LetA:Rn→Rn be a linear operator. It is important to know vectors of the space that are mapped to a parallel vector by a linear operator. A numberλ∈R is aneigenvalueof A, if there exists a non-zero n-vectorv such thatA(v) =λv. In this case, v is an eigenvector that belongs toλ. We can collect all eigenvectors belonging to the same eigenvalueλin a set H(λ). That is,H(λ) ={v∈Rn :A(v) =λv}.
Notice that here the zero vector ofRn is also included. It is not difficult to prove thatH(λ) is a vector space called the eigenspace ofλ. The geometric multiplicity ofλis simply the dimension of H(λ).
Another property that we can deduce from the definitions is that eigenvectors that belong to different eigenvalues are linearly independent. It implies that a linear operatorA:Rn→Rn can have at mostn different eigenvalues.
As we learnt in the previous Section, every linear transformation can be represented by a matrix multiplication. Therefore, it is immediate that we can define eigenvalues and eigenvectors of a matrix as follows:
Let M be an n×n matrix. The number λ∈ R is aneigenvalue of M if and only if there exists a non-zeron-vectorv such thatM v=λv. In that casev is aneigenvectorthat belongs toλ.
Observe that M v =λv can be written as M v−λv = (M −λIn)v = 0. We deduce the following equivalences buildt on our previous theorems: The vectorv is an eigenvector belonging to λif and only if the homogeneous system of linear equations (M −λIn)x= 0 has a non-zero solution v. This latter happens if and only if det(M −λIn) = 0.
It follows from the definition of the determinant that det(M −λIn) is a polynomial inλof degree at mostn. Therefore, ifAis ann×nmatrix, thenp(λ) = det(A−λIn) is thecharacteristic polynomialofA andp(λ) = 0 is thecharacteristic equation. The roots of the characteristic equation are the eigenvalues of A. If the rootλi is a multiple root of multiplicitym, then thealgebraic multiplicityofλi ism. It follows from the Fundamental Theorem of Algebra thatp(λ) can have at mostnreal roots. Another fact is that the geometric multiplicity of an eigenvalue is always smaller than or equal to the algebraic multiplicity.
Building on the results of this Section, we can use the following recipe to determine the eigenspaces of a given linear operator inRn.
1. We determine the transformation matrixA.
2. We write the characteristic equation det(A−λIn) = 0.
3. Solvingp(λ) = 0, we get the eigenvalues λ1, . . . and their algebraic multiplicities.
4. We substitute each eigenvalueλi separately into the matrixA−λiIn.
5. We solve the homogeneous system of linear equations (A−λiIn)x = 0. The solutions are the eigenvectors inH(λi). Its dimension is the geometric multiplicity ofλi.
Example 8.1. Let A : R2 → R2, (x1, x2) 7→ (x1+ 3x2,2x1−4x2). Determine the eigenvalues and eigenspaces ofA. Calculate the algebraic and geometric multiplicity of each eigenvalue.
Solution: We determine the transformation matrix by calculating the images of the standard basis vectors ofR2. We write the images as column vectors of the transformation matrix.
A=
1 3 2 −4
.
Next we write the characteristic polynomial: det(A−λIn) = (1−λ)(−4−λ)−2∗3 =λ2+ 3λ−10.
We have to find the roots of this polynomial. Sinceλ2+ 3λ−10 = (λ+ 5)(λ−2), the roots are−5 and 2. We deduce the algebraic multiplicity of each eigenvalue is 1.
Next we substitute −5 in place of λ in A−λIn to get
6 3 2 1
. Therefore we want the non-trivial solutions of the system
6 3 2 1
x = 0. We see that 2x1 +x2 = 0 must hold. Therefore, H(−5) ={(x1, x2) :x2=−2x1, x1∈R}. We deduce the dimension ofH(−5) is 1, since there was 1 free variable. This is the geometric multiplicity of the eigenvalue−5.
Now we substitute 2 in place of λ in A−λIn to get
−1 3 2 −6
. Therefore we want the non-trivial solutions of the system
−1 3 2 −6
x= 0. We see that −x1+ 3x3 = 0 must hold. Therefore,
H(2) = {(x1, x2) : x1 = 3x2, x2 ∈ R}. We deduce the dimension of H(2) is 1, since there was 1 free variable. This is the geometric multiplicity of the eigenvalue 2.
Excercises
1. Determine the eigenvalues and eigenspaces of the following linear transformation! Calculate the algebraic and geometric multiplicity of each eigenvalue.
A:R2→R2, (x1, x2)7→(4x1+ 2x2,3x1−x2).
2. Determine the eigenvalues and eigenspaces of the following linear transformation! Calculate the algebraic and geometric multiplicity of each eigenvalue.
A:R2→R2, (x1, x2)7→(5x1−2x2,−4x1+ 3x2).
3. Determine the eigenvalues and eigenspaces of the following linear transformation! Calculate the algebraic and geometric multiplicity of each eigenvalue.
A:R2→R2, (x1, x2)7→(−3x1+ 2x2,−4x1+ 6x2).
4. Determine the eigenvalues and eigenspaces of the following linear transformation! Calculate the algebraic and geometric multiplicity of each eigenvalue.
A:R3→R3, (x1, x2, x3)7→(3x1,−x1+x2,−3x1−2x2+ 3x3).
5. True or false?
A linear operator can have an eigenvalue for which there is only one associated eigenvector.
A linear operatorA:Rn→Rn can have at mostndifferent eigenvectors.
Solutions
their scalar product is zero.(f) |v|v = (0,√−1
3. Only with the trivial combination 0a+ 0b+ 0c. ThereforeH is linearly independent.
4. Yes. The coordinates ofvwith respect toa1,a2,a3 are 1,2,3.
5. Only with the trivial combination 0a+ 0b+ 0c. ThereforeH is linearly independent. No.
6. The two vectors of H1 are independent. Any basis of R3 has 3 elements. Therefore,H1 is not a basis and not a generating set either.
The vectors ofH2 are independent, they form a basis and therefore a generating set as well.
SinceH3 containsH2, thereforeH3is a generating set, but not a basis. ThereforeH3 is dependent.
7. The rankr(H) = 2. Use the linear combination of the two new basis vectors to add a vector that is dependent on them.
8. Hint: Show, using the basis transformation process, that the union of the basis of V1 and V2 is a basis ofR3. v1= (4,4,4) andv2= (−3,6,−2).
2. matrixX, other than the identity can have the property thatAX=A.
3. AB=
5. Matrix multiplication is associative. A(BC) = (AB)C= −179 . 7. The determinant is 0, hence Ainverse does not exist.
8. A−1= 12
10. B−1= 191
11 −3 −5
−14 9 15 10 −1 −8
11. We only define inverse of a square matrix, soAT does not have an inverse.
B−1= 18
3 1 −2
−1 −3 6
−2 2 4
Section 5
1. Positive, positive.
2. In the calculation of a 5×5 determinant, we have to multiply 5 terms in each summand such that the terms lie in different rows and columns. In our case, there are two rows and two columns consisting of non-zero elements. If we select 5 items, one in each row and column, then one of them must be a 0.
Therefore each summand is 0, hence the determinant is 0 as well.
3. In a 4×4 determinant, we have to calculate (4! times) the product of 4 entries that lie in different columns and rows. In our case, the determinant will be a polynomial of degree at most 4. There is only one quadruple of entries such that each of them containsx. Therefore, the coefficient ofx4 is 2.
Notice the following: If we select a number from the matrix, then thex’s in its row and column cannot be used for the product of four entries. Therefore, the only way to get a product that contains threex’s is to select the 1 in the second row and first column. Now, we have to decide its sign! It is negative. So the coefficient ofx3 is -1.
4. Use the elementary row operations for matrices! Try to eliminate like in Gauss-Jordan and factor out common terms.
5. Same as in the previous exercise.
6. -18, 0, 160, 48, 1, -33, 230.
7. abc−ab+a−1.
8. -696, 62, -174.
9. 2a−b−c−d.
Section 5.1
1. x1= 2,x2= 5,x3=−3.
2. x1= 0,x0= 0,x3= 1.
3. x1= 1,x2= 2,x3=−1,x4=−2.
4. x1=−2,x2= 2, x3=−3,x4= 3.
Section 6
1. The parametric equations are x = 3 + 4t, y = 1 + 5t, z = −4. The symmetric equations are:
x−3
4 =y−15 andz=−4.
2. The parametric equations are x= 1 + 2t, y = 4 + 2t, z = 5−6t. The symmetric equations are:
x−1
2 =y−42 = z−5−6.
3. For the linee: we find a pointP = (2,−1,5) and the direction vectorv= (3,2,−4). The symmetric equations are x−23 =y+12 =z−5−4.
For the line f: we find a pointP = (0,−2,4) and the direction vectorv = (5,7,0). The symmetric equations are x5 = y+27 andz= 4.
For the line g: we find a point P = (6,1,0) and the direction vector v = (0,3,0). There is no symmetric equation and the linex= 6,z= 0 is parallel to they-axis.
4. The parametric equation is: x= 3−5t,y= 4 andz=−1 + 7t.
5. The parametric equation is: x= 1 + 2t,y=−1 + 3tandz= 4 +t.
6. The normal vector is (2,−3,5). For instance the following points belong toπ: (1,1,6/5), (0,0,1), (1,−1,0). We see by substitution that (−8,3,6) belongs to the plane, and (1,4,−3) does not.
7. The equation of the plane is: 2x+ 3y−z+ 3 = 0.
8. The equation of the plane is: 2x+y+ 3z−9 = 0.
9. The equation of the plane is: −30x+ 76y−50z+ 292 = 0.
10. The equation of the plane is: 31x−59y−6z= 101.
Section 6.1
1. The lines eand f intersect each other in point (0,2,−2).
The lineseandg, and similarly the linesf andg do not intersect.
2. They intersect in a point (−3,−3,0).
3. They intersect in a point (−9,−18,−1).
4. The two planes have a common point, for instance (6,5,2/3) is a solution of both equations. We get the direction of the intersection line using the vector product: (2,−5,1)×(−3,1,−2) = (9,1,−13).
Therefore the parametric equations of the intersection line is the following: x= 6 + 9t, y = 5 +t and z= 2/3−13t.
5. The two planes have a common point, for instance (2,1,0). We get the direction of the intersection line using the vector product: (−5,1,8)×(2,5,8) = (−32,56,−27). Therefore the parametric equations of the intersection line is the following: x= 2−32t, y= 1 + 56tandz=−27t.
Section 7
1. The map is A :R3→ R3, (x, y, z)7→(x, y,0). It is easy to check the additive and homogeneous property.
The kernel consists of vectors with first two components being zero. ker(A) = {(x, y, z)∈R3 :x= y= 0, z∈R}. It has dimension 1.
Any vector in thex−y coordinate plane can be the image of some vector in 3D. Therefore,Im(A) = {v∈R3:v3= 0}.
The transformation matrix is M(A) =
2. The map Ais linear. Its transformation matrix isM(A) =
2 3 0 1 1 −3
. The mapB is non-linear.
The mapC is linear again. Its transformation matrix isM(C) =
(c) Only B◦Aexists. Its transformation matrix isM(A)M(B) =
2 16 10
ForB, we get a contradiction using any method solving the equations, so (3,4,6) does not belong to the range.
For C, we get that vectors (x1, x2, x3, x4) satisfying x1 = 2−x3−x4 and x2 = 1−x3−2x4 are mapped to (2,4,3).
Section 8
1. The eigenvalues are 5 and−2, both have algebraic multiplicity 1. The eigenspaceH(5) ={(x1, x2) : x1 = 2x2, x2 ∈ R}. The eigenspace H(−2) = {(x1, x2) : x2 = −3x1, x1 ∈ R}. Therefore both eigenvalues have geometric multiplicity 1.
2. The eigenvalues are 1 and 7, both have algebraic multiplicity 1. The eigenspaceH(1) ={(x1, x2) : x2 = 2x1, x1 ∈ R}. The eigenspace H(7) = {(x1, x2) : x2 = −x1, x1 ∈ R}. Therefore both eigenvalues have geometric multiplicity 1.
3. The eigenvalues are 5 and−2, both have algebraic multiplicity 1. The eigenspaceH(5) ={(x1, x2) : x2 = 4x1, x1 ∈ R}. The eigenspace H(−2) = {(x1, x2) : x1 = 2x2, x2 ∈ R}. Therefore both eigenvalues have geometric multiplicity 1.
4. The eigenvalues are 1 and 3. The eigenvalue 3 has algebraic multiplicity 2, the eigenvalue 1 has algebraic multiplicity 1. The eigenspace H(1) = {(x1, x2, x3) : x1 = 0, x3 = x2, x2 ∈ R}. The eigenspace H(3) ={(x1, x2, x3) : x1 =x2 = 0, x3 ∈ R}. Therefore both eigenvalues have geometric multiplicity 1.
5. False, if there is one eigenvector, then all its scalar multiples are eigenvectors too.
False, this is true for eigenvalues not for eigenvectors.