CS6015: Linear Algebra and Random Processes

Lecture 14:  Properties of determinants

Learning Objectives

What are the properties of determinants?

(for today's lecture)

What is the formula for computing a determinant?

What are co-factors?

Where are we?

Linear transformation

A\mathbf{x} = \mathbf{b}
A\mathbf{x} = \mathbf{0}
A\mathbf{x} = \lambda\mathbf{x}
matrix
vector
vector
matrix
vector
vector
matrix
vector
vector
scalar

Linear equations

(why? we will see soon)
(why? we will see soon)

Linear combinations

Linear independence

What have we mainly focussed on?

Pivot columns

Free columns

rank < n,m

Non-zero pivots

zero pivots

rows with all 0s after GE

1~solution
\infty~solutions
0~or~1~solution
0~or~\infty~solutions
\begin{bmatrix} ~~~&~~~&~~~\\ ~~~&~~~&~~~\\ ~~~&~~~&~~~\\ \end{bmatrix}
rank=m=n
rank=m < n
\begin{bmatrix} ~~~&~~~&~~~&~~~&~~~\\ ~~~&~~~&~~~&~~~&~~~\\ ~~~&~~~&~~~&~~~&~~~\\ \end{bmatrix}
\underbrace{~~~~~~~~~~~~~~~~~~}
\underbrace{~~~~~~~~~~~~~~~}
\begin{bmatrix} ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~& \end{bmatrix}

No Free columns

rank=n < m
\begin{bmatrix} ~~~&~~~&~~~&~~~&~~~\\ ~~~&~~~&~~~&~~~&~~~\\ ~~~&~~~&~~~&~~~&~~~\\ \end{bmatrix}

Pivot columns

\underbrace{~~~~~~~~~~~~~}
\underbrace{~~~~~~~~~~~~~~~~~~~~}

Free columns

\begin{bmatrix} ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~&\\ ~~~&~~~& \end{bmatrix}
\underbrace{~~~~~~}

Free

Pivot

\underbrace{~~~~~~~~~}

Rectangular matrices!

What will we focus on now?

Square matrices!

invertible (non-singular)

non-invertible (singular)

One test for invertibility: \( det(A) \neq 0 \)

A formula for determinant

We will follow Prof. Strang's approach

We will discuss properties of determinants

And then arrive at a formula based on these properties

Properties of determinants

Property 1.  

Property 2. row exchanges reverse the sign of the determinant

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc
det(I)=1
P = \begin{bmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{bmatrix}
det(P) = -1
P = \begin{bmatrix} 0&0&1\\ 1&0&0\\ 0&1&0\\ \end{bmatrix}
det(P) = -(-1) = 1
det(A_{permute}) = (-1)^k det(A)

\(k=\) number of row exchanges required to get from \(A\) to \(A_{permute}\)

Properties of determinants

Property 3a.  

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc
\begin{vmatrix} ta&tb\\ kc&kd\\ \end{vmatrix}
= kt\begin{vmatrix} a&b\\ c&d\\ \end{vmatrix}

Property 3b. 

\begin{vmatrix} a+a'&b+b'\\ c&d\\ \end{vmatrix}
= \begin{vmatrix} a&b\\ c&d\\ \end{vmatrix}
+ \begin{vmatrix} a'&b'\\ c&d\\ \end{vmatrix}
\begin{vmatrix} ta&tb\\ c&d\\ \end{vmatrix}
= t\begin{vmatrix} a&b\\ c&d\\ \end{vmatrix}
det(2A)=
2^ndet(A)

(With these two properties we have covered linear combinations of rows)

det(A+B) \neq det(A) + det(B)
(generally not equal)

Properties of determinants

Property 4.  If \(A\) has 2 equal rows, \(det(A) = 0 \)

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Exchange the two rows

Proof:

A = A_{exchange} \implies det(A) = det(A_{exchange})

But from property 2

det(A) = -det(A_{exchange})
\therefore det(A) = det(A_{exchange}) = 0

Properties of determinants

Property 5.  Elementary row operations on a matrix (as in GE) do not change the determinant of the matrix

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

(by example)

Informal Proof:

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}
A'=\begin{bmatrix} a & b\\ c - la & d - lb \end{bmatrix}
r2 = r2 - l*r1
det(A')=\begin{vmatrix} a & b\\ c - la & d - lb \end{vmatrix}
=\begin{vmatrix} a & b\\ c & d \end{vmatrix} -\begin{vmatrix} a & b\\ la & lb \end{vmatrix}
=\begin{vmatrix} a & b\\ c & d \end{vmatrix} -l\begin{vmatrix} a & b\\ a & b \end{vmatrix}
=det(A) - l * 0 = det(A)
property 3b
property 3a

Implication: If \(A = LU\)  then \(det(A) = det(U) \)

property 4

Properties of determinants

Property 6.  If \(A\) has a row of zeroes then \( det(A) = 0 \)

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

A=\begin{bmatrix} a & b&\dots\\ c & d&\dots\\ \dots&\dots&\dots \end{bmatrix}
det(A') = t\cdot det(A)
property 3b

Implication: If after GE, \(U\) has a row of 0s then \(det(A) = det(U) = 0\)

from properties 5 & 6
if~t=0
Let~A'=\begin{bmatrix} ta & tb&\dots\\ c & d&\dots\\ \dots&\dots&\dots \end{bmatrix}
A'=\begin{bmatrix} 0 & 0&\dots\\ c & d&\dots\\ \dots&\dots&\dots \end{bmatrix}
det(A') = 0\cdot det(A)
= 0

Properties of determinants

Property 7.  If \(U\) is an upper triangular matrix

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

U=\begin{bmatrix} a_{11} & * & * & \dots\\ 0 & a_{22} & * & \dots\\ 0 & 0 & a_{33} &\dots \\ \dots&\dots&\dots&\dots \\ 0 & 0 & 0 & a_{nn} \\ \end{bmatrix}
continue elimination to get row reduced form - this does not change the diagonal elements
R=\begin{bmatrix} a_{11} & 0 & 0 & \dots\\ 0 & a_{22} & 0 & \dots\\ 0 & 0 & a_{33} &\dots \\ \dots&\dots&\dots&\dots \\ 0 & 0 & 0 & a_{nn} \\ \end{bmatrix}
det(U) = det(R)
by property 5
det(U) = a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}

Properties of determinants

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

Implication: Once you do GE, you have an easy way of computing \(det(A)\)

R=\begin{bmatrix} a_{11} & 0 & 0 & \dots\\ 0 & a_{22} & 0 & \dots\\ 0 & 0 & a_{33} &\dots \\ \dots&\dots&\dots&\dots \\ 0 & 0 & 0 & a_{nn} \\ \end{bmatrix}
det(R) = a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}\cdot det(I)
by property 3a
I=\begin{bmatrix} 1 & 0 & 0 & \dots\\ 0 & 1 & 0 & \dots\\ 0 & 0 & 1 &\dots \\ \dots&\dots&\dots&\dots \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}
= a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}
from properties 5 & 7

Property 7.  If \(U\) is an upper triangular matrix

det(U) = a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}

Properties of determinants

Property 8.  \(det(A) = 0\) when \(A\) is singular

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

Go from \(A\) to \(U\) using GE

by property 7
(determinant does not change - by property 5)

\(A\) is not invertible if there will be a 0 pivot

If there is a 0 pivot (diagonal element) in \(U\), \( det(U) = 0 \)

(remember, A is square)

Hence, proved

Properties of determinants

Property 9.  \(det(AB) = det(A) det(B)\) 

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

Implications:

HW4

A^{-1}A=I
\therefore det(A^{-1}A)=det(I)
\therefore det(A^{-1})det(A)=1
\therefore det(A^{-1})=\frac{1}{det(A)}
det(A^2)=det(AA)
=det(A)det(A)
=det(A)^2

Properties of determinants

Property 10.  \(det(A^\top) = det(A)\) 

A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ det(A) = ad - bc

Proof:

Implications:

HW4

All the properties that we saw for rows are applicable for columns too

exchanging two columns reverses the sign of the determinant

a column of zeroes \(\implies\) the determinant is 0

elementary column operations do not change the determinant

Learning Objectives

(achieved)

What are the properties of determinants?

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