9.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

λ2−10λ+16= 0.

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

3 5−λ∣∣∣∣= (5−λ)2−9.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

λ2−10λ+16= 0.

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

3 5−λ∣∣∣∣= (5−λ)2−9.

(5−λ)2 = 25−10λ+λ2.

(5−λ)2−9= 25−10λ+λ2−9= 16−10λ+λ2.

λ2−10λ+16= 0.

(λ−8)(λ−2)= 0.

Thusλ= 8,λ= 2

Comments

1. The equation det(A−λI )= 0 is called the characteristicequation of A.

2. The polynomial involving λ you get after computing thedeterminant (and simplifying) in step 1 is called thecharacteristic polynomial of A.

3. Thus the characteristic polynomial of A is a quadratic equationif A is a 2×2 matrix, a cubic equation if A is a 3×3 matrix etc.

4. The characteristic polynomial may not be nicely factorizable inall cases. In that case you may need your high schoolquadratic formula.

Comments

Back to High School

If you have a quadratic equation

ax2+bx +c = 0,

the 2 roots (values of x that solve the above equation) are given by

−b+pb2−4ac

and−b−

pb2−4ac

This works for any quadratic equation. If you cannot �gure out thefactorization immediately, this is a safe option. (provided you won'tmake mistakes of course).

Back to High School

If you have a quadratic equation

ax2+bx +c = 0,

the 2 roots (values of x that solve the above equation) are given by

−b+pb2−4ac

and−b−

pb2−4ac

This works for any quadratic equation. If you cannot �gure out thefactorization immediately, this is a safe option. (provided you won'tmake mistakes of course).

Trace of a Matrix

De�nition

Trace of any square matrix A is the sum of the diagonal elementsof A. It is denoted by tr A.

Why is the trace useful? The trace of any square matrix is equal tothe sum of the eigenvalues of A (irrespective of the size of A).Thus, it is a good check to your eigenvalue calculations.

The product of the eigenvalues of A = detA.

Trace of a Matrix

De�nition

Trace of a Matrix

De�nition

How many eigenvalues?

A quadratic equation has exactly 2 roots, if complex roots areincluded. (could be the same root repeated)

A cubic equation has exactly 3 roots, if complex roots are included(one or more roots could be repeated)

In general, a square matrix of size n×n will have exactly n

eigenvalues.

Example

Find the characteristic equation, characteristic polynomial and theeigenvalues of the matrix

5 3 7 90 5 9 10 0 2 80 0 0 2

Observe that this is a triangular matrix (and hence convenient towork with). The characteristic equation is found by solving theequation det(A−λI )= 0 which is∣∣∣∣∣∣∣∣∣

5−λ 3 7 90 5−λ 9 10 0 2−λ 80 0 0 2−λ

∣∣∣∣∣∣∣∣∣= 0.

Example

5 3 7 90 5 9 10 0 2 80 0 0 2

Observe that this is a triangular matrix (and hence convenient towork with). The characteristic equation is found by solving theequation det(A−λI )= 0 which is

∣∣∣∣∣∣∣∣∣5−λ 3 7 90 5−λ 9 10 0 2−λ 80 0 0 2−λ

∣∣∣∣∣∣∣∣∣= 0.

Example

5 3 7 90 5 9 10 0 2 80 0 0 2

Observe that this is a triangular matrix (and hence convenient towork with). The characteristic equation is found by solving theequation det(A−λI )= 0 which is∣∣∣∣∣∣∣∣∣

5−λ 3 7 90 5−λ 9 10 0 2−λ 80 0 0 2−λ

∣∣∣∣∣∣∣∣∣= 0.

Example Contd.

Since A−λI is a triangular matrix, we have

(5−λ)2︸︷︷︸(2−λ)2︸︷︷︸= 0

This is the characteristic equation. To �nd the char. polynomial,we have to expand both terms

(25−10λ+λ2)︸︷︷︸(4−4λ+λ2)︸︷︷︸=⇒ 100−100λ︸︷︷︸+︷︸︸︷

25λ2−40λ︸︷︷︸+︷︸︸︷40λ2−10λ3+

︷︸︸︷4λ2 −4λ3+λ4

=⇒ 100−140λ︸︷︷︸+︷︸︸︷69λ2−14λ3+λ4

Example Contd.

(5−λ)2︸︷︷︸(2−λ)2︸︷︷︸= 0

(25−10λ+λ2)︸︷︷︸(4−4λ+λ2)︸︷︷︸

=⇒ 100−100λ︸︷︷︸+︷︸︸︷25λ2−40λ︸︷︷︸+︷︸︸︷

40λ2−10λ3+︷︸︸︷4λ2 −4λ3+λ4

=⇒ 100−140λ︸︷︷︸+︷︸︸︷69λ2−14λ3+λ4

Example Contd.

(5−λ)2︸︷︷︸(2−λ)2︸︷︷︸= 0

(25−10λ+λ2)︸︷︷︸(4−4λ+λ2)︸︷︷︸=⇒ 100−100λ︸︷︷︸+︷︸︸︷

25λ2−40λ︸︷︷︸+︷︸︸︷40λ2−10λ3+

︷︸︸︷4λ2 −4λ3+λ4

=⇒ 100−140λ︸︷︷︸+︷︸︸︷69λ2−14λ3+λ4

Example Contd.

(5−λ)2︸︷︷︸(2−λ)2︸︷︷︸= 0

(25−10λ+λ2)︸︷︷︸(4−4λ+λ2)︸︷︷︸=⇒ 100−100λ︸︷︷︸+︷︸︸︷

25λ2−40λ︸︷︷︸+︷︸︸︷40λ2−10λ3+

︷︸︸︷4λ2 −4λ3+λ4

=⇒ 100−140λ︸︷︷︸+︷︸︸︷69λ2−14λ3+λ4

Example Contd.

What about the eigenvalues of A? Since A is triangular, thediagonal entries are its eigenvalues.

Here the eigenvalues are 5, 5, 2 and 2.

We say that 5 has multiplicity 2 and 2 has multiplicity 2. In otherwords, we have repeated eigenvalues.

Example Contd.

Example

The characteristic polynomial of a 5×5 matrix is

λ5−4λ4−12λ3.

Find the eigenvalues and their multiplicities.

Solution: This polynomial (in spite of being of degree 5) can befactorized easily as follows.

λ3(λ2−4λ−12)= λ3(λ−6)(λ+2).

Thus we have λ3 = 0=⇒ λ= 0, λ= 6 and λ=−2.

Zero has multiplicity 3 (repeated roots), others have multiplicity 1each.

Example

λ5−4λ4−12λ3.

λ3(λ2−4λ−12)= λ3(λ−6)(λ+2).

Example

λ5−4λ4−12λ3.

λ3(λ2−4λ−12)= λ3(λ−6)(λ+2).

Example

λ5−4λ4−12λ3.

λ3(λ2−4λ−12)= λ3(λ−6)(λ+2).

Find the char. polynomial and eigenvalues of the matrix[5 −3−4 3

Solution: The characteristic equation is found by solving theequation det(A−λI )= 0 which is∣∣∣∣ 5−λ −3

−4 3−λ∣∣∣∣= 0.

(5−λ)(3−λ)−12= 0

15−8λ+λ2−12= 0

3−8λ+λ2 = 0=⇒ λ2−8λ+3= 0

λ2−8λ+3 is the char.polynomial.

−4 3−λ∣∣∣∣= 0.

(5−λ)(3−λ)−12= 0

15−8λ+λ2−12= 0

3−8λ+λ2 = 0=⇒ λ2−8λ+3= 0

−4 3−λ∣∣∣∣= 0.

(5−λ)(3−λ)−12= 0

15−8λ+λ2−12= 0

3−8λ+λ2 = 0=⇒ λ2−8λ+3= 0

−4 3−λ∣∣∣∣= 0.

(5−λ)(3−λ)−12= 0

15−8λ+λ2−12= 0

3−8λ+λ2 = 0=⇒ λ2−8λ+3= 0

−4 3−λ∣∣∣∣= 0.

(5−λ)(3−λ)−12= 0

15−8λ+λ2−12= 0

3−8λ+λ2 = 0=⇒ λ2−8λ+3= 0

Use the quadratic formula (factorization will not work here)

λ=8±

√82−4(1)(3)

2(1)= 8±p

Since 52=4× 13 we have

λ= 8±p4.13

2= 8±2

2= 4±

The sum of these eigenvalues is

4+p13+4−

p13= 8

and the trace of the given matrix is 5+3=8.

λ=8±

√82−4(1)(3)

2(1)= 8±p

λ= 8±p4.13

2= 8±2

2= 4±

4+p13+4−

p13= 8

λ=8±

√82−4(1)(3)

2(1)= 8±p

λ= 8±p4.13

2= 8±2

2= 4±

4+p13+4−

p13= 8

Example 8, section 5.2Find the char. polynomial and eigenvalues of the matrix[

7 −22 3

2 3−λ∣∣∣∣= 0.

(7−λ)(3−λ)+4= 0

21−10λ+λ2+4= 0

25−10λ+λ2 = 0=⇒ λ2−10λ+25= 0

λ2−10λ+25 is the char.polynomial. What about eigenvalues? Tryit yourself!!!!

7 −22 3

2 3−λ∣∣∣∣= 0.

(7−λ)(3−λ)+4= 0

21−10λ+λ2+4= 0

25−10λ+λ2 = 0=⇒ λ2−10λ+25= 0

7 −22 3

2 3−λ∣∣∣∣= 0.

(7−λ)(3−λ)+4= 0

21−10λ+λ2+4= 0

25−10λ+λ2 = 0=⇒ λ2−10λ+25= 0

7 −22 3

2 3−λ∣∣∣∣= 0.

(7−λ)(3−λ)+4= 0

21−10λ+λ2+4= 0

25−10λ+λ2 = 0=⇒ λ2−10λ+25= 0

Find the char. polynomial of the matrix 0 3 13 0 21 2 0

For a 3×3 matrix, �nding the char equation (and the charpolynomial) is more involved.You could do a cofactor expansion or the Sarru's mnemonic rule(where you repeat the �rst 2 rows and strike through) of A−λI .

NEVER do a reduction to echeleon form. The eigenvalues of theechelon form are totally di�erent from that of the original matrix.

Solution: The characteristic equation is found by solving theequation det(A−λI )= 0 which is∣∣∣∣∣∣

0−λ 3 13 0−λ 21 2 0−λ

∣∣∣∣∣∣= 0.

−λ∣∣∣∣ −λ 2

2 −λ∣∣∣∣︸︷︷︸

λ2−4

−3∣∣∣∣ 3 21 −λ

∣∣∣∣︸︷︷︸−3λ−2

+1∣∣∣∣ 3 −λ1 2

∣∣∣∣︸︷︷︸6+λ

−λ(λ2−4)−3(−3λ−2)+6+λ= 0

−λ3+4λ+9λ+6+6+λ= 0=⇒−λ3+14λ+12= 0

−λ3+14λ+12 is the char.polynomial.

0−λ 3 13 0−λ 21 2 0−λ

∣∣∣∣∣∣= 0.

−λ∣∣∣∣ −λ 2

2 −λ∣∣∣∣︸︷︷︸

λ2−4

−3∣∣∣∣ 3 21 −λ

∣∣∣∣︸︷︷︸−3λ−2

+1∣∣∣∣ 3 −λ1 2

∣∣∣∣︸︷︷︸6+λ

−λ(λ2−4)−3(−3λ−2)+6+λ= 0

−λ3+4λ+9λ+6+6+λ= 0=⇒−λ3+14λ+12= 0

0−λ 3 13 0−λ 21 2 0−λ

∣∣∣∣∣∣= 0.

−λ∣∣∣∣ −λ 2

2 −λ∣∣∣∣︸︷︷︸

λ2−4

−3∣∣∣∣ 3 21 −λ

∣∣∣∣︸︷︷︸−3λ−2

+1∣∣∣∣ 3 −λ1 2

∣∣∣∣︸︷︷︸6+λ

−λ(λ2−4)−3(−3λ−2)+6+λ= 0

−λ3+4λ+9λ+6+6+λ= 0=⇒−λ3+14λ+12= 0

0−λ 3 13 0−λ 21 2 0−λ

∣∣∣∣∣∣= 0.

−λ∣∣∣∣ −λ 2

2 −λ∣∣∣∣︸︷︷︸

λ2−4

−3∣∣∣∣ 3 21 −λ

∣∣∣∣︸︷︷︸−3λ−2

+1∣∣∣∣ 3 −λ1 2

∣∣∣∣︸︷︷︸6+λ

−λ(λ2−4)−3(−3λ−2)+6+λ= 0

−λ3+4λ+9λ+6+6+λ= 0=⇒−λ3+14λ+12= 0

Example 14, section 5.2Find the char. polynomial of the matrix 5 −2 3

0 1 06 7 −2

5−λ −2 30 1−λ 06 7 −2−λ

∣∣∣∣∣∣= 0.

Expand along second row, we get

(1−λ)

∣∣∣∣ 5−λ 36 −2−λ

∣∣∣∣︸︷︷︸(5−λ)(−2−λ)−18=−10−3λ+λ2−18=λ2−3λ−28

Example 14, section 5.2Find the char. polynomial of the matrix 5 −2 3

0 1 06 7 −2

5−λ −2 30 1−λ 06 7 −2−λ

∣∣∣∣∣∣= 0.

Expand along second row, we get

(1−λ)

∣∣∣∣ 5−λ 36 −2−λ

∣∣∣∣︸︷︷︸(5−λ)(−2−λ)−18=−10−3λ+λ2−18=λ2−3λ−28

(1−λ)(λ2−3λ−28)= 0

λ2−3λ−28−λ3+3λ2+28λ= 0

4λ2+25λ−28−λ3 = 0

The char polynomial is thus 4λ2+25λ−28−λ3.

(1−λ)(λ2−3λ−28)= 0

λ2−3λ−28−λ3+3λ2+28λ= 0

4λ2+25λ−28−λ3 = 0

(1−λ)(λ2−3λ−28)= 0

λ2−3λ−28−λ3+3λ2+28λ= 0

4λ2+25λ−28−λ3 = 0

Similarity

De�nition

Let A and B be 2 square matrices. We say that A is similar to B ifwe can �nd an invertible matrix P such that

P−1AP =B

Theorem

If n×n matrices A and B are similar, they have the same char

polynomial and hence the same eigenvalues (with same

multiplicities).

finding eigenvalues, char poly

Education

nd eigenvalues

eigenvalue checkwhether

dened eigenvalues

given eigenvalue row

matrix ai

given number

given vector

triangular matrix