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
Ax=b
A\mathbf{x} = \mathbf{b}
Ax=0
A\mathbf{x} = \mathbf{0}
Ax=λx
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
rank < n,m
Non-zero pivots
zero pivots
rows with all 0s after GE
1 solution
1~solution
∞ solutions
\infty~solutions
0 or 1 solution
0~or~1~solution
0 or ∞ solutions
0~or~\infty~solutions
\begin{bmatrix}
~~~&~~~&~~~\\
~~~&~~~&~~~\\
~~~&~~~&~~~\\
\end{bmatrix}
rank=m=n
rank=m=n
rank=m<n
rank=m < n
\begin{bmatrix}
~~~&~~~&~~~&~~~&~~~\\
~~~&~~~&~~~&~~~&~~~\\
~~~&~~~&~~~&~~~&~~~\\
\end{bmatrix}
\underbrace{~~~~~~~~~~~~~~~~~~}
\underbrace{~~~~~~~~~~~~~~~}
\begin{bmatrix}
~~~&~~~&\\
~~~&~~~&\\
~~~&~~~&\\
~~~&~~~&\\
~~~&~~~&\\
~~~&~~~&
\end{bmatrix}
No Free columns
rank=n<m
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)=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=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
det(I)=1
det(I)=1
P=010100001
P = \begin{bmatrix}
0&1&0\\
1&0&0\\
0&0&1\\
\end{bmatrix}
det(P)=−1
det(P) = -1
P=010001100
P = \begin{bmatrix}
0&0&1\\
1&0&0\\
0&1&0\\
\end{bmatrix}
det(P)=−(−1)=1
det(P) = -(-1) = 1
det(Apermute)=(−1)kdet(A)
det(A_{permute}) = (-1)^k det(A)
k= number of row exchanges required to get from A to Apermute
Properties of determinants
Property 3a.
A=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
takctbkd
\begin{vmatrix}
ta&tb\\
kc&kd\\
\end{vmatrix}
=ktacbd
= kt\begin{vmatrix}
a&b\\
c&d\\
\end{vmatrix}
Property 3b.
a+a′cb+b′d
\begin{vmatrix}
a+a'&b+b'\\
c&d\\
\end{vmatrix}
=acbd
= \begin{vmatrix}
a&b\\
c&d\\
\end{vmatrix}
+a′cb′d
+ \begin{vmatrix}
a'&b'\\
c&d\\
\end{vmatrix}
tactbd
\begin{vmatrix}
ta&tb\\
c&d\\
\end{vmatrix}
=tacbd
= t\begin{vmatrix}
a&b\\
c&d\\
\end{vmatrix}
det(2A)=
det(2A)=
2ndet(A)
2^ndet(A)
(With these two properties we have covered linear combinations of rows)
det(A+B)=det(A)+det(B)
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=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
Exchange the two rows
Proof:
A=Aexchange⟹det(A)=det(Aexchange)
A = A_{exchange} \implies det(A) = det(A_{exchange})
But from property 2
det(A)=−det(Aexchange)
det(A) = -det(A_{exchange})
∴det(A)=det(Aexchange)=0
\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=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
(by example)
Informal Proof:
A=[acbd]
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
A′=[ac−labd−lb]
A'=\begin{bmatrix}
a & b\\
c - la & d - lb
\end{bmatrix}
r2 = r2 - l*r1
det(A′)=ac−labd−lb
det(A')=\begin{vmatrix}
a & b\\
c - la & d - lb
\end{vmatrix}
=acbd−alablb
=\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
-\begin{vmatrix}
a & b\\
la & lb
\end{vmatrix}
=acbd−laabb
=\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
-l\begin{vmatrix}
a & b\\
a & b
\end{vmatrix}
=det(A)−l∗0=det(A)
=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=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
Proof:
A=ac…bd…………
A=\begin{bmatrix}
a & b&\dots\\
c & d&\dots\\
\dots&\dots&\dots
\end{bmatrix}
det(A′)=t⋅det(A)
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
if~t=0
Let A′=tac…tbd…………
Let~A'=\begin{bmatrix}
ta & tb&\dots\\
c & d&\dots\\
\dots&\dots&\dots
\end{bmatrix}
A′=0c…0d…………
A'=\begin{bmatrix}
0 & 0&\dots\\
c & d&\dots\\
\dots&\dots&\dots
\end{bmatrix}
det(A′)=0⋅det(A)
det(A') = 0\cdot det(A)
=0
= 0
Properties of determinants
Property 7. If U is an upper triangular matrix
A=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
Proof:
U=a1100…0∗a220…0∗∗a33…0…………ann
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=a1100…00a220…000a33…0…………ann
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)
det(U) = det(R)
by property 5
det(U)=a11⋅a22⋅a33…ann
det(U) = a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}
Properties of determinants
A=[acbd]det(A)=ad−bc
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=a1100…00a220…000a33…0…………ann
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)=a11⋅a22⋅a33…ann⋅det(I)
det(R) = a_{11}\cdot a_{22}\cdot a_{33}\dots a_{nn}\cdot det(I)
by property 3a
I=100…0010…0001…0…………1
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}
=a11⋅a22⋅a33…ann
= 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)=a11⋅a22⋅a33…ann
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=[acbd]det(A)=ad−bc
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=[acbd]det(A)=ad−bc
A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
det(A) = ad - bc
Proof:
Implications:
HW4
A−1A=I
A^{-1}A=I
∴det(A−1A)=det(I)
\therefore det(A^{-1}A)=det(I)
∴det(A−1)det(A)=1
\therefore det(A^{-1})det(A)=1
∴det(A−1)=det(A)1
\therefore det(A^{-1})=\frac{1}{det(A)}
det(A2)=det(AA)
det(A^2)=det(AA)
=det(A)det(A)
=det(A)det(A)
=det(A)2
=det(A)^2
Properties of determinants
Property 10. det(A⊤)=det(A)
A=[acbd]det(A)=ad−bc
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 ⟹ the determinant is 0
elementary column operations do not change the determinant
Learning Objectives
(achieved)
What are the properties of determinants?
CS6015: Linear Algebra and Random Processes Lecture 14: Properties of determinants
CS6015: Lecture 14
By Mitesh Khapra
CS6015: Lecture 14
Lecture 14: Properties of determinants
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