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Matrices are essentially boxes filled with numbers organized in rows and columns. Hence matrices are easily represented on an Excel worksheet. Capital Letters such as A, B, C, D, etc. are generally used as names for matrices. Most statistical notation uses x, y & z to denote sample data; similarly X, Y & Z will denote a matrix of sample data with each column representing a variable and each row representing a sampled unit. The size of a matrix is determined by the number of rows and number of columns. In reporting the size of a matrix, the numbers of rows is 1st and number of columns is 2nd. 2 by 3 denotes a matrix with 2 rows and 3 columns. 5 by 4 denotes a matrix with 5 rows and 4 columns. 3 by 1 denotes a matrix with 3 rows and 1 column and is called a column matrix or column vector 1 by 5 denotes a matrix with 1 row and 5 columns and is called a row matrix or row vector. If the (number of rows) = (number of columns), then the matrix is considered to be a Square Matrix. I will place a box around a set of cells to denote a matrix in Excel. Standard mathematical notation is to put numbers in brackets. Matrix A Col. 1 Col. 2 Col. 3 Col. 4 é -1 - 2 3 4 ù A = ê 11 12 13 14 ú Row 1 -1 -2 3 4 Row 2 11 12 13 14 ê ú Row 3 24 23 21 20 ê24 ë 23 21 20ú û 3 by 4 matrix of sample data umn vector. s in brackets. Special Matrices Column Matrix - Column Vector - Matrix with one column. Row Matrix - Row Vector - Matrix with one row. Square Matrix - Matrix with (number of rows) = (number of columns). Zero Matrix - Matrix with only zeroes for all number values or entries. Matrix Transpose - A matrix created from another matrix by reversing rows and columns. Transpose notation - the transpose of A can be denoted as AT or A'. Symmetric Matrix - A square matrix such that A = A', that is aij = aji for all values of i & j. Diagonal Matrix - A symmetric matrix with all values off of the main diagonal = 0; a ij=0 for i≠j. Identity Matrix - A diagonal matrix will all diagonal values = 1; aij=1 for i=j & aij=0 for i≠j. A Transpose Matrix A Col. 1 Col. 2 Col. 3 Col. 4 AT or A' Col. 1 Col. 2 Col. 3 Row 1 -1 -2 3 4 Row 1 -1 11 24 Row 2 11 12 13 14 Row 2 -2 12 23 Row 3 24 23 21 20 Row 3 3 13 21 3 by 4 Row 4 4 14 20 4 by 3 ws and columns. l values of i & j. onal = 0; a ij=0 for i≠j. =0 for i≠j. To perform the operations of addition or subtraction for matrices the size of the matrices must be the same. Number of rows must be the same for both matrices and the number of columns must also be the same. The result is a matrix of the same exact size. The result is determined by performing the appropriate operation (addition or subtraction) on all corresponding elements or cells. A B A+B -1 -2 3 4 9 8 7 5 8 6 10 9 11 12 13 14 + -1 0 -2 0 = 10 12 11 14 24 23 21 20 0 -3 1 -5 24 20 22 15 3 by 4 3 by 4 3 by 4 A B A-B -1 -2 3 4 9 8 7 5 -10 -10 -4 -1 11 12 13 14 - -1 0 -2 0 = 12 12 15 14 24 23 21 20 0 -3 1 -5 24 26 20 25 3 by 4 3 by 4 3 by 4 Excel has several functions that perform operations on matrices and some can produce a result that is a matrix displayed in an array. To use one of the functions that places a matrix result in an array of cells one MUST hold down the Ctrl & Shift keys simultaneously while clicking on OK or pressing the Enter key. These functions appear in two primary locations in Excel functions. TRANSPOSE in the Lookup & Reference category of functions. MMULT in the Math & Trig category of functions. MINVERSE in the Math & Trig category of functions. The above functions allow for a matrix as an input and can produce matrix output. For the above functions one MUST hold down the Ctrl & Shift keys simultaneously while clicking on OK or pressing the Enter key to obtain matrix output. MDETERM in the Math & Trig category of functions calculates the DETERMINANT of a square matrix.. The above function allows for a matrix as an input and will have a single number rather than a matrix as output. If the value of the determinant for a covariance or correlation matrix is zero then at least one variable in the data set is a linear combination of other variables. can produce a MUST OK or matrix output. ift keys key to obtain of a gle number s zero then at A B 0 1 The -1 0 1 Two 1 2 Must 2 3 4 2 by 2 # Columns for < be > # Rows for the 2 by 3 the 1st matrix Equal 2nd matrix If the number of columns for the A*B 1st matrix does not equal the number of rows for the 2nd matrix Col1 Col2 Col3 then multiplication of the two ROW1 0*(-1)+1*2 0*0+1*3 0*1+1*4 matrices is impossible. ROW2 1*(-1)+2*2 1*0+2*3 1*1+2*4 Result = 2 by 3 2 3 4 These are calculated from 3 6 9 the cells above 2 3 4 This result was obtained 3 6 9 by using the MMULT function f columns for the not equal the s for the 2nd matrix ion of the two ulated from s obtained MULT function Identity Inverse Inverse Element Operation Element Operation Element a Addition 0 Subtraction -a a Subtraction 0 Addition -a -1 a Multiplication 1 Division a = 1/a a Division 1 Multiplication a-1 = 1/a A Matrix Identity Multipy by Inverse 1 Multiplication Matrix Inverse Matrix = A-1 0 2 by 2 Identity To find an inverse of a matrix A 1 0 1. A must be a square matrix 0 1 2. The determinant of A must not = 0 0 0 3 by 3 Identity Steps for finding an inverse using Excel is on the next workbook. 0 1 2 by 2 Identity 0 0 1 by 3 Identity A-1 is used to denote the inverse of the matrix A |A| is used to denote the determinant of the matrix A To find an inverse of a matrix To find an inverse of a matrix A A 1. A must be a square matrix 1. A must be a square matrix 2. The determinant of A must not be equal to 0, |A|≠0. 2. The determinant of A must not = 0 To find an inverse, the matrix MUST be square. If the number of rows A does not equal the number of columns, an inverse cannot be found. 1 2 3 Also the determinant must not be zero. Placing the cursor on cell B13 0 1 0 click on function button, go to Math & Trig and then select MDETERM. 2 -1 4 Select the cells in B8 through D10 and click on OK. Since the 3 by 3 determinant of -2 is not zero then an inverse can be found. Highlight the cells B16 through D18,click on function button, go to |A| = -2 = Determinant of A Math & Trig and then select MINVERSE. Select the cells in B8 through D10 and simultanedously hold down both the Ctrl & Shift A-1 keys while clicking on OK. The inverse will appear in the -2 5.5 1.5 highlighted cells B16 through D18. 0 1 0 1 -2.5 -0.5 To confirm that this is the inverse multipy the original matrix times the inverse. Highlight the cells B22 through D24,click on function button, 3 by 3 go to Math & Trig and then select MMULT. Select the cells B8 through D10 for Array1 and cells B22 through D24 for Array2 then A * A-1 hold down the Ctrl & Shift keys simultaneously while clicking on 1 0 0 OK. The resulting 3 by 3 identity matrix provides confirmation that it is 0 1 0 the inverse. 0 0 1 3 by 3 Determinant of a 2 by 2 matrix éa b ù B=ê , then B = a d - b c ë c dú û mber of rows on cell B13 MDETERM. Ctrl & Shift x times the ion button, clicking on tion that it is The system of 3 equations below is coverted to matrix form. The inverse matrix is found and used to obtain solution. 2x + 5y + 4z = 4 x + 4y + 3z = 1 x - 3y - 2z = 5 (Example 6, page 263 of 6th edition of Harshbarger & Reynolds C Determinant(C) = -1 < hence an inverse can be found 2 5 4 x 4 To obtain C-1, the Matrix Inverse for C, Highlight a 3 by 3 area, Select the function MINVERSE, 1 4 3 * y = 1 Indicate the location of the matrix C as below. 1 -3 -2 z 5 Hold down the Ctrl & Shift keys OK. simultaneously while clicking on 3 by 3 C-1 -1 2 1 -5 8 2 7 -11 -3 x 4 3 To find this matrix of Solution values, use the MMULT -1 function to multiply C-1 times the column matrix of right y = C * 1 = -2 hand side values (See Below). Hold down the Ctrl & z 5 2 Shift keys simultaneously while clicking on OK. se can be found A I (2by2) I*A 5 2 4 1 0 5 2 4 6 4 7 0 1 6 4 7 (2 by 3) B A' or A transpose I (3by3) 1 3 5 6 1 0 0 -3 0 2 4 0 1 0 -3 5 4 7 0 0 1 Scalar Multiplication B + A' A*I 2*A 6 9 5 2 4 10 4 8 -1 4 6 4 7 12 8 14 1 12 Data matrix for values of observations of m variables (m columns) for n sampling units (n rows). x1= variable 1 x2= variable 2 x3= variable 3 … xm= variable m Sampling unit 1 x11 x12 x13 … x1m Sampling unit 2 x21 x22 x23 … x2m Sampling unit 3 x31 x32 x33 … x3m Sampling unit 4 x41 x42 x43 … x4m … … … … … … Sampling unit n xn1 xn2 xn3 … xnm Each variable will be measured using the same unit of measure (inches, kilograms, test grade, etc.) for all sampling units but each variable may have a unit of measure that is different from all of the other variables. Mean Vector - a vector containing the sample mean of each variable Variance Vector - a vector containing the sample variance of each variable Standard Deviation Vector - a vector containing the sample standard deviation of each variable Outlier An Outlier may have An observation with a unique combination of characteristics large or small value identifiable as distinctly different from the other observations. (Hair 6th text definition page 73) variable or it may ha An observation that is so extreme that it stands apart from the rest of combination of value the observations; that is, it differs so greatly from the remaining the dark red point in observations that it gives rise to the question whether it is from the below. same population or involves measurement error. (Pocket Dictionary of Statistics, 2002) Reasons for an Outlier: 25 Procedural or measurement error that is a result of data entry error, 20 measurement coding or measuring the wrong thing. A rare or extraordinary event has occured for the phenomenon being observed and the observation(s) come from this event. 15 (Rain attributing to a 100 year flood.) The data are for an observation that may be from a different 10 phenomenon than the one being studied. 5 Should an outlier be included or excluded for an analysis? 0 Is it truly indicative of the phenomenon being studied? 0 5 Does its inclusion mean that the sample data are no longer truly representative of the phenomenon? I like to use the terms Extreme point for a data value set apart from from the rest and Nonindicative point for a data value that is deemed to not be representative of the phenomenon being studied. X1 X2 sum diff An Outlier may have an extremely 1 18 18 36 0 small value for a single 2 2 2 4 0 ariable or it may have an unusual 3 11 13 24 -2 4 7 10 17 -3 ombination of values as shown by 5 18 18 36 0 he dark red point in the X-Y scatter 6 19 21 40 -2 7 6 9 15 -3 8 0 0 0 0 9 17 20 37 -3 10 19 19 38 0 11 5 7 12 -2 12 11 14 25 -3 13 2 3 5 -1 14 7 8 15 -1 15 15 19 34 -4 16 17 18 35 -1 17 2 3 5 -1 18 1 1 2 0 19 11 14 25 -3 20 19 19 38 0 outlier 18 4 22 14 Extreme Point that seems to be Nonindicative. Mean 10.7 11.4 22.14 -0.714 10 15 20 Variance 49.1 52.4 189.8 13.11 Correlation 0.87 -0.065 Total Variance 101 202.9 49.1 eems to be Nonindicative. x1, x2 through xm denote m variables with observations for n sampling units. Sample Mean for the jth variable Sample Variance for the jth variable Sample Standard Deviation is x - xj n n the square root of the sample xij 2 variance. ij The Sample Standard Deviatio for the jth variable is xj = i =1 s2 = i =1 sj. n -1 j n AVERAGE function in Excel VAR is a 2007 function and VAR.S is an Excel 2010 function Sample Covariance between the jth and kth variables Excel 2007 does not have a fu x - x j xik - x k n sample covariance. The COVAR function ij population covariance or the m COV ( x j , x k ) = s jk = i =1 estimator rather than the minim n -1 unbiased estimator that is gen calculate the sample covarianc COVARIANCE.S in Excel 2010 calculates the sample covariance. The Sample Covariance =COVAR(data)*n/(n Excel 2007 or earlier does not find a sample covariance. COVARIANCE.S(data) covariance in Excel 2010. e Standard Deviation is uare root of the sample mple Standard Deviation variable is denoted by el 2010 function 2007 does not have a function to find a e covariance. OVAR function formula finds the ation covariance or the maximum likelihood ator rather than the minimum variance sed estimator that is generally used to ate the sample covariance. ample Covariance can be calculated by AR(data)*n/(n-1) 2007 or earlier does not have a function to sample covariance. ARIANCE.S(data) calculates the sample ance in Excel 2010. Measurement Statistic Original Units The Mean is a measure of the location of the middle of a distribution. (Original Units)2 The Variance is a measure of the dispersion or spread for a distribution. Original Units The Standard Deviation is a measure of the dispersion or spread for a distributi Product of two sets of The Covariance is a measure of the linear relationship between two variables. Original Units Unit Free The Correlation is a unit free measure of the linear relationship between two var Sample Correlation between the jth and kth variables s jk COV ( x j , x k ) r jk = = s j sk Std .Dev.( x j ) Std .Dev.( x k ) n xij - x j xik - x k s s = i =1 j k n -1 a distribution. for a distribution. n or spread for a distribution. between two variables. ationship between two variables. Data matrices for values of observations of m variables (m columns) for n sampling units (n rows). Original Data matrix X x1= variable 1 x2= variable 2 x3= variable 3 … xm= variable m Sampling unit 1 x11 x12 x13 … x1m Sampling unit 2 x21 x22 x23 … x2m Sampling unit 3 x31 x32 x33 … x3m … … … … … … Sampling unit n xn1 xn2 xn3 … xnm Xd is the matrix of deviation scores from the Xs is the matrix of standardized scores for each mean of each variable. It is often referred to as variable. The mean is zero and the standard mean-corrected or mean-adjusted or centered deviation is one for each variable in the mean data matrix. The mean is zero for each variable in corrected data set. a mean-corrected or centered data set. é x11 - x1 x12 - x 2 x1m - x m ù ê s ú é x11 - x1 x12 - x 2 ... x1m - x m ù s2 sm ê 1 ú ê x 21 - x1 x 22 - x 2 x2 m - xm ú ê x 21 - x1 x 22 - x 2 x2m - xm ú ... Xd =ê ú X s = ê s1 s2 sm ú ê ... ... ... ... ú ê ú ê ú ê x n1 - x1 xn 2 - x2 x nm - x m ú ë x n1 - x1 xn 2 - x2 ... x nm - x m û ê ú n ë s1 s2 sm û xij j n sj = standard x ij - x 2 i =1 xj = = Mean for the j variable th deviation of the j i =1 n s2 = variable. n -1 j sampling units (n rows). tandardized scores for each n is zero and the standard each variable in the mean- 2 - x2 x1m - x m ù ú s2 sm ú 2 - x2 x2m - xm ú s2 sm ú ú 2 - x2 x nm - x m ú ú s2 sm û sj = standard deviation of the jth variable. For a set of data with observation values for m variables and n sampling units one can calculate a covariance matrix and a correlation matrix. Both are m by m matrices. - xk n xij - x j xik S = Covariance Matrix R = Correlation Matrix x1 x2 x3 … xm x1 x2 x3 … xm i =1 x 1 s1 2 s12 s13 … s1m x1 1 r12 r13 … r1m s jk = x2 s21 s22 s23 … s2m x2 r21 1 r23 … r2m n -1 x3 s31 s32 s32 … s3m x3 r31 r32 1 … r3m s jk … … … … … … … … … … … … rjk = xm sm1 sm2 sm3 … sm2 xm rm1 rm2 rm3 … 1 s j sk sj2 = sample variance of the jth variable. é xi1 - x1 n 2 sjk = sample covariance between the jth and kth variables. ê i =1 rjk = sample correlation between the jth and kth variables. ê n SSCP = ê i1 xi1 - x1 xi 2 - x SSCP = Martrix of Sum of Squares and Cross Products SSCP ê = SSCP = Xd' * Xd SSCP = X d X d ' S= ên n -1 ê xi1 - x1 xim - x In Excel the covariance calculated by COVARIANCE.S or by ë =1 adjusting the COVAR or by Data Analysis value is for the Population Covarianceiwhich divides by n rather than by (n-1). Hence to get the Sample Covariance one must correct the calculations by multiplyi dividing by (n-1). Data Analysis can be found on the Data tab in 2010 or 2007 and under tools for previous ve covariance in Data Analysis calculates both the covariances and the variances using the population formula t meaning that all values must be corrected to obtain the sample Covariance Matrix. The CORREL function in Data Analysis do not need to be corrected. 1. Total Variance of a matrix = Trace of the matrix = Sum of the diagonal elements 2. Generalized Variance of a matrix = Determinant of the matrix units one can xij - x j xik - xk n -1 xi1 - x1 n 2 xi 2 - x2 xi1 - x1 n xim - xm xi1 - x1 n ù i =1 i =1 i =1 ú ú xi1 - x1 xi 2 - x2 xi 2 - x2 xim - xm xi 2 - x2 ú n n 2 n =1 i =1 i =1 ú ú xi1 - x1 xim - xm xi 2 - x2 xim - xm xim - xm n n n 2 ú =1 which divides i =1 i =1 û he calculations by multiplying by n and then under tools for previous versions. The g the population formula that divides by n, The CORREL function and the correlation A covariance or correlation matrix, S, can be used to determine eigenvalues (or Characteristic Roots). An eigenvalue is a value λ that is a solution to the equation |S - λ*I|=0. For a matrix A, |A| denotes the determinant of the matrix A. For a value λ there is a vector V (m by 1) that satifies the matrix equation S*V = λ*V. This vector V is an eigenvector. (For each eigenvalue there is an associated eigenvector.) Important properties: For an m by m matrix S, there are m eigenvalues. The sum of these m eigenvalues is equal to the total variance of S, the trace of S. The product of these m eigenvalues is equal to the generalized variance of S, the determinant of S. é9 4 ù 9- 4 For = 11 S =ê ú , Solve =0 ë 4 3û 4 3- é9 4ù é v1 ù év ù ê ú =11 ê 1 ú ê 4 3ú v (9 - ) (3 - ) - 4 4 = 0 ë û ë 2û ëv2 û This yields 2 equations with 2 unknowns 27 -12 -16 =11-12 = (11 - ) (1 - ) = 0 2 2 9 v1 4 v2 =11 v1 11 - = 0, or 1 - = 0 4 v1 3 v2 =11 v2 = 11, or = 1 or The following was copied on 1-30-06 from Wikipedia, the free encyclopedia http://en.wikipedia.org/wiki/Eigenvectors - 2 v1 4 v2 = 0 The German word eigen was first used in this context by Hilbert in 1904 (there was an earlier related usage by Helmholtz). "Eigen" can be translated as "own", "peculiar to", 4 v1 - 8 v2 = 0 "characteristic" or "individual" emphasizing how important eigenvalues are to defining the unique nature of a specific transformation. In English mathematical jargon, the closest See the next tab for eigenvectors translation would be "characteristic"; and some older references do use expressions like "characteristic value" and "characteristic vector", or even "Eigenwert", German for eigenvalue. In the past, the standard translation used to be "proper". Today the more distinctive term "eigenvalue" is standard. distinctive term "eigenvalue" is standard. For each eigenvalue λ there is an associated eigenvector V (m by 1) that satifies the equation S*V = λ*V. é9 4 ù For S = ê ú, = 11 & = 1 ë 4 3û For = 11 For = 1 é9 4ù é v1 ù év ù é9 4 ù é v1 ù é v1 ù ê ú =11 ê 1 ú ê 4 3ú v =1 ê ú ë û ë 2û ëv2 û ê4 ë 3 ú êv2 ú û ë û ëv2 û This yields 2 equations with 2 unknowns This yields 2 equations with 2 unknowns 9 v1 4 v2 =11 v1 9 v1 4 v 2 = 1 v1 4 v1 3 v2 =11 v2 4 v1 3 v 2 = 1 v 2 or or - 2 v1 4 v2 = 0 8 v1 4 v 2 = 0 4 v1 - 8 v2 = 0 4 v1 2 v 2 = 0 There is no unique solution to the above system There is no unique solution to the above system v1 = 2 v2 Any two values of v1 and v2 satisfying the equation define an eigenvector. An v 2 = - 2 v1 eigenvector is a line or direction in the data space. A general procedure is to give a vector of length one as the eigenvector. Find the eigenvalues and eigenvectors for the covariance matrix below Matrix 1. To solve for using Excel select a 10 -2 -2 4 cell and put in an initial value for . Use this cell to create a new matrix Determinant = 36 (Matrix -*I) and use MDETERM to find = 10.60555 its determinant. Matrix - *I 2. Use Solver to find a value of -0.60555 -2 that will make the determinant = 0. -2 -6.60555 Determinant = 5.44E-07 3. Select initial values for Vector Matrix * Vector the eigenvector for the -1.19338 -12.6564 calculated eigenvalue (Vector 0.361325 3.83205 to the left), then use MMULT to 1.246876 = Length multiple the matrix times the Vector * Vector difference 4. Use scalar multiplication to multiply l -12.6564 4.01E-07 times the Vector. Take the difference 3.832049 1E-06 between this scalar product and the matrix product. 5. Use Solver to find the vector values that will make both differences = 0. There will be more than one eigenvalue and associated eigenvector so you can find the other eigenvalues by trying different starting values and for each eigenvalue the same process can be used to find the associated eigenvector. Data can be transformed from the original measurement to another variable. A new variable y1 can be defined as a linear transformation of the variable x1 with y1 = a1*x1 + b1. A new variable y2 can be defined as a linear transformation of the variable x2 with y2 = a2*x2 + b2. The coefficients a1 and a2 must not be zero. y1 = a1 x1 b1 ( a1 0) & y 2 = a2 x2 b2 ( a2 0 y1 = a1 x1 b1 & y 2 = a2 x2 b2 VAR ( y1 ) = s 21 = a1 VAR ( x1 ) or s y1 = a1 s x1 y 2 VAR ( y 2 ) = s 22 = a2 VAR ( x2 ) or s y2 = a2 s x2 y 2 COV ( y1 , y 2 ) = s y1 y2 = a1 a2 COV ( x1 , x2 ) = a1 a2 ry1 , y2 = rx1 , x2 é a1 ù é b1 ù X = x1 x 2 , A = ê ú , B = ê ú , Y = y1 y 2 , Y = X A ë a2 û ëb2 û X = x1 x2 , Y = y1 y2 , Y = X A B é sx 2 sx1 x2 ù é s2 s y1 y2 ù SX = ê 1 2 ú , SY = ê y1 2 ú , SY = A'S X A êsx1 x2 ë s x2 úû ês y1 y2 ë s y2 ú û 1 with y1 = a1*x1 + b1. 2 with y2 = a2*x2 + b2. x2 b2 ( a2 0) = a1 s x1 = a 2 s x2 ( x1 , x2 ) = a1 a2 s x1 x2 y 2 , Y = X A B Let a ' = a1 a2 ... am A vector describes a direction in space. The Length of the vector a = ||a|| m Suppose m=3 & a' = [10 2 11] a = a 2 a = 102 22 112 = 100 4 121 = 225 =15 j j A vector is normalized if its length is 1. To normalize a vector, multiply the vector by the inverse o [10/15 2/15 11/15] is the normalized vector for the vector a. Euclidean Distance between the two points a & b = ||a - b|| a -b = a1 - b1 2 a2 - b2 2 ... am - bm 2 Suppose m=3 & b' = [8 3 13] a -b = 10-8 2 -3 11-13 = 4 1 4 = 9 = 3 2 2 2 Two Vectors a and b are ORTHOGONAL if and only if a'•b=0 a' b a'•b 10 2 11 8 229 =10*8+2*3+11*13 3 Hence, a & b are not orthogonal. 13 escribes a direction in multiple dimensional vector by the inverse of its length. #1 in Chapter 2 of the 2002 edition of the Stevens multivariate text A B C D E X 2 4 1 1 2 1 3 5 4 2 1 -1 2 1 2 3 -2 5 2 1 6 2 1 2 6 -1 3 1 3 1 3 4 2 1 10 4 6 u' v 5 7 1 3 2 7 Find a) A + C b) A + B c) A*B d) A*C e) u' * D * u f) u' * v g) (A + C)' h) 3 * C i) Determinant of D j) Inverse of D = D-1 k) Determinant of E l) Inverse of E m) u' * D-1 * u n) B*A (compare result to c above) o) X' * X Find the eigenvalues & eigenvectors for each covariance matrix. a. 4 -4 1 = 12 eigenvector = [v1, v2]', where -2*v1 = v2 -4 10 2 = 2 eigenvector = [v1, v2]', where v1 =2*v2 Product = 24 24 = Generalized Variance b. 21 12 1 = 29 eigenvector = [v1, v2]', where 2*v1 = 3*v2 12 11 2 = 3 eigenvector = [v1, v2]', where -3*v1 =2*v2 Product = 87 87 = Generalized Variance c. 1 -0.632456 1 = 1.632456 eigenvector = [v1, v2]', where -v1 = v2 -0.632456 1 2 = 0.367544 eigenvector = [v1, v2]', where v1 =v2 Product = 0.6 0.6 = Generalized Variance Find the distance between these two points in 3-dimensional space Variable 1 Variable 2 Variable 3 Point 1 1 2 3 2.236068 =SQRT((1-3)^2+(2-2)^2+(3-4)^2) Point 2 3 2 4 = 12 Matrix Matrix - *I 10 -4 -2 -4 -4 4 -4 -8 Determinant = 0 difference Vector Matrix * Vector * Vector 2 24 24 0 -1 -12 -12 0 2.236068 = Length re -2*v1 = v2 re 2*v1 = 3*v2 re -3*v1 =2*v2 2-2)^2+(3-4)^2) X1 X2 X3 10 1 7 Use the last six digits of your student 5 2 7 number for X3. The values in X3 are for 7 4 1 0 4 9 someone with 771906 as the last 6 digits 2 5 0 of their student number. 6 2 6 a. Find the covariance matrix for the variables X1 & X2 b. Find the correlation matrix for the variables X1 & X2 c. Find the mean corrected or centered values for the variables X1 & X2 d. Find the covariance matrix for the mean corrected or centered values for the variables X1 & X2 e. Find the correlation matrix for the mean corrected or centered values for the variables X1 & X2 f. Find the standardized values for the variables X1 & X2 g. Find the covariance matrix for the standardized for the variables X1 & X2 h. Find the correlation matrix for the standardized values for the variables X1 & X2 i. Find the eigenvalues & an eigenvector for each eigenvalue for the covariance matrix for the variables X1 & X2 j. Find the eigenvalues & an eigenvector for each eigenvalue for the correlation matrix for the variables X1 & X2 k. Find the covariance matrix for the variables X1, X2 & X3 l. Find the correlation matrix for the variables X1, X2 & X3 bles X1 & X2 les X1 & X2 ix for the variables X1 & X2 x for the variables X1 & X2 last 6 digits = 771906 X1 X2 X3 digit Xd1 Xd2 Xd3 Z1 Z2 10 1 7 1 5 -2 2 1.39754 -1.291 5 2 7 2 0 -1 2 0 -0.6455 7 4 1 3 2 1 -4 0.55902 0.6455 0 4 9 4 -5 1 4 -1.3975 0.6455 2 5 0 5 -3 2 -5 -0.8385 1.29099 6 2 6 6 1 -1 1 0.27951 -0.6455 Mean 5 3 5 0 0 0 0 0 Variance 12.8 2.4 13.2 12.8 2.4 13.2 1 1 Std. Dev. 3.57771 1.54919 3.63318 3.57771 1.54919 3.63318 1 1 Covariance Matrix Covariance Matrix Covariance Matrix 12.8 -4 -0.4 12.8 -4 -0.4 1 -0.7217 -4 2.4 -3.4 -4 2.4 -3.4 -0.7217 1 -0.4 -3.4 13.2 -0.4 -3.4 13.2 -0.0308 -0.6041 Correlation Matrix Correlation Matrix Correlation Matrix 1 -0.72169 -0.03077 1 -0.7217 -0.0308 1 -0.7217 -0.72169 1 -0.60407 -0.7217 1 -0.6041 -0.7217 1 -0.03077 -0.60407 1 -0.0308 -0.6041 1 -0.0308 -0.6041 X1 X2 X3 X1 1 X2 -0.72169 1 X3 -0.03077 -0.60407 1 1.2 -0.86603 1.2 -0.03693 -0.72488 1.2 Z3 Xd1*Xd2 Xd1*Xd3 Xd2*Xd3 0.55048 -10 10 -4 0.55048 0 0 -2 -1.101 2 -8 -4 1.10096 -5 -20 4 -1.3762 -6 15 -10 0.27524 -1 1 -1 0 Sum -20 -2 -17 1 Cov -4 -0.4 -3.4 1 ovariance Matrix -0.0308 Covariance Matrix of the -0.6041 Standardized Data is the 1 Correlation Matrix of the Original orrelation Matrix Data. -0.0308 -0.6041 1 Covariance Matrix Correlation Matrix 12.8 -4 1 -0.72169 -4 2.4 -0.72169 1 Trace = 15.2 = Total Variance Trace = 2 Determinant = 14.72 = Generalized Variance Determinant = 0.479167 = 2 = 0.5 Matrix - *I Matrix - *I 10.8 -4 0.5 -0.72169 -4 0.4 -0.72169 0.5 Determinant = -11.68 Determinant = -0.27083 Eigenvalue 1 = 14.16049 15.2 = sum of eigenvalues Eigenvalue 1 = 1.721688 Eigenvalue 2 = 1.039512 14.72 = product of eigenvalues Eigenvalue 2 = 0.278313 Eigenvalues found by varying starting value & using solver Eigenvalues found by varying starting value & u = Total Variance = Generalized Variance 2.000001 = sum of eigenvalues 0.479167 = product of eigenvalues varying starting value & using solver Singular Value Decomposition X = data matrix W = orthogonal rotation, where W'=W-1 D-1 = diagonal stretching/shrinking matrix Zs = Matrix of standardized data Zs = X•W•D-1 then Zs•D = X•W•(D-1•D) = X•W•I = X•W Zs•D = X•W then Zs•D•W' = X•W•W' = X Data Matrix = (Standardized Data)*(Scaling Matrix)*(Orthogonal Rotation Matrix) on Matrix)