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FINS4779/5579

Week 2Matrix Algebra

Min Kim

March 5, 2012

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1 Matrix de…nition and notation

Estimator = f (data) ) estimate

year S&P500 (%)

2000 -9.102001 -11.89

2002 -22.10

2003 28.68

2004 10.88

2005 4.91

2006 15.79

year S&P500 (%) NASDAQ (%)

2000 -9.10 64.42001 -11.89 6.39

2002 -22.10 -7.09

2003 28.68 -3.74

2004 10.88 7.21

2005 4.91 -2.13

2006 15.79 -39.05

one-dim data (time-series) two-dim data (panel)

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Data1 =

266666666664

9:10

11

:89

22:10

28:68

10:88

4:91

15:79

377777777775

Data2 =

266666666664

9:10 64:4

11

:89 6

:39

22:10 7:09

28:68 3:74

10:88 7:21

4:91 2:13

15:79 39:05

377777777775

7 rows and 1 column (size 71) 7 rows and 2 columns (size 72)

matrix A of size m n =

26664

a11 a12 ::: a1n

a21 a22 ::: a2n

:: ::am1 am2 ::: amn

37775 = (aij)i=1;:::;m; j=1;:::;n

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vector a =264a1

:::am

375; (column) vecor or

ha1 a2 ::: an

i; row vector

square matrix: m = n (ex)

"1 00 1

#

diagonal elements of a square matrix = aii where i = 1; 2; :::m

diagonal matrix: aij = 0 where i 6= j

identity matrix (I m): diagonal matrix with aii = 1 where i = 1; 2; :::m

zero matrix 0: aij = 0 for all i; j

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2 Matrix addition and multiplication

A + = (aij) + = + (aij) = + A where is a number (scalar)

A + B = (aij) + (bij) = (aij + bij) where both sizes are m n

A = (aij) = (aij) = (aij) = A

(ex) = 2; A =

"1 00 1

#; B =

"1 11 1

#

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c = ab where a is 1 p and B is p 1

c =h

a1 a2 ::: a p

i26664

b1

b2::

b p

37775 = a1b1 + a2b2 + ::: + a pb p

(ex) a =h

1 1 1 1i

; b =

266641111

37775

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C  = AB = (cij) = (A)i row(B) j column where A is m  p and B is p n

26664c11 c12 ::: c1nc21 a22 ::: c2n

:: ::

cm1 cm2 ::: cmn

37775 =

!26664a11 a12 ::: a1 pa21 a22 ::: a2 p

:: ::

am1 am2 ::: amp

37775 #

26664b11 b12 ::: b1nb21 b22 ::: b2n

:: ::

b p1 b p2 ::: b pn

37775

c11 =h

a11 a12 ::: a1 pi

26664

b11b21

::b p1

37775

= a11b11 + a12b21 + ::: + a1 pb p1 = pX

k=1

a1kbk1

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26664

c11 c12 ::: c1n

c21 a22 ::: c2n

:: ::

cm1 cm2 ::: cmn

37775 =

!26664

a11 a12 ::: a1 p

a21 a22 ::: a2 p

:: ::

am1 am2 ::: amp

37775 #

26664

b11 b12 ::: b1n

b21 b22 ::: b2n

:: ::

b p1 b p2 ::: b pn

37775

cij =h

ai1 ai2 ::: aip

i

26664b1 j

b2 j

::

b pj

37775

= ai1b1 j + ai2b2 j + ::: + aipb pj = p

Xk=1

aikbkj

(ex) A =

"1 00 1

#; B =

"1 11 1

#; B =

"21

#; A =

"0 21 1

#

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Theorem 1 (assume the sizes are well-speci…ed)

A + B = B + A

(A + B) + C  = A + (B + C )

(A + B) = A + B

( + )A = A + A

A A = A + (A) = 0

A(B + C ) = AB + AC 

(A + B)C  = AC  + BC 

(AB)C  = A(BC )

Let the LHS be a matrix L and the RHS be a matrix R. To show L = R;

simply show (L)ij = (R)ij

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3 Transpose

A = (aij)0s transpose: B = (bij) = (a ji)

A =

26664

a11 a12 ::: a1n

a21 a22 ::: a2n

:: ::am1 am2 ::: amn

37775) B =

26664

a11 a21 ::: am1a12 a22 ::: am2

:: ::a1n a2n ::: amn

37775

The transpose of  A is denoted by A0:

(ex) A =

"1 00 1

#;

"1 11 1

#;

2641

23

375 ;

h1 2 3

i;

2641 2 3 4

7 0 8 97 6 6 5

375

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symmetric matrix : A = A0, i.e., aij = a ji

(ex) A =

"1 00 1

#;

"1 22 1

#;

2641

23

375

Theorem 2

(A)0 = A0

(A0)0 = A

(A + B)

0

= A

0

+ B

0

(AB)0 = B0A0

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4 Inverse

Assume that a square matrix A of size m m is nonsingular (the inverse of 

A exists). The inverse of  A is a matrix, A1; such that

AA1 = A1A = I m =26664

1 0 ::: 0

0 1 0::: ::: ::: :::

0 0 ::: 1

37775

A ="

a bc d

#) A = 1

ad bc

"d bc a

#

(ex) A =

"1 00 1

#;

"1 20 1

#;

"0 11 7

#

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Is the inverse unique?

Theorem 3.

(A)1 = 1A1 =1

A1

(A0)1 = (A1)0

(A1)1 = A

(AB)1 = B1A1

If  A is a diagonal matrix with diagonal elements (aii),

then A1

is a diagonal matrix with (a

1

ii ) or (

1

aii)

If  A is symmetric, then A1is also symmetric.