Successfully reported this slideshow.
Upcoming SlideShare
×

# Notes

1,177 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### Notes

1. 1. Don't Panic! (Here's how you should use these notes. The next page has a list of terms that you should be familiar with so you should look at the denition of each term and write down some concrete examples. My examples should help you get started but if they are confusing just ignore them and focus on understanding the denitions by coming up with some of your own examples. p.s. I got a little tired towards the end so some examples are missing. Please forward your comments and complaints to DaveCo Inc. PO Box 1.)
2. 2. Keywords: vector, vector space, subspace, linear combination, vector addition, scalar multi- plication, spanning set, linearly independent, linearly dependent, Wronskian, basis, dimension, rowspace, colspace, nullspace, rank, nullity, inner product, norm, orthogonal complement, Gram- Schmidt process, linear map, domain, codomain, kernel, range, one-to-one, onto, isomorphism, matrix representation, eigenvalue, eigenvector, characteristic polynomial, eigenspace, geometric multiplicity, algebraic multiplicity, defective, non-defective. ⇑Vector spaces: These are the important examples to keep in mind Rn , Cn , Mn (R), Pn , rowspace(A), colspace(A), nullspace(A), W ⊥ , span{v1 , . . . , vk } ................................................................................................ Advice: I strongly suggest you attempt problems 12-16 on pages 249-250. I would start with problem 16 and work backwards because the later problems are easier. ⇑Subspaces: Verifying that some set V satises the axioms of a vector space is a lot of work but once we know V is a vector space it is really easy to verify if some subset W ⊆ V of vectors is also a vector space. Think of Russian Matroska dolls, the big doll V contains a smaller doll W inside. Subspace Criterion i) Closure under addition ii) Closure under scalar multiplication ................................................................................................ Example : Let V = Mn (R), the set of n × n matrices with real entries, and let W = {A ∈ V | A = A}, the set of symmetric matrices. To verify that W is a vector space T we just need to verify i) A, B ∈ W ⇒ A + B ∈ W and ii) s ∈ R, A ∈ W ⇒ sA ∈ W . Both conditions are easy to verify. ? ? A, B ∈ W ⇒ A + B ∈ W s ∈ R, A ∈ W ⇒ sA ∈ W ? ? (A + B)T = A+B (sA)T = sA (A + B)T = AT + B T (sA)T = sAT = A+B = sA A, B ∈ W ⇒ A + B ∈ W s ∈ R, A ∈ W ⇒ sA ∈ W ................................................................................................ Example : Let V = Mn (R), the set of n × n matrices with real entries, and let W = {A ∈ V | det(A) = 0}, the set of matrices with 0 determinant. In this case W fails to [ ] [ ] 1 0 0 0 be closed under addition so it can not be a subspace. Let A = and B = then 0 0 0 1
3. 3. [ ] 1 0 both A and B are in W because they have determinant 0 but A + B = has determinant 0 1 1 and so is not in W . ⇑Spanning sets: There are only two operations we can perform in a vector space, i) scalar multiplication and ii) vector addition. So the most general way of combining a set of vectors is to scalar multiply them and add, i.e. take v 1 , v2 , . . . , vn and form c1 v 1 +. . .+cn v n for arbitrary scalars c1 , . . . , cn . This is called taking a linear combination. The scalars can be real or complex numbers but let's assume the scalars are real. When we take a set of vectors {v1 , . . . vk } and consider all possible linear combinations of those vectors we call that set the span of {v1 , . . . vk } and denote it by span{v1 , . . . vk } = {c1 v1 + . . . + ck vk | c1 , . . . , ck ∈ R}. Spanning sets are useful because vector spaces contain lots of vectors but often times we need very few vectors to recover the entire vector space. Note that the span of a set of vectors is always a subspace because it is closed under addition and scalar multiplication. ................................................................................................ Example : Let V = R3 and consider S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. It is not hard to see that every vector in R3 can be written as a linear combination of vectors from S . So we say that S spans R3 and write that as R3 = span{(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Note that S is not the only spanning set for R3 , {(1, 1, 1), (1, 1, 0), (1, 0, 0)} and {(1, 2, 3), (1, 1, 0), (2, 0, 0), (1, 1, 1)} would work as well but it is a little more work to see that these are spanning sets for R3 . ................................................................................................ [ ] [ ] [ ] [ ] 1 0 0 0 0 0 0 1 Example : Let V = M2 (R) and S = { , , , }, then S is a span- 0 0 1 0 0 1 0 0 ning set for M2 (R), i.e. every 2 × 2 matrix can be written as a linear combination of matrices from S. [ ] [ ] [ ] [ ] [ ] a b 1 0 0 0 0 0 0 1 =a +c +d +b c d 0 0 1 0 0 1 0 0 ................................................................................................ Non-example : Let V = P2 and let S = {1 + 3x, x + x2 }. If we look at the span of S it might not be obvious that it is not equal to P2 . So let's see what the span of S is explicitly span{1 + 3x, x + x2 } = {p ∈ P2 | p(x) = c1 (1 + 3x) + c2 (x + x2 ), c1 , c2 ∈ R} = {p ∈ P2 | p(x) = c1 + (3c1 + c2 )x + c2 x2 , c1 , c2 ∈ R} Now consider p(x) = 1 + x + x2 ∈ P2 and note that p(x) is not in span{1 + 3x, x + x2 } because if it was we would be able to nd c1 and c2 such that p(x) = 1 + x + x2 = c1 (1 + 3x) + c2 (x + x2 ) = c1 + (3c1 + c2 )x + c2 x2 but this is not possible. Comparing terms we get the equations c1 = 1, c2 = 1 and 3c1 + c2 = 1 but this system is incosistent so there is no way to write 1 + x + x2 as a linear combination of 1 + 3x and x + x2 .
4. 4. ⇑Linear independence/dependence: Some sets of vectors have special properties and one such special property is that of linear independence. A set of vectors {v1 , . . . , vk } is said to be linearly independent if c1 v1 + . . . + ck vk = 0 ⇒ c1 = c2 = . . . = ck = 0, i.e. the only way to get the zero vector from v1 , . . . , vk as a linear combination is to make sure all the coecients are 0. Otherwise, we can nd non-zero coecients and get the zero vector as a linear combination of v1 , . . . , vk and in this case we say {v1 , . . . , vk } is linearly dependent. ................................................................................................ Example : The set {(1, 1), (0, 2)} is linearly independent because if c1 (1, 1) + c2 (0, 2) = (0, 0) then c1 and c2 must be 0. To show this we just need to perform a small computation. c1 (1, 1) + c2 (0, 2) = (0, 0) ⇔ (c1 , c1 ) + (0, 2c2 ) = (0, 0) ⇔ (c1 , c1 + 2c2 ) = (0, 0) ⇔ c1 = 0, c1 + 2c2 = 0 ⇔ c1 = 0, c2 = 0 ................................................................................................ Non-example : The set {(1, 1), (0, 2), (1, 2)} is linearly dependent because 1 1 (1, 1) + (0, 2) = (1, 2) ⇔ (1, 1) + (0, 2) − (1, 2) = (0, 0) 2 2 In general saying a set is linearly dependent is another way of saying some vectors are a linear combination of other vectors from the same set, just like in this example. ................................................................................................ Non-example : Lets nd out the status of the set {(−1, 1, 2), (0, 2, −1), (3, 1, 2), (−1, −1, 1)}. c1 (−1, 1, 2) + c2 (0, 2, −1) + c3 (3, 1, 2) + c4 (−1, −1, 1) = (0, 0, 0) (−c1 , c1 , 2c1 ) + (0, 2c2 , −c2 ) + (3c3 , c3 , 2c3 ) + (−c4 , −c4 , c4 ) = (0, 0, 0) (−c1 + 3c3 − c4 , c1 + 2c2 + c3 − c4 , 2c1 − c2 + 2c3 + c4 ) = (0, 0, 0)     c1   −c1 + 3c3 − c4 = 0 −1 0 3 −1   0   c2    c1 + 2c2 + c3 − c4 = 0 ⇔  1 2 1 −1    = 0    c3  2c1 − c2 + 2c3 + c4 = 0 2 −1 2 1 0 c4     −1 0 3 −1 0 1 0 0 2/5 0 c1 + (2/5)c4 = 0   ∗    1 2 1 −1 0  →  0 1 0 −3/5 0  ⇔ c2 − (3/5)c4 = 0 2 −1 2 1 0 0 0 1 −1/5 0 c3 − (1/5)c4 = 0 The above calculations show that c4 is a free variables so we can choose it to be some non-zero number. In this case c4 = 5 is a good choice because then we get c1 = −2, c2 = 3, c3 = 1 and it is easy to check that −2(−1, 1, 2) + 3(0, 2, −1) + (3, 1, 2) + 5(−1, −1, 1) = (0, 0, 0)
5. 5. which means that the set of vectors we started with was linearly dependent. Note that this isn't the only linear relation because we could have chosen c4 to be any non-zero number. ⇑Wronskian: Suppose we have some dierentiable functions f1 , . . . , fn on some interval I and we would like to know if they are linearly independent on that interval then there is a test that uses determinants that will give us an answer in many cases. The Wronskian of a set of functions   f1 f2 ··· fn    f1 f2 ··· fn  {f1 , . . . , fn }, denoted by W [f1 , . . . , fn ], is the determinant of the matrix   . . . . .   . . . . . . . .  (n−1) (n−1) (n−1) f1 f2 ··· fn If we have a set of functions {f1 , . . . , fn }, dened and dierentiable on an interval I , and if W [f1 , . . . , fn ] = 0 at some point in the interval then {f1 , . . . , fn } is linearly independent on I . Note that this does not say if W [f1 , . . . , fn ] = 0 then {f1 , . . . , fn } is linearly dependent. The test only works if W [f1 , . . . , fn ] = 0. ................................................................................................ Example: Consider the functions f1 (x) = sin x and f2 (x) = cos x on the interval [−π, π]. Using the Wronskian we can nd out if {f1 , f2 } is linearly independent. sin x cos x W [f1 , f2 ] = W [sin, cos] = = − sin2 x − cos2 x = −1 = 0 cos x − sin x Since W [sin, cos] is always -1 it is non-zero at every point of [−π, π] so {sin x, cos x} is a linearly independent set of functions on [−π, π]. ⇑Basis/Dimension: A set of vectors {v1 , . . . , vk } in a vector space V is called a basis of V if it is linearly independent and spans V . The number of vectors in a basis for a vector space is called the dimension of V . ................................................................................................ Example : {(1, 0), (0, 1)} is a basis of R2 and it has 2 elements so the dimension of R2 is 2. Note, any other basis of R2 will also have exactly 2 linearly independent vectors. ................................................................................................ Non-example : {(1, 1, 1), (1, 2, 2), (0, 1, 1)} is not a basis of R3 because it is linearly dependent, i.e. (1, 2, 2) = (1, 1, 1) + (0, 1, 1) ⇔ (1, 2, 2) − (1, 1, 1) − (0, 1, 1) = (0, 0, 0) ................................................................................................ Non-example : {(1, 0, 0), (0, 1, 0)} is not a basis of R3 because it does not span R3 , i.e. there is a vector in R3 that is not a linear combination of (1, 0, 0) and (0, 1, 0). Take the vector (0, 0, 10) ∈ R3
6. 6. for example and you can see that there is no way to write it as a linear combination of (1, 0, 0) and (0, 1, 0), i.e. there are no scalars c1 and c2 such that c1 (1, 0, 0) + c2 (0, 1, 0) = (0, 0, 10) ................................................................................................ Example : {(1, 1, 1), (0, 1, 1), (0, 0, 1)} is a basis of R3 because it is linearly independent and any vector (x, y, z) ∈ R3 can be written as a linear combination of (1, 1, 1), (0, 1, 1) and (0, 0, 1). To see why just write down the equation c1 (1, 1, 1) + c2 (0, 1, 1) + c3 (0, 0, 1) = (x, y, z) in matrix form as      1 0 0 c1 x       1 1 0   c2  =  y  1 1 1 c3 z and solve for c1 , c2 , c3 . You should get the solution c1 = x, c2 = y − x, c3 = z − y , i.e. (x, y, z) = x(1, 1, 1) + (y − x)(0, 1, 1) + (z − y)(0, 0, 1) ................................................................................................ Using the notion of dimension we can write down two important criteria for sets of vectors in a vector space. Basis Criterion If dim V = n and S contains exactly n linearly independent vectors then S is automatically a basis of V . There is no need to check the spanning condition. Linear Dependence Criterion If dim V = n and S has more than n vectors then S is automatically linearly dependent so it can not be a basis of V . ................................................................................................ Example : S = {1, 1 + x, x + x2 } is a basis of P2 because it is easy to check S is linearly independent and since the dimension of P2 is 3 we don't need to check that S is a spanning set, we can just apply the basis criterion. To see why dim P2 = 3 consider the set {1, x, x2 } and note that it is linearly independent and spans P2 . ................................................................................................ Example : S = {(1, 1, 1), (0, 1, 1), (0, 0, 1), (1, 2, 3)} can not be a basis of R3 because it has more than 3 vectors so it must be linearly dependent by the linear dependence criterion since it has more vectors than the dimension of R3 .
7. 7. ⇑Rowspace/Colspace: Suppose we have an m × n matrix A. The row space of A is the span of the row vectors of A and is denoted by rowspace(A) and the column space of A is the span of the column vectors of A and is denoted by colspace(A). Almost always we are interested in nding a basis of the row space or the column space of A and there is a very simple way of nding both at the same time. ................................................................................................ Example : Let's determine a basis for the column space and the row space of   1 2 −1 −2 −1    2 4 −2 −3 −1  A=    5 10 −5 −3 −1   −3 −6 3 2 1     1 2 −1 −2 −1 1 2 −1 0 0      2 4 −2 −3 −1  ∗  0 0 0 1 0   →    5 10 −5 −3 −1   0 1     0 0 0  −3 −6 3 2 1 0 0 0 0 0 Note that the leading 1's appear in columns 1,4 and 5 so a basis for the column space of A consists of the columns 1,4 and 5 from the non-reduced form of A, i.e.       1 −2 −1        2   −3   −1  Bc = {  ,   ,   }   5   −3   −1  −3 2 1 and the basis for the row space consists of the non-zero rows of the reduced form of A, i.e. Br = {(1, 2, −1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0, 0, 1)} ⇑Rank/Nullity: If we have an m × n matrix A then there is another vector space other than rowspace(A) and colspace(A) that we can associate to A. That other vector space is called the null space of A and it consists of all vectors in R that A maps to the zero vector. Another way of n saying all that is nullspace(A) = {x ∈ Rn | Ax = 0}. There is a theorem that says the dimension of the column space of A and the dimension of the row space of A are the same number so we call that common number rank(A). We also give a name to the dimension of the null space of A. The dimension of the null space of A is called nullity(A). There is a very nice theorem relating rank(A) and nullity(A). Rank-Nullity Thm If A is an m×n matrix then rank(A) + nullity(A) = n
8. 8. Advice : If you want to get a good handle on the rank-nullity theorem then looking at the true/false questions on page 311 is a really good idea. ⇑Inner product/Norm/Orthogonality: An inner product associates a real or complex number to two vectors. The inner product of u and v is denoted by u, v and it satises the following properties if we assume that the scalars for our vector space are real numbers: 1. u, u ≥ 0 and u, u = 0 if and only if u = 0 2. u, v = v, u 3. ku, v = k u, v 4. u + v, w = u, w + v, w ................................................................................................ Inner products are useful because we can use them to extend the notion of angle, orthogonality and magnitude from Rn to other vector spaces that don't necessarily look like Rn . So we say two vectors u, v from a vector space V are orthogonal if u, v = 0 and we dene the norm of a vector u √ from V to be ||u|| = u, u . With the help of inner products we can also nd the angle between u, v two vectors in a real inner product space because we can dene cos θ = . ||u|| · ||v|| ................................................................................................ Example : Consider the vector space of continuous functions on the interval [−1, 1] and dene ˆ 1 f, g = f (x) · g(x) dx −1 Let f (x) = cos x and g(x) = sin x then f and g are orthogonal because ˆ 1 f, g = cos, sin = cos x · sin x dx = 0 −1 I'll let you calculate ||f || and ||g||. ⇑Gram-Schmidt Process: You can think of the Gram-Schmidt process as a kind of factory that takes a linearly independent set of vectors as raw materials and produces another set of vectors that has the same span as the original set but satises the extra property of orthogonality, i.e. the vectors in the new set are orthogonal to each other. Geometrically it's really easy to understand, we project each vector onto all the previous vectors and subtract.
9. 9. input:linearly independent set ({x1 , x2 , . . . , xk }) output: orthogonal linearly independent set ({v1 , v2 , . . . , vk }) process: v1 = x1 x2 , v1 v2 = x2 − v1 ||v1 ||2 x3 , v1 x3 , v2 v3 = x3 − v − 2 1 v2 ||v1 || ||v2 ||2 . . . xk , v1 xk , v2 xk , vk−1 vk = xk − v − 2 1 v − ... − 2 2 vk−1 ||v1 || ||v2 || ||vk−1 ||2 ................................................................................................ Example : Let's use the Gram-Schmidt process on the linearly independent set {x1 = (1, 0, −1, 0), x2 = (1, 1, −1, 0), x3 = (−1, 1, 0, 1)} and see what we get. v1 = x1 = (1, 0, −1, 0) x2 , v1 2 v2 = x2 − v = (1, 1, −1, 0) − (1, 0, −1, 0) = (0, 1, 0, 0) 2 1 ||v1 || 2 x3 , v1 x3 , v2 −1 1 1 1 v3 = x3 − v − 2 1 v = (−1, 1, 0, 1) − 2 2 (1, 0, −1, 0) − (0, 1, 0, 0) = (− , 0, , 1) ||v1 || ||v2 || 2 1 2 2 1 1 So the new set of vectors is {(1, 0, −1, 0), (0, 1, 0, 0), (− , 0, , 1)} and it's easy to check that this 2 2 new set of vectors is orthogonal, i.e. v1 , v2 = v1 , v3 = v2 , v3 = 0. ⇑Linear maps/transformations: Linear transformations are the links between vector spaces and since a vector space is really all about addition and scalar multiplication we require that linear transformations play nice with those operations, i.e. a function T : V → W between two vector spaces is called a linear transformation if it satises i) T (v + w) = T (v) + T (w) and ii) T (cv) = cT (v) for all v, w ∈ V, c ∈ R. The vector space V is called the domain and the vector space W is called the codomain. ................................................................................................ Example : The transformation T : R2 → R given by T (x1 , x2 ) = x1 + x2 is linear. ? ? T (x + y) = T (x) + T (y) T (cx) = cT (x) T (x + y) = T (x1 + y1 , x2 + y2 ) T (cx) = T (cx1 , cx2 ) = (x1 + y1 ) + (x2 + y2 ) = cx1 + cx2 = (x1 + x2 ) + (y1 + y2 ) = c(x1 + x2 ) = T (x1 , x2 ) + T (y1 , y2 ) = cT (x1 , x2 ) = T (x) + T (y) = cT (x) T (x + y) = T (x) + T (y) T (cx) = cT (x)
10. 10. ................................................................................................ Non-example : Let T : P3 → R2 be given by T (p(x)) = 1. So no matter what polynomial is given to T the answer is always 1, i.e. T (1 + x) = 1, T (x3 ) = 1, T (2 + x + 11x3 ) = 1,etc. T fails to satisfy both properties of a linear transformation. ? ? T (p(x) + q(x)) = T (p(x)) + T (q(x)) T (cp(x)) = cT (p(x)) T (p(x) + q(x)) = 1 T (cp(x)) = 1 T (p(x)) + T (q(x)) = 1 + 1 cT (p(x)) = c · 1 = 2 = c T (p(x) + q(x)) = T (p(x)) + T (q(x)) T (cp(x)) = cT (p(x)) ................................................................................................ Zero vector criterion All linear transformation satisfy T (0) = 0 so if a transformation fails to satisfy this property then you don't need to check anything else and you can conclude right away that T is not linear. If you suspect that a map is not linear then this is a really easy test to use. ................................................................................................ Example : Suppose T : R3 → R2 is given by T (x1 , x2 , x3 ) = (1, 1) then you can see right away that T (0, 0, 0) = (1, 1) = (0, 0) so by the zero vector criterion T is not a linear map. This same trick would have worked for the example right before the zero vector criterion. ................................................................................................ Non-example : Suppose T : R3 → R2 is given by T (x1 , x2 , x3 ) = (x2 + x2 + x3 , 0) then we can 1 not use the zero vector criterion to conclude that T is not linear because T (0, 0, 0) = (02 + 0 + 0, 0) = (0, 0), i.e. T (0) = 0. So T satises the zero vector criterion but T is not linear because T ((1, 0, 0) + (1, 0, 0)) = T (1, 0, 0) + T (1, 0, 0). The lesson here is that failing to satisfy the zero vector criterion is enough evidence to disqualify a transformation from being linear but satisfying the zero vector criterion is not enough evidence to conclude that T is linear. ⇑Kernel/Range: (See page 361 for some helpful pictures) The idea of kernel and range of a linear map is similar to the nullspace and colspace of a matrix. For a linear map T : V → W we call the set {v ∈ V | T (v) = 0} the kernel of T and denote it by Ker(T ). Similarly we call the set {w ∈ W | w = T (v), v ∈ V } the range of T and denote it by Rng(T ). Note that Ker(T ) is a subspace of V and Rng(T ) is a subspace of W . ................................................................................................ Example : Let's nd Ker(T ) and Rng(T ) for T : R4 → R2 given by
11. 11. T (x1 , x2 , x3 , x4 ) = (x1 − x2 , x3 − x4 ). x ∈ Ker(T ) ⇔ T (x) = 0 ⇔ T (x1 , x2 , x3 , x4 ) = (0, 0) ⇔ (x1 − x2 , x3 − x4 ) = (0, 0) ⇔ x1 − x2 = 0, x3 − x4 = 0 Looking at the last set of equations we see that they are satised whenever x1 = x2 , x3 = x4 so x1 and x3 are bound variables and x2 = s and x4 = t are free variables. This means x = (x1 , x2 , x3 , x4 ) ∈ Ker(T ) ⇔ x = (s, s, t, t) ⇔ x = s(1, 1, 0, 0) + t(0, 0, 1, 1) ⇔ x ∈ span{(1, 1, 0, 0), (0, 0, 1, 1)} Now let's gure out the range. w ∈ Rng(T ) ⇔ w = T (x) ⇔ (w1 , w2 ) = T (x1 , x2 , x3 , x4 ) ⇔ (w1 , w2 ) = (x1 − x2 , x3 − x4 ) ⇔ w1 = x1 − x2 , w2 = x3 − x4 Again if we look at the last line we have a system of two equations in four unknowns that has rank 2 so no matter how w1 and w2 are chosen this system always has a solution which means Rng(T ) = R2 . ................................................................................................ If a linear map T is dened by T (x) = Ax for some matrix A then Ker(T ) = nullspace(A) and Rng(T ) = colspace(A). [ ] 1 1 0 0 Example : Suppose T : R4 → R2 is given by T (x) = Ax where A = . Find Ker(T ) 0 0 1 1 and Rng(T ). Ker(T ) = nullspace(A) = {x ∈ R4 | Ax = 0} = span{(−1, 1, 0, 0), (0, 0, −1, 1)} Rng(T ) = colspace(A) = span{(1, 0), (0, 1) = R2 ................................................................................................ General Rank-Nullity theorem For a linear map T : V → W between nite dimensional vector spaces V and W we have dim[Ker(T )] + dim[Rng(T )] = dim[V ] This is a very handy theorem that allows us to answer some questions without doing too much work.
12. 12. ................................................................................................ Example: Is it possible for a linear map T : V → W to have dim[Ker(T )] = 0 if dim V = 4 and dim W = 2? The answer is No. The general rank-nullity theorem tells us that dim[Ker(T )] + dim[Rng(T )] = dim[V ], i.e. dim[Ker(T )] + dim[Rng(T )] = 4 So if dim[Ker(T )] = 0 we must have dim[Rng(T )] = 4 and this is impossible because Rng(T ) is a subspace of W which has dimension 2 so the dimension of Rng(T ) must be less than or equal to 2 and 4 is not less than or equal to 2. ⇑One-to-one transformation: A linear map T : V → W is called one-to-one if v1 = v2 ⇒ T (v1 ) = T (v2 ). Kernel criterion A linear map T : V → W is one-to-one if and only if Ker(T ) = {0}, i.e. dim[Ker(T )] = 0. ⇑Onto transformation: A linear map T : V → W is said to be onto if Rng(T ) = W , i.e. for every vector w ∈ W we can nd some vector v ∈ V so that T (v) = w. ⇑Isomorphisms: A linear map T : V → W is called an isomorphism if it is both one-to-one and onto. Note that if V and W are nite dimensional then V and W must have the same dimension. ⇑Matrix representations: Whenever we have a linear map T : V → W and we choose a basis B for V and a basis C for W then we can write down a matrix [T ]C that represents the B linear transformation T . We call [T ]C the matrix representation of T relative to the bases B and B C . This is useful because we can use the matrix [T ]C to answer questions about the linear map B T and sometimes working with matrices is much easier than working with a general linear map T especially if the bases B and C are chosen in some clever way to simplify the representation. ⇑Eigenvalue/Eigenvector: For an n × n matrix A the value λ for which Av = λv has a non- trivial solution is called an eigenvalue of A and the solution v is called the eigenvector corresponding to λ. ................................................................................................ [ ] 3 −1 Example : For the matrix A = λ = 4 is an eigenvalue with corresponding eigenvector −5 −1 v = (−1, 1), i.e. Av = 4v . [ ][ ] [ ] [ ] 3 −1 −1 −4 −1 = =4 −5 −1 1 4 1
13. 13. ⇑Characteristic polynomial/Algebraic multiplicity: You might wonder how we go about nding eigenvalues and eigenvectors without using magic like in the previous example and the answer is that we use the characteristic polynomial of a matrix. For an n × n matrix A the polynomial p(λ) = det(A − λI) is called the characteristic polynomial of A and the roots of that polynomial are exactly the eigenvalues of A. Suppose p(λ) = (λ − 1)2 (λ − 3)3 (λ + 1) for some matrix A then the eigenvalues of A are λ1 = 1, λ2 = 3, λ3 = −1. Notice how some of the factors are raised to a power, e.g. (λ − 1)2 . When that happens we say λi appears with algebraic multiplicity mi . So λ1 = 1 has algebraic multiplicity 2, λ2 = 3 has algebraic multiplicity 3 and λ3 = −1 has algebraic multiplicity 1. ................................................................................................ [ ] 1 4 Example : Let's nd all the eigenvalues and eigenvectors for the matrix A = with the 2 3 help of the characteristic polynomial. det(A − λI) = |A − λI| 1−λ 4 = 2 3−λ = (1 − λ)(3 − λ) − 8 = λ2 − 4λ − 5 = (λ − 5)(λ + 1) So the characteristic polynomial tells us that the eigenvalues are λ1 = 5, λ2 = −1. Now let's nd the eingevector corresponding to λ1 = 5. We are looking for a solution to (A − 5I)v = 0. ([ ] [ ]) [ ] [ ] 1 4 1 0 v1 0 −5 = 2 3 0 1 v2 0 [ ][ ] [ ] −4 4 v1 0 = 2 −2 v2 0 Now write down the augmented matrix and row reduce to get the matrix [ ] [ ] [ ] 1 −1 0 v1 t = A ⇔ v1 − v2 = 0 ⇔ v1 = v2 ⇔ = 0 0 0 v2 t Now choosing a value of t gives us an eigenvector. Let t = 1 and we get the eigenvector v = (1, 1). I'll let you nd the other eigenvector that corresponds to λ2 = −1, just follow the same steps as above but for the matrix A − λ2 I = A + I . ⇑Eigenspaces/Geometric multiplicity: The set of all vectors v satisfying Av = λi v for a specic eigenvalue λi of A is called the eigenspace of A corresponding to λi and is denoted by Ei .
14. 14. The dimension of Ei is called the geometric multiplicity of λi .