Lecture 5: Matrix Algebra

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1 Lecture 5: Matrix Algebra In Song Kim September 7, Matrix Algebra 11 Definition Matrix: A matrix is an array of mn real numbers arranged in m rows by n columns a 11 a 12 a 1n a 21 a 22 a 2n A = a m1 a m2 a mn Vectors are special cases of matrices; a column vector of length k is a k 1 matrix, while a row vector of the same length is a 1 k matrix You can also think of larger matrices as being made up of a collection of row or column vectors For example, 12 Properties of Matrices A = ( a 1 a 2 a m ) Matrix Addition: Let A and B be two m n matrices Then a 11 + b 11 a 12 + b 12 a 1n + b 1n a 21 + b 21 a 22 + b 22 a 2n + b 2n A + B = a m1 + b m1 a m2 + b m2 a mn + b mn Matrices A and B must be the same size, allowing them to be conformable for addition Example A =, B = A + B = Please do not distribute without permission PhD candidate, Department of Politics, Princeton University, Princeton NJ

2 Scalar Multiplication: Given the scalar k, the scalar multiplication of ka is a 11 a 12 a 1n ka 11 ka 12 ka 1n a 21 a 22 a 2n ka = k = ka 21 ka 22 ka 2n a m1 a m2 a mn ka m1 ka m2 ka mn Example k = 3, A = ka = Matrix Multiplication: If A is an m k matrix and B is a k n matrix, then their product C = AB is the m n matrix where Example a c e c ij = a i1 b 1j + a i2 b 2j + + a ik b kj b aa + bc d A B = ca + dc C D f ea + fc = ab + bd cb + dd eb + fd 1( 2) + 2(4) 1(2) 1(5) + 2( 3) 1(1) = 3( 2) + 1(4) + 4(2) 3(5) + 1( 3) + 4(1) The number of columns of the first matrix must equal the number of rows of the second matrix If so they are conformable for multiplication The sizes of the matrices (including the resulting product) must be (m k)(k n) = (m n) NOTE HOW THE INSIDE DIMENSIONS ARE THE SAME!! Example 4 [ ] Example 5 Example 6 [ ] [ ]

3 13 Matrix Algebra Laws Laws of Matrix Algebra: 1 Associative (A + B) + C = A + (B + C) (AB)C = A(BC) 2 Commutative A + B = B + A 3 Distributive A(B + C) = AB + AC (A + B)C = AC + BC Commutative law for multiplication does not hold the order of multiplication matters: AB BA Have class calculate the following, which is the reverse of above example [ ] [ ] Example A =, B = AB =, BA = Transpose Transpose: The transpose of the m n matrix A is the n m matrix A T (sometimes written A ) obtained by interchanging the rows and columns of A Examples: A =, A T = B = 3, B T = ( ) As class exercize: A = , transpose: The following rules apply for transposed matrices: 1 (A + B) T = A T + B T 2 (A T ) T = A 3 (sa) T = sa T 3

4 Example 8 4 (AB) T = B T A T Example of (AB) T = B T A T : A = , B = T 1 (AB) T = B T A T = = = Proposition 21 (AB) T = B T A T It is important to understand that Ax combines the columns of A while x T A T combines the rows of A T (show example) Many times in statistics we will make use of this sort of fact If we have a mathematical object x T A T that is is multiplied by other terms, we might want to express it as Ax in order to simplify it with some other part of the equation Example 9 Calculate A + 2B, AB, and BA using the following matrices [ ] A =, B = 4 3, Square Matrices Square matrices have the same number of rows and columns; a k k square matrix is referred to as a matrix of order k The diagonal of a square matrix is the vector of matrix elements that have the same subscripts If A is a square matrix of order k, then its diagonal is [a 11, a 22,, a kk ] Trace: The trace of a square matrix A is the sum of the diagonal elements: tr(a) = a 11 + a a kk Properties of the trace operator: If A and B are square matrices of order k, then 1 tr(a + B) = tr(a) + tr(b) 2 tr(a T ) = tr(a) 4

5 3 tr(sa) = str(a) 4 tr(ab) = tr(ba) Example , trace: Here are some examples of the square matrix: 1 Symmetric Matrix: A matrix A is symmetric if A = A T ; this implies that a ij = a ji for all i and j Examples: A = = A T, B = = B Diagonal Matrix: A matrix A is diagonal if all of its non-diagonal entries are zero; formally, if a ij = 0 for all i j Examples: A = , B = Triangular Matrix: A matrix is triangular one of two cases If all entries below the diagonal are zero (a ij = 0 for all i > j), it is upper triangular Conversely, if all entries above the diagonal are zero (a ij = 0 for all i < j), it is lower triangular Examples: A LT = 4 2 0, A UT = Identity Matrix: The n n identity matrix I n is the matrix whose diagonal elements are 1 and all off-diagonal elements are 0 Examples: I 2 =, I = The Inverse of a Matrix From elementary algebra, we define an inverse of a as 1 a such that a 1 a = 1 THIS IS DIFFERENT FROM THE INVERSE OF A FUNCTION!! A similar property is important in matrix algebra Definition 1 Identity Matrix: An n n with all values off the diagonal equal to 0 and all on diagonal values equal to Example 11 I 2 = 0 1 5

6 Definition 2 Inverse Matrix: An n n matrix A is invertible if there exists an n n matrix A 1 such that AA 1 = A 1 A = I n A 1 is the inverse of A If there is no such A 1, then A is noninvertible (another term for this is singular ) Example 12 Let Since A = , B = AB = BA = I n we conclude that B is the inverse, A 1, of A and that A is nonsingular Theorem 41 (Uniqueness of Inverse) The inverse of a matrix, if it exists, is unique denote the unique inverse of A by A 1 We Theorem 42 (Properties of Inverse) Let A and B be nonsingular n n matrices 1 AB is nonsingular and (AB) 1 = B 1 A 1 2 A 1 is nonsingular and (A 1 ) 1 = A 3 (A ) 1 = (A 1 ) One application of inverting a matrix is to solve a system of linear equations In fact, matrices can be motivated in terms of linear equations Consider a set of m linear equations of the form y 1 = a 11 x 1 + a 12 x a 1n x n y 2 = a 21 x 1 + a 22 x a 2n x n y m = a m1 x 1 + a m2 x a mn x n Then, its matrix representation is Y = AX where a 11 a 12 a 1n a 21 a 22 a 2n A =, X = a m1 a m2 a mn x 1 x 2 x m, Y = We call A a coefficient matrix With this notation, we can see that A 1 (provided that A is nonsingular) solves this system since we obtain X = A 1 Y by premultiplying the equation by A 1 y 1 y 2 y m 6

7 Example 13 Confirm that [ ] 1 = [ solve the following system of linear equations by using the inverse of matrix Class examples: Show B is inverse of A: A = [ ] 1 2, B = 1 1 2x 1 + 3x 2 = 1 2x 1 + 2x 2 = 2 [ ] [ ] 1 4 Find the inverse of the matrix A, A = 1 3 [ ] a b It turns out that there is a shortcut for 2x2 matrices that are invertible: A =, then c d [ ] A 1 = 1 d b ad bc Show this c a 5 Finding the inverse of a matrix We know that if B is the inverse of A, then AB = BA = I n Looking only at the first and last parts of this AB = I n Solving for B is equivalent to solving for n linear systems, where each column of B is solved for the corresponding column in I n In performing Gauss-Jordan elimination for each individual system, the same row operations will be performed on A regardless of the column of B and I n Hence, we can solve the systems simultaneously by augmenting A with I n and performing Gauss-Jordan elimination on A Note that for the square matrix A, Gauss-Jordan elimination should result in A becoming row equivalent to I n Therefore, if Gauss-Jordan elimination on [A I n ] results in [I n B], then B is the inverse of A Otherwise, A is singular To summarize: To calculate the inverse of A 1 Form the augmented matrix [A I n ] 2 Using elementary row operations, transform the augmented matrix to reduced row echelon form 3 The result of step 2 is an augmented matrix [C B] (a) If C = I n, then B = A 1 (b) If C I n, then C has a row of zeros A is singular and A 1 does not exist Exercise 1 Find the inverse of A = ] 7

8 6 Properties of matrices In our discussion of systems of equations we said that there were three possible cases: one solution (intersecting lines or planes), no solutions (parallel lines or planes), or infinite solutions (identical lines or planes) In matrix algebra we can identify which case a matrix fall into by identifying certain mathematical properties of the matrix There are three key pieces of information 1 the number of equations m, 2 the number of unknowns n, 3 the rank of the matrix representing the linear system Definition 3 Rank: The rank of a matrix is the number of nonzero rows in its row echelon form The rank corresponds to the maximum number of linearly independent row or column vectors in the matrix Properties of rank: For an i, j matrix, rank(a) min(i, j); rank(ab) min(ranka, rankb) Existence of Solutions: 1 One solution: Necessary condition for a system to have a unique solution is that there are exactly as many equations as unknowns 2 Infinite solutions: If a system has a solution and has more unknowns than equations, then it has infinitely many solutions 3 No solution: There is a row with all 0 s in RREF form ASIDE: You will often hear the expression degrees of freedom problem in statistical analyses At the extreme, this is a problem where you have more independent variables than observations In this case, there does not exist a solution and you will not get anything out of your computer software that performs the estimation More generally, consider the equation x+y = 7 This has one degree of freedom Once we fix x then y is now no longer free to vary Find the rank and solution to each system of equations 1 x 3y = 3 2x + y = 8 2 x + 2y + 3z = 6 2x 3y + 2z = 14 3x + y z = 2 8

9 3 x + 2y 3z = 4 2x + y 3z = 4 4 x + 2y + 3z + 4w = 5 x + 3y + 5z + 7w = 11 x z 2w = 6 7 Solving and Simplifying Linear Systems with Inverses In elementary algebra we often multiplied by inverses to simplify equations We will also do this often with matrix algebra Matrix representation of a linear system Ax = b If A is an n n matrix,then Ax = b is a system of n equations in n unknowns If A is invertible then A 1 exists To solve this system, we can left multiply each side by A 1 (must be left multiplication on both sides!!) A 1 (Ax) = A 1 b (A 1 A)x = A 1 b I n x x = A 1 b = A 1 b Hence, given A and b and given that A is nonsingular, then x = A 1 b is a unique solution to this system Notice also that the requirements for A to be nonsingular correspond to the requirements for a linear system to have a unique solution: ranka =rowsa =columnsa 8 Determinants Finding the inverse of a particular matrix can be a pain, though there are some short cuts you could use that we won t cover here Alternatively, we can make use of another property, the determinant If the determinant is non-zero then the matrix is invertible For a 1 1 matrix A = a the determinant is simply a We want the determinant to equal zero when the inverse does not exist Since the inverse of a, 1/a, does not exist when a = 0 a = a a11 a For a 2 2 matrix A = 12, the determinant is defined as a 21 a 22 a 11 a 12 a 21 a 22 = det A = a 11 a 22 a 12 a 21 == a 11 a 22 a 12 a 21 Notice that with matrices the verticle bars mean the determinant, not the absolut value 9

10 81 Extension to larger matrices Things get a bit more cumbersome once we start dealing with larger matrices In particular: 1 Deteriminant of 3 3 matrix: a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = a 11 a 22 a 23 a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 2 Deteriminant of n n matrix Let A ij be the (n 1) (n 1) submatrix of A obtained by deleting row i and column j Let the (i, j)th minor of A be Then for any n n matrix A M ij = A ij A = a 11 M 11 a 12 M ( 1) n+1 a 1n M 1n Example: Does the following matrix have an inverse? A = Calculate its determinant A = = 1(2 15) 1(0 15) + 1(0 10) = = 8 2 Since A 0, we conclude that A has an inverse There is nothing complicated here, just keep track of all the addition and substraction, etc Definition 4 Let A be an n n matrix 1 Let M ij be the (n 1) (n 1) submatrix of A obtained by deleting the ith row and jth column of A Then, M ij is called the (i, j)th minor of A 2 The cofactor of A ij of A is defined as C ij = ( 1) i+j M ij Now, the following theorem gives us a new method to compute determinants Theorem 81 (Cofactor Expansion) Let A be an n n matrix Then, for any i and j, A = n j=1 a ijc ij, where a ij is the (i, j) the element of A 10

11 82 Special matrices and their determinants Then Triangular or Diagonal Matrices: For any upper-triangular, lower-triangular, or diagonal matrix, the determinant is just the product of the diagonal terms Suppose we have the following square matrix in row echelon form (ie, upper triangular) r 11 r 12 r 13 R = 0 r 22 r r 33 r R = r 22 r r 33 = r 11r 22 r 33 Notice how the 0 terms let us cancel things out Think about this in the diagonalized case!! 83 Properties of determinants Properties of Determinants: 1 A = A T 2 Interchanging rows changes the sign of the determinant 3 If two rows of A are exactly the same, then det A =0 4 If a row of A has all 0 then det A =0 5 If c 0 is some constant then A = ca 6 Performing elementary row operations on a matrix does not change the determinant other than by a scalar 7 A square matrix is nonsingular iff its determinant is 0 8 AB = A B 84 Bringing it together: Relationship between rank, inverse, determinant, and solutions: Consider the following matrix equations that uses a non square matrix which has more rows than columns Ie, we have more observations than we have variables a 11 a 12 b 1 [ ] a 21 a 22 X = b 2 x1 where X = x a 31 a 32 b 2 3 We want to find X = [ x1 x 2 ] to make this work This matrix could be reduced using elementary row operations to zero out the last row Thus, we would get a new matrix where the b 3 would have been transformed to something because of the application of the row operations to zero out a 31, a 32 Lets call this new value of b 3, s + b 3, where s is just some scalar 11

12 a 11 a 12 b 1 Then we have: a 21 a 22 X = b s + b 3 Performing matrix multiplication we obtain the following system of equations: a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 0x 1 + 0x 2 = sb 3 Now notice that no matter what we put into the equation for x 1, x 2 we could never satisfy the third equation (this could happen if b 3 = 0, which means that no information at all is in the row and it should have just been dropped from the matrix in the first place) Thus we have no solutions to this system of equations Consider now the case of when we have more variables than observations, ie, m < n [ ] [ ] x a11 a 12 a 1 13 b1 X = where X = x a 21 a 22 a 23 b 2 2 x 3 Matrix multiplication gives us: a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 Elementary row operations will allow us to zero out two of the entries in the matrix, and will in doing so put new coefficients on the a terms and will add something to the b terms So lets just use new variables to represent these transformed variables c 11 x 1 + 0x 2 + c 13 x 3 = d 1 0x 1 + c 22 x 2 + c 23 x 3 = d 2 Now notice that there is nothing we could do to zero out an entry in the third column without losing the zero we obtained in the row we were trying to put a zero in the third column Thus the best we can do is to rewrite things as: Which we can first manipulate to give us: c 11 x 1 = d 1 c 13 x 3 c 22 x 2 = d 2 c 23 x 3 12

13 x 1 = d 1 c 11 c 13x 3 c 11 x 2 = d 2 c 22 c 23x 3 c 22 And here we see why we have an infinite number of solutions Consider the first equation, x 1 = d 1 c 11 c 13x 3 c 11 We have c 13 c 11 in this expression Lets say the value of this that we needed to solve the equation was c 13 c 11 = 5 But, we can get a 5 from this ratio in an infinite number of ways Eg, c 11 = 2 and c 13 = 10, 10 2 Thus there are an infinite number of values for c 11 and c 13 that would solve our system of equations Assuming that our matrix A is in reduced echelong form, When m > n then we have no solution and our rank is less than the number of rows, ie, rank(a)<rows(a) When m < n we have an infinite number of solutions and rank(a)<columns(a) Only when m = n do we have a single solution, and rank(a)=row(a)=row(b) Furthermore, only when rank(a)=row(a)=row(b) is the matrix invertible 13

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