1 Matrices and Systems of Linear Equations

Transcription

1 March 3, Systems of Linear Equations Matrices and Systems of Linear Equations An m n matrix is an array A = a ij of the form a a n a 2 a 2n... a m a mn where each a ij is a real or complex number. We sometimes call such an array A an m by n matrix. The matrix has m rows and n columns. The numbers m and n are called the row dimension and column dimension of A. For i m, j n,, the m matrix a j. a mj is called the j th column of A and the n matrix a i... a in is called the i th row of A. If A = a ij is an m n matrix, then its transpose, denoted A T is the n m matrix defined by A T = a t ij = a ji for each i, j. If the row and column dimensions of the matrix A are equal, then we call A a square matrix, and we call the common value of its row and column dimensions, its dimension. We will see that square matrices have special properties. We can add m n matrices as follows. If A = a ij and B = b ij, then C = A + B is the matrix c ij defined by c ij = a ij + b ij. We can only multiply matrices A and B if A is m n and B is n p. That is, the number of columns of A is the same as the number of rows of B.

2 March 3, In that case, if A = a ij and B = b jk, then C = A B is the m p matrix C = c ik defined by n c ik = a ij b jk. j= Thus the element c ik is the dot product of the ith row of A and the jth column of B. Both the operations of matrix addition and matrix multiplication are associative. That is, A + B + C = A + B + C, ABC = ABC. Multiplication of matrices is not always commutative, even for square matrices. For instance, if [ ] [ ] 0 A = and B =, 0 then, AB = [ 2 ] and BA = [ 2 ]. Let us consider some matrices A, B and illustrate these concepts. Example [ ] [ ] A =, B = [ ] C = A B = Example 2 A = B A not defined [ ] 2 A T = , B = 3 2

3 March 3, A T = C = A B = 7 5, A.B T = [ 3 2 ] Fact. A B T = B T A T. Let e i be the n vector with zeroes everywhere except in the i th position and a there. This is called the standard i th unit n vector. The n n matrix whose i th row consists of a in the i th position and zeroes elsewhere is called the n n identity matrix, and is denoted I n or simply I if the context makes the size clear. For any m n matrix A we have I m A = AI n = A. 2 Multiplication of matrices by row and column vectors Let p and n be positive integers. Let u, u 2,..., u n be n vectors in R p, and let a, a 2,..., a n be n real numbers. The expression u = a u + a 2 u a n u n is called the linear combination of the vectors {u, u 2,..., u n } with coefficients {a, a 2,..., a n }. Any expression of the above form is called a linear combination of vectors in R p. It is useful to note the following properties of matrix multiplication.

4 March 3, Let A = a ij be m n and let B = b jk be n p. Then, of course, C is m p. Let Ci r be the i th row of C and A r i be the i th row of A, then C r i = A r i B Similarly, if Cj c B, then is the j th column of C and B c j is the j th column of C c j = A B c j We wish to write these matrix expressions as certain linear combinations. From the definitions, it follows that and n Ci r = a i B r + a i2 B2 r a in Bn r = a ij Bj r, j= n Cj c = b j A c r + b 2j A c b nj A c n = b ij A c i 2 i= Thus, we see that The i th row of A B is the linear combination of the rows of B with coefficients given by the i th row of A and The j th column of A B is the linear combination of the columns of A with coefficients given by the j th column of B 3 Some properties of square matrices An n n matrix A is invertible if there is another n n matrix B such that AB = BA = I. We also call A non-singular. A singular matrix is one that is not invertible. The matrix B is unique and called the inverse of A. It is usually written A. It is a fact that, if A and B are two n n invertible matrix, then their produce A B is also invertible, and we have the formula A B = B A

5 March 3, There is a useful number which we can associate to a square matrix, called its determinant. This is often written deta. For a [ matrix] A = a, we set deta = a. a a If A = 2, we set a 2 a 22 deta = a a 22 a 2 a 2 We define deta inductively for matrices of higher dimension. Assuming that we know detb for all n n matrices, let A be an n + n + matrix. Define n+ deta = i+ a i detai i= where Ai is the n n matrix obtained by deleting the i th row and first column of A. This is called expansion by minors along the first column of A. One gets the same answer by expanding by minors along any row or column. As an example of expansion along the second row assuming that it exists, take n+ deta = 2+j deta2 j. j= An easy way to remember the signs in the preceding summation is to make an n S matrix whose entries are either + or - as follows. Let S = +. Make S 2 =, S 3 = +, and continue along the first row. For the second row, start with - and alternate along this row. Continue in this way to obtain the whole matrix S. Examples of S for 2 2 and 3 3 matrices are [ + + ], Let 0 denote the n vector all of whose entries are 0.

6 March 3, A collection u, u 2,..., u k of vectors in R n is called a linearly independent set of vectors in R n, if whenever we have a linear combination α u α k u k = 0, with the α is constants scalars we must have α i = 0 for every i. Fact. The following conditions are equivalent for an n n matrix A to be invertible.. the rows of A form a linearly independent set of vectors 2. the columns of A form a linearly independent set of vectors 3. for every vector b, the system has a unique solution Ax = b Here we are writing b as a column vector or as an n matrix. 4. deta 0 The function det maps the set of square matrices with real entries to the real numbers. If the matrix A has complex entries, then deta is a complex number. In this case, the determinant function has similar properties. For simplicity, we emphasize real matrices. Additional properties of determinants:. For any square matrix A, deta = deta T here A T is the transpose of A. 2. For any two square matrices A and B of the same dimension, deta B = deta detb. 3. deti n = for any n. 4. For an n n matrix A, and, for each i n, let A i be its i th row. We may consider the determinant function as a function of the rows. That is, we may write

7 March 3, deta = deta, A 2,..., A n. Now, fix some i with i n. Suppose that A and B are n n matrices which only differ in their i th rows. That is, for j i, A j = B j. Let C be the matrix whose i row is the sum aa i + bb i, where a and b are constants, and whose other rows also equal the corresponding rows of A and B. Then, detc = a deta + b detb This property of determinants is called multi-linearity as a function of the rows. Replacing rows by columns, we also get the deta is multi-linear as a function of the columns. 5. If B is the matrix obtained from A by interchanging two rows then detb = deta. One refers to this property by saying that deta is skew-symmetric as a function of the rows of A. The function a deta is also skew-symmetric as a function of the columns of A. Let us compute some determinants. A = [ ], deta = 22 3 = 7 A = , deta = = 2 4 Systems of Linear Equations We will write vectors x = x,..., x n in R n both as row vectors and column vectors. Matrices are useful for dealing with systems of linear equations. Suppose we are given a system of m equations in n unknowns

8 March 3, We can write the system as a single vector equation a x + a 2 x a n x n = b a 2 x + a 22 x a 2n x n = b 2 a m x + a m2 x a mn x n = b m. 3 Ax = b 4 where A in the m n matrix a ij, x is an unknown n vector, and b is a known m vector. Comment: Strictly speaking, the left and right sides of 4 are m matrices. It is customary to ignore this and simply refer to these as m vectors. Sometimes, we refer to n matrices as n vectors. The context will make it clear what is being done. We wish to determine whether the system 3 has a solution, and, if so, how many solutions does it have. Also, we seek a convenient way to represent the solutions. Before we consider this in detail, we note that there is a nice geometric description of the expression Linear Maps and Matrices Consider the map T x = A x. This defines a map from R n to R m. To solve the equation 4, where A and b are given, we are looking for a vector x R n such that T x = b. That is, we want b to be in the set-theoretic image of T. The map T has very special properties. It is what is called a linear map or a linear transformation from R n to R m. The definition is the following. Definition 4. The map T : R n R m is called linear if it preserves the operations of vector addition and scalar multiplication. That is, for any two vectors x, y R n and any real number α, we have T x + y = T x + T y,

9 March 3, and T αx = αt x One can easily see that maps from R n to R m defined using multiplication of vectors x by m n matrices A as above are linear maps. Conversely, it is relatively easy to prove that every linear map T : R n R m is defined by multiplication by some m n matrix. To be actually correct in this case, it is easier to ignore our convention of writing vectors as either row or column vectors, and to stick to row vectors. That is, we write x = x, x 2,..., x n as a n matrix. Then, we have the following Proposition 4.2 If T : R n R m is a linear map, then there is an n m matrix A such T x = x A for all x R n. Proof. Let {e, e 2,..., e n } denote the standard unit vectors in R n, and let {f, f 2,..., f m } denote the standard unit vectors in R m. Observe that writing a vector x R n as x = x, x 2,..., x n amounts to same as n x = x i e i. i= as Similarly, if y R m is written as y = y,..., y m, then this is the same m y = y j f j. j= Let T : R n R m be a linear map. By linearity, we have n n T x = T x i e i = x i T e i 5 i= i= Now, each T e i is a vector in R m, so there are real numbers a ij such that m T e i = a ij f j j=

10 March 3, Inserting these expressions into 5 we get n m n m m n T x = x i a ij f j = x i a ij f j = x i a ij f j i= j= i= j= j= i= Let A be the n m matrix given by A = a ij. Thinking of x = x, x 2,..., x n as a n matrix, and y = T x = y, y 2,..., y m as a m matrix, we have x A = y. 6 QED. Remark To keep things in column vector format, we would only have to replace the matrix A by its transpose A T and write T x = A T x. 4.2 Matrix methods for systems of linear equations Let us return to the system of linear equations 3. Our goal is to develop simple and effective ways of obtaining the solution set of the system. We first observe that the set of solutions is unchanged it we do any combination of the following operations on the system 3 including both the left and right sides of the equation.. interchange two rows 2. multiply a row by a non-zero constant 3. add a mutltiple of one equation to another. These are called elementary row operations on the matrix. Because the solutions don t change under these operations, we can use them to try to simplify the equations i.e., put them into a form where we can determine the solutions. First, we observe that it is not necessary to keep the variables in manipulation of the equations. From the system 3, we write the following matrix

11 March 3, a a 2... a n a 2 a a 2n. a m a m2... a mn This is called the augmented matrix of the system. We manipulate this matrix with the operations above to put the part corresponding to A in a better form so that we can read off the solutions. Example. Consider the system 2x + 3x 2 = 4 x x 2 = 6 Of course, we can solve this by Cramer s rule. We first do this. Afterward, we use the method of row operations. This latter method provides a systematic way to handle general linear systems of equations. Using Cramer s rule, we first write the system in terms of matrices: [ 2 3 ] [ x x 2 ] = [ 4 6 b b 2. b m ] 7 Then, we get x = [ ] 4 3 det 6 [ ] = 22/5, x 2 = 2 3 det [ ] 2 4 det 6 [ ] = 8/5 2 3 det Next, let s do this with row operations. We denote row i by R i. Write the augmented matrix: Replace R with 2R 2 + R

12 March 3, We can now read off that 5x 2 = 8, x x 2 = 6, so we can read the solution as x 2 = 8/5, x = 6 + x 2 = 22/5. Example 2 Use row operations to solve The augmented matrix is: 2x + 3x 2 + x 3 = 2 x x 2 x 3 = 3x 2x 2 = Replace R with 2R 2 + R Replace R 3 with 3R 2 + R Replace R 2 with R 3 + R 2 Now read off the solutions as: x 3 = 0 x 2 = 3x 3 = 0 x = 2x 3 = or, x =, x 2 = 0, x 3 = 0. Example 3. Use row operations to solve the system.

13 March 3, x + 3x 2 + x 3 = 2 x x 2 x 3 = 3x + 2x 2 = 3 Notice that this system only differs from that in the preceding example by the a change of sign in the multiplier of x 2 in the third equation. The solutions, however, will change considerably. In Example 2, there was a unique solution. In this example, it will be seen that there are infinitely many solutions. We begin with the augmented matrix, and proceed to use the row operations. Instead of writing each changed matrix, we combine several steps R 2 + R followed by 3R 2 + R Now, this reduces to the two equations x = + x 2 + x 3 5x 2 + 3x 3 = 0 x 2 = 3/5x 3 x = 3/5x 3 + x 3 x = + 2/5x 3 So we see that solutions have the form + 2/5x 3 3/5x 3 = 0 + x 3 x 3 0 2/5 3/5

14 March 3, This shows that the set of solutions is a line in R 3 running through the point, 0, 0 with the direction 2/5, 3/5,. In general, the set of solutions of a system of 3 equations in 3 unknowns will be a subset of R 3 of one of the following types:. empty i.e., there are no solutions 2. a point there is a unique solution 3. a line there are infinitely many solutions 4. a plane there are infinitely many solutions 5. all of R 3 the coefficients a ij are all 0. Example 4. Solve the system 2x 2x 2 + x 3 = 3 x x 2 x 3 = 3x + 2x 2 + 2x 3 = 2 We write the augmented matrix and do some row operations R2 + R, 3R 2 + R x 3 = /3 x 2 = /5 x 3 = /5 /3 3/5 5/5 = 8/5 x = + x 2 + x 3 = 8/5 + 5/5 = 2/5 = 4/5.

15 March 3, Finding the inverse of a matrix The 2 2 matrix A is invertible if and only if deta 0. Letting A = [ a a 2 a 2 a 22 we can compute A from the formula A = Example 5. Find the matrix X such that [ 2 2 [ deta ] X = ], a 22 a 2 a 2 a [ Here the matrix X is itself 2 2. This has the form A X = B for 2 2 matrices. If A is non-singular, then we get the answer from X = A B. We check non-singularity by computing deta. We get deta = 5, so the matrix is non-singular and its inverse is so, X = A B = /5 /5 [ 2 2 [ 2 2 ] [ ] ] ] = ] [. 8/5 /5 /5 2/5 For higher dimensional matrices, the formula for the inverse is harder, so it is convenient to find another method. Suppose that the dimension of A is n. Then, we are looking for an n n matrix B such that A B = I where I is the n n identity matrix. ]

16 March 3, Thus, we can form a big augmented matrix of the form A I. This is an n 2n matrix with the vertical line in the middle. Applying the row operation method for solving linear systems to this matrix, we do a sequence of successive row modifications. If A is actually invertible, then, at the end of doing these operations, we will actually get an n 2n matrix of the form I B. It turns out that the resulting matrix B is actually A. The proof of this is not difficult. It amounts to the following observation. Let us use the notation op to denote any of the elementary row operations. Write A op B to mean that B is obtained from A by applying the elementary row operation op. As above, let I be the n n identity matrix. Then the following is true. Proposition Let I denote that n n identity matrix. Let op be any elementary row operation, and assume that A op B and I op D. Then, B = DA This means that we get B from A doing a left multiplication by D where D is gotten from I via the same elementary row operation used to get B from A. Let us describe this in more detail. It will be convenient to have specific notations for the matrices gotten by applying elementary row operations to I. There are of three types: Let i, j be any two integers with i n and j n.. Interchange of rows Let P ij denote the matrix obtained by interchanging the i th and j th rows of I. This is usually called a permutation matrix since it permutes the rows of I.

17 March 3, Multiplication of a row by a non-zero constant. Let c be a non-zero constant, and let E c,i denote the matrix whose j th row is the unit vector i j if j i, and whose i th row is ce i. 3. Addition of a multiple of row i to row j. Let E ci+j denote the matrix obtained from I by adding c times row i to row j. Let us consider some examples. Example 6 Consider an interchange of the first two rows. A = , B = The associated permutation matrix is Then, one sees that P 2 = B = P 2 A Example 7. Consider the replacing row by row plus 2 times row 3 A = , B = The associated matrix is E 23+, with and E 23+ = B = E 23+ A ,

18 March 3, Try a few more examples yourself to get comfortable with this. Now, notice that each of the matrices P ij, E c,i, E ci+j is invertible. Indeed, Eij = E ij, Ec,i = E /c,i, and Eci+j = E ci+j. It follows that if we do a sequence of k row modifications to a matrix A this amounts to successively multiplying it on the left by k matrices obtained from one of the three types P ij, E c,i, E ci+j. This means that after k such modifications, the matrix A is replaced by D D 2... D k A where each D k is one of the P ij, E c,i or E ci+j we just discussed. If we had an augmented matrix A I and we do the same row modifications on it, then, and the end, we get D D 2... D k A D D 2... D k I If we end up with I on the left, then D D 2... D k = A. On the right we end up with D D 2... D k I = D D 2... D k. So we have written down A. 5. A simple method to compute the inverse of a 3 3 matrix Let A = a ij be an invertible n n matrix. We can use determinants and Cramer s rule to get a simple way to compute A using Cramer s rule. We first observe that Cramer s rule holds for every invertible square matrix: The i th coordinate of the solution to A x = b has the form x i = deta i deta where A i is the matrix obtained by replacing the i th column of A by the i th unit column vector e T i. Since A is the solution X of the matrix equation A X = I, the i th column of A is the solution to the matrix equation A x = e T i. We apply Cramer s rule to the case of a 3 3 invertible matrix

19 March 3, A = a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 Let u i denote the i th column of A. Then, u = deta a 22 a 23 a 32 a 33 a 2 a 23 a 3 a 33 a 2 a 22 a 3 a 32, u 2 = deta a 2 a 3 a 32 a 33 a a 3 a 3 a 33 a a 2 a 3 a 32 u 3 = deta a 2 a 3 a 22 a 23 a a 3 a 2 a 23 a a 2 a 2 a 22 In practice, this easy to compute. After computing the determinant of A, it simply involves computing determinants. 5.2 The inverse of an n n matrix A Let A be an n n matrix. For a given pair i, j of indexes i.e., i n, j n, define the n n matrix C = AdjA by C ij = i+j detaj i

20 March 3, where Aj i is the n n matrix obtained by deleting the j th row and the i column of A. The matrix C = AdjA is called the classical adjoint of A. The following theorem can be found in most books on Matrix Theory or Linear Algebra. Theorem For any square real or complex matrix A, we have A AdjA = deta I where I is the n n identity matrix. If follows that if deta 0, then deta = 6 Eigenvalues and eigenvectors AdjA 8 deta Let A be an n n matrix real or complex. An eigenvector is a non-zero vector ξ such that there is a scalar r such that Aξ = rξ. Note that the scalar r may be 0. When one can find such a ξ and a scalar r, one calls r an eigenvalue of the matrix A, and one calls ξ an eigenvector of A associated to r or for r. If ξ is a non-zero eigenvector, then the unit vector in the direction of ξ, ξ namely is called a unit eigenvector. ξ It turns out that if ξ is an eigenvector associated to the eigenvalue r, then so is any non-zero scalar mutliple of ξ. It is sometimes useful to have a unique way to specify a particular vector in the set of non-zero scalar multiples of ξ. Accordingly we define the scaled version of a non-zero vector v to be the vector v = v i v where v i is the first non-zero entry in v. We use the sharp symbol from music to denote the scaled version of v. For instance, we have [ 2 3 ] = 3 2, =

21 March 3, To understand the notion of eigenvector better, observe that the equation Aξ = rξ is equivalent to either of the equations or ri Aξ = 0 A riξ = 0 where I is the n n identity matrix. If either of these equations had a non-zero vector ξ as a solution, then it would follow that detri A = 0 0 The expression detri A is actually a polynomial of degree n of the form zr = r n + a n r n a 0 and the above equation can be written as zr = 0. The polynomial zr is called the characteristic polynomial of the matrix A, and its roots are the eigenvalues of A. The existence of these roots is provided by the Theorem Fundamental Theorem of Algebra Let pr = a n r n + a n r n a 0 be a polynomial of degree n i.e. a n 0 with complex coefficients a 0, a,..., a n. Then, there is a complex number α such that zα = 0. Remark. The Euclidean algorithm for positive integers states that given two positive integers p > q there are integers k > 0 and 0 s < q such that p = kq + s There is an analogous result for polynomials. Let us denote by degzr the degree of the polynomial zr with real of complex coefficients.

22 March 3, Euclidean Algorithm for Polynomials: Let pr and qr be two polynomials with degqr < degpr. Then there are polynomials kr and sr such that pr = krqr + sr and degsr < degqr. From this we have, for any complex polynomial zr, and any complex number α, there exist a complex polynomial qr with degzr = degqr + and a complex complex number c such that zr = r αqr + c Note that if α is a root of zr i.e., zα = 0, then c = 0. That is, the polynomial z α is a factor of zr. Let us make repeated use of the Fundamental Theorem of Algebra on zr. There exists a root r of zr and a polynomial z r such that zr = r r z r Next, there exists a root r 2 of z r and a polynomial z 2 r such that zr = r r r r 2 z 2 r as Continuing this way, we obtain all of the roots of zr as can express it zr = a n r r r r 2 r r n Note that the roots need not be distinct. So, we refer to this expression for zr as its factorization with multiplicities. We also say that the polynomial zr of degree n has n roots with multiplicity. Remarks.. If zr is a polynomial of degree n with real coefficients, then its roots may be real or complex. If r = a + bi is such a complex root i.e., b 0, the complex conjugate r = a bi is also a root of zr.

23 March 3, The Fundamental Theorem of Algebra is an existence theorem. It gives no information about how to find the roots of a given polynomial zr. For degree two, one can explicitly find the roots via that quadratic formula i.e., involving square roots of expressions involving the coefficients. For degrees three and four, there also are explicit formulas to find the roots in terms of taking expressions involving various roots of the coefficients. For degree greater than four, there are no such formulas that work in all cases. This surprising result often called the unsolvability of the quintic was proved by Abel in 823. For more information look up the Abel-Ruffini Theorem on Wikipedia. Note that some texts call the polynomial z r = deta ri the characteristic polynomial of A. Since z r = n zr, these two polynomials have the same roots, so to solve problems concerning eigenvalues, it really does not matter which definition is used. Of course, even if A is real, the characteristic polynomial zr may have real or complex roots. A real eigenvalue will have associated eigenvectors which are also real, and a complex eigenvalue will only have associated complex eigenvectors i.e. written as u + iu 2 with u and u 2 both real vectors and u 2 non-zero. 6. Simple formulas for the characteristic polynomials of 2 2 and 3 3 matrices For any matrix A = a ij, define the trace of A to be n tra = a ii i= Thus, tra is simply the sum of the diagonal entries. If a b A =, c d then

24 March 3, zr = detri A = r 2 trar + deta. In the 3 3 case, formula for zr is a bit more complicated: If a b c A = d e f, g h i then zr = r 3 trar 2 + det e f h i + det a c g i + det a b d e r deta Remark. There are also simple formulas for eigenvalues and eigenvectors for two dimensional square matrices. These are described in Section 7 of the Notes. In order to find eigenvalues and eigenvectors for matrices A of dimension higher than 2, it is necessary to do more work on solving the associated linear systems. For instance, if A = then, the characteristic polynomial will be Its roots are r = zr = r 3 2r 2 8r 5, r 2 = 3 29, r 3 = 2 The associated three eigenvectors are found by solving the three linear systems A riξ = 0 for each of the three values r = r, r = r 2, r = r 3.

25 March 3, Subspaces of R n, Null Space and Range of a Linear Map It is convenient to have name for subsets of R n which behave well under vector addition and scalar multiplication. Definition. A subset W of R n is called a linear subspace or, simply a subspace if it satisfies the following two properties.. For any two vectors v and w in W, we have v + w W. 2. For any vector v W and scalar α, we have αv W. We often say that W is closed under vector addition and scalar multiplication. Let W be a subspace of R n which contains at least one non-zero vector. A basis for W is a maximal linear independent set B = {v, v 2,..., v k } of vectors in W. This means that, for any vector w in W the set B = {w} B is no longer linear independent. That is, we cannot increase B inside W and keep it a linearly independent set. It is a fact that any two bases of a subspace W have the same number of elements. This common number is called the dimension of W. Note the single element subset {0} is a subspace of R n. We define its dimension to be 0. Let B be any subset of R n. Then, the subspace spanned by B, denoted spb is defined to the set of finite linear combinations a v + a 2 v 2 + a j v j where a i is a scalar and v i is a vector in B for all i. If B = {v, v 2,..., v k } is a finite set, then we write spv, v 2,..., v k for spb. Note that if dimspb = d and k > d, then the set B cannot be linearly independent. Examples.. The subspaces of R 2 consist of a the set {0 } consisting of the zero vector. dimension 0 b the lines through the origin dimension

26 March 3, c all of R 2 dimension 2 2. The subspaces of R 3 consist of a the set {0 } consisting of the zero vector. dimension 0 b the lines through the origin dimension c the planes through the origin dimension 2 d all of R 3 dimension 3 3. Let 0 < k < n and consider the set of vectors x, x 2,..., x k, 0,... 0 of vectors in R n whose coordinates x i are zero for i > k. This is a subspace of dimension k spanned by the vectors e, e 2,..., e k where e i is the standard unit vector whose only non-zero entry is a in the i th position. Definition. Let T : R n R p be a linear map with associated matrix A whose j th column A c j is T e j. The null-space or kernel of T is the set of vectors v in R n such that T v = 0. The Range of T is the set of vectors w R p such that there is a v R n such that T v = w. It is easy to show that the kernel of T and the range of T are subspaces of R n and R p, respectively. 6.3 Some properties of eigenvalues and eigenvectors First, we discuss some general facts about eigenvalues and eigenvectors which are valid in any dimension.. Let A be an n n matrix and let r be an eigenvalue of A. Let ξ and η be eigenvectors associated to r. Then, for arbitrary scalars α, β, we have that αξ + βη is also an eigenvector associated to r provided that it is not the zero vector. Proof. Let v = αξ + βη and assume this is not 0. We have

27 March 3, Av = Aαξ + βη = αaξ + βaη = αr ξ + βr η = r αξ + βη = r v Therefore v is also an eigenvector as required. 2. Let r r 2 be distinct eigenvalues of A with associated eigenvectors ξ, η, respectively. Then, ξ is not a multiple of η. Proof. Assume that ξ = αη for some α. Since both vectors are not 0, we must have α 0. Now, Aξ = r ξ = r αη, Aξ = Aαη = αaη = αr 2 η, So, r αη = r 2 αη. Since α 0, and η 0, we get r = r 2 which is a contradiction. 3. A real matrix may not have any real eigenvalues, but always has complex eigenvalues. In the language of subspaces, if we consider all of the eigenvectors for r and add the zero vector, then we get a subspace of R n. This is called the eigenspace of r.

28 March 3, A simple method to find eigenvectors for 2 2 matrices Let a b A = c d be a 2 2 matrix with characteristic polynomial zr = r 2 a + br + ad bc, and let r be a root of zr. We wish to find a vector v = v v 2 such that Av = r v or A r Iv = 0 with I equal to the 2 2 identity matrix. That is, we wish to solve the system of equations a r v + bv 2 = 0 cv + d r v 2 = 0 This is a homogeneous system of linear equations, and, since it has a solution, the two equations must be multiples of each other. Thus, we only need to solve the first equation. Case : b 0 Since b 0, we can let v = and get v 2 = r a to solve the equation. b Hence, the vector v = r a b is an eigenvector for r. Note that this works whether the root is real or complex. In the complex case, we get an associated complex eigenvector there is no associated real eigenvector. If the two roots of zr are r, r 2 with both real and distinct, then the same formula works for each of them.

29 March 3, That is, an eigenvector for r is v = r 2 a. b r a b and one for r 2 is v 2 = If the roots are real and equal, then this gives one eigenvector v There are two possibilities that can occur: Either all eigenvectors are non-zero multiples of v or there is second eigenvector v 2 which is not a multiple of v. In this latter case all non-zero vectors in R 2 in fact are eigenvectors. Case 2: b = 0 but c 0. In this case, we use the second equation in a similar way and get the eigenvector v = r d c are similar as well. Examples.. The cases of real and equal or complex eigenvalues. 3 8 A = 6 Characteristic polynomial: zr = r 2 + 3r 0 = r + 5r 2, roots: r = 5, 2 Eigenvalues and Eigenvectors: 2. r = 5, v = r = 2, v = r 3 8 r 3 8 A = = = = = /8 Characteristic polynomial: zr = r 2 8r + 7,

30 March 3, roots: r = 8 ± = 4 ± i Eigenvalues and Eigenvectors: r = 4 + i, v = r = 4 i, v = 4+i 4 4 i 4 = = i i 3. A = Characteristic polynomial: zr = r 2 2r + 36 = r 6 2, roots: 6, 6 Eigenvalues and Eigenvectors: r = 6, v = 0