Determinants
𝑎 𝑏
A=[
]
𝑐 𝑑
A is invertible iff 𝑎𝑑 − 𝑏𝑐 ≠ 0
1
𝑑 −𝑏
[
]
𝑎𝑑 − 𝑏𝑐 −𝑐 𝑎
𝑎𝑑 − 𝑏𝑐 Is the determinant of A det 𝐴 = det(𝐴) = 𝑎d-bc
𝐴−1 =
Defintion:
If A is an nxn matrix, it’s determinant is defined as follows:
1
1) If n=1, 𝐴 = [𝛼], then det A=𝛼, A is invertible iff 𝛼 ≠ 0. 𝐴−1 = 𝛼 [𝛼]
2) Otherwise if n>1, and 𝐴 = 𝑎𝑖𝑗 then for any i
𝑛
det 𝐴 = ∑((−1)𝑖+𝑗 𝑎𝑖𝑗 ∙ det(𝑀𝑖𝑗 ))
𝑗=1
where 𝑀𝑖𝑗 is the matrix obtained from A by deleting row i and column j.
𝑎
det 𝑏
𝑐
Also good for rows.
+
−
∗
+
− −
+
− = 𝑎 det ( ∗ ∗ ) − 𝑏𝑑𝑒𝑡 (
∗
∗
+
+ +
) + 𝑐𝑑𝑒𝑡 (
)
∗
− −
Thm
The determinant function det: [nxn matrix] -> ℝ
has the following properties:
1) Det(AB) =(det A)(det B)
2) det(𝐴−1 ) = 1/ det 𝐴
3) 𝐷𝑒𝑡 𝐴𝑇 = det 𝐴
4) det(𝑘𝐴) = 𝑘 𝑛 (det 𝐴)
Elemantair row operation marix determinants
Swap (E1) -- det 𝐸1 = λ
Multiply 1 row by (E2() -- det 𝐸2 = λ
add multiple of one row to row (E3) -- det 𝐸3 = 1
Thm
is a square matrix is in upper triangular then it’s determinant is the product of the diagonal entries of
the matrix.
Corrallory: a matrix in echelon form is automatically upper triangular.
so if matrix is in echelon form then its determinant is the product of the diagonals.
Finding the derteminant of uning row reductuion
0 2 3
2 7 4 has det =d
1 1 5
1 1 5
2 7 4 R1→ R3 (type 1)
0 2 3
1 1 5
0 5 −6
0 2 3
has det =-d
R2—> R2-2R1 (type 3) has det =-d
1 1 5
6
𝑑
0 1 − 5 R2—> R2/5 (type 2) had det=−
5
0 2 3
1 1 5
6
0 1 − R3-2R2 (type 3) had det=− 𝑑
5
5
27
0 0
5
−
𝑑
27
=1 ×1×
⇒ 𝒅 = −𝟐𝟕
5
5
Thrm
A matrix is invertible iff it’s determinate is not zero.
Proof:
let A’ be the echeloen form of A
𝐷𝑒𝑡 𝐴′ = 𝛼 det 𝐴 ,
𝛼 ≠ 0 α is product of all determinants of the row operations done on it
If A is invertible A’ has non-zero entries on the diagonals.
Geometric interpretation of determinant
𝑎
(
𝑐
𝑏
)
𝑑
Imagine a parellagram. Coners on orign, (a,b) ,(a+c,b+d),(c,d)
The determinate is the area of that parrellagram.
The area is zero when the matrix is not invertible
In 3D we ger a parallel pyperd (3D solid with faces all parallelograms) where the volume is the
determinant.
In n-dimensions, the determinant is the nD-hypervolume
If the matix is not invertible the (hyper) solid is clapseds so it does have a (hyper) volume
EigenVector:
Definition;
Let A be an nxn matrix
A nonzero vertor 𝑣̃ ∈ ℝ𝑛 is called an eigenvector of A with eigen values λ if
𝐴𝑣̃ = 𝜆𝑣̃
A marix of size nxn can have atmost n eigenvalues.
Finding Eigenvectors and egen values
If
𝐴𝑣̃ = 𝜆𝑣̃
(𝜆𝐼 − 𝐴)𝑣̃ = 0̃
(𝜆𝐼 − 𝐴) is a null space for v.
For A to be invertible det (𝜆𝐼 − 𝐴)=0
1) Solve3 (𝜆𝐼 − 𝐴) = 0 to find possible eigenvalues (λ)
2) Solve
(𝜆𝐼 − 𝐴)𝑣̃ = 0̃
to find v.
Intro:
Matrix A, we are theyin iot find vectors such that the matrix just scales the vector.
ie: 𝐴𝑣̃ = 𝜆𝑣̃
where A is a matrix, λ is a scalar, and 𝑣̃ is a vector.
2 Steps:
1) f9ind possible eigenvalues:
𝑒𝑔 𝐴𝑣̃ = 3𝑣̃
therefore (𝐴 − 3𝐼)𝑣̃ = 0̃
λ is a eigen values iff: |𝜆𝐼 − 𝐴| = 0
remember: |𝜆𝐼 − 𝐴| = det( 𝜆𝐼 − 𝐴)
Example: 𝐴 =
7 −4
𝜆
, 𝜆𝐼 =
2 1
0
𝜆−7
𝜆𝐼 − 𝐴 = [
−2
0
𝜆
4
]
𝜆−1
|𝜆𝐼 − 𝐴| = (𝜆 − 7)(𝜆 − 1) + 8 = (𝜆 − 3)(𝜆 − 5)
Eigenvalues for A are 3 and 5
The chartaceristic polynomial of A, ϕ(A,λ ), 𝐶𝐴 (𝜆)= (𝜆 − 3)(𝜆 − 5)
2
Solve 𝐴𝑣̃ = 𝜆𝑣̃
As Eigenvalues for A are 3 and 5l. λ =3,5
=
For Eigenvalue λ=3
Solve [
𝑣1
3𝑣
7 −4 𝑣1
] [ ] = 3 [𝑣 ] = [ 1 ]
3𝑣2
2 1 𝑣2
2
Solution: (use guaisan elimination after rearaging to put variables on left side)
𝑣2 𝑖𝑠 𝑓𝑟𝑒𝑒
𝑣1 = 𝑣2
eigenspace of A,3 is 𝐸(𝐴, 3) = 𝑠𝑝𝑎𝑛 {(1,1)} = {(𝑣2 , 𝑣2 )|𝑣2 ∈ ℝ}
For Eigenvalue λ=5
𝑣1
5𝑣
7 −4 𝑣1
solve [
] [ ] = 5 [𝑣 ] = [ 1 ]
5𝑣2
2 1 𝑣2
2
Eigenspace: span{(2,1)}
Check: [
7 −4 2
10
2
] [ ] = [ ] = 5 [ ] correct
2 1 1
5
1
Example 2:
6
8
4
𝐴 = [−2 −2 −2]
−1 −2 1
What are the eigenvalues and eigen vectors?
1) find the charascteristic polyniomial. When it is zero the siolutioons to it are the only possible
eiogen values:
𝜆 − 6 −8
−4
|𝜆𝐼 − 𝐴| = | 2
𝜆−2
2 | = (𝜆 − 1)1 (𝜆 − 2)2
1
2
𝜆−1
has eigenvalues λ = 1,2
the indexes 1,2 are called Alegbraix multiplicity 1,2 as eigenvalues.
that means that the \lamda=1 root offucrs onxe anf λ =2 occurs twice sortof
Solve: 𝐴𝑣 = 1𝑣 and Av=2v
𝐴𝑣 = 1𝑣 has eigenspace of dimension 1.
Av=2v has eigenspace of dimension 2.
The dimensions of the eigenspace is the geomantic Multiplicities.
FINAL THEORM
For any eigenvalue the
(geometric multiplicity) ≤(algebraic multiplicity)
1
2
(𝑥 + ) (𝑥 − 1) + = 0
6
6
1
2
1
𝑥=
3
𝑥=