Definition. Given n vectors v1 , · · · , vn ∈ F n , the determinant det(v1 , . . . , vn ) is a
scalar in F such that
(1) Multilinearity: det is linear in each argument
det(v1 , . . . , vi−1 , u + w, vi , . . . , vn ) =
det(v1 , . . . , vi−1 , u, vi , . . . , vn )
+det(v1 , . . . , vi−1 , w, vi , . . . , vn )
det(v1 , . . . , vi−1 , cu, vi , . . . , vn ) = c det(v1 , . . . , vi−1 , u, vi , . . . , vn )
(2) Alternating: det changes sign when you switch neighboring arguments
det(v1 , . . . , vi , vi+1 , . . . , vn ) = −det(v1 , . . . , vi+1 , vi , . . . , vn )
(3) Normalization: if {e1 , . . . , en } is the standard basis then
det(e1 , . . . , en ) = 1
If A is an n × n-matrix with columns v1 , . . . , vn , we define det(A) = det(v1 , . . . , vn ).
Proposition. The determinant is uniquely defined.
a b
Example. det
= ad − bc. We need to check that the axioms for the
c d
determinant hold:
(1) Multilinearity
det
a1 + a2
c1 + c2
b
d
= (a1 + a2 )d − b(c1 + c2 )
a1
c1
b
d
a2
c2
b
d
= (a1 d − bc1 ) + (a2 d − bc2 ) = det
+ det
;
a b 1 + b2
det
= a(d1 + d2 ) − (b1 + b2 )c
c d1 + d2
a b1
a b2
= (ad1 − b1 c) + (ad2 − b2 c) = det
+ det
;
c d1
c d2
sa b
a b
det
= (sa)d − b(sc) = s(ad − bc) = s det
;
sc d
c d
a sb
a b
det
= a(sd) − (sb)c = s(ad − bc) = s det
.
c sd
c d
(2) Alternating
c d
a b
det
= cb − ad = −(ad − bc) = −det
a b
c d
(3) Normalization
det
1
0
0
1
=1·1−0·0=1
Proposition. Given n × n-matrices A, B, det(AB) = det(A)det(B)
Proof. Suppose B has columns v1 , . . . , vn . Then AB has columns Av1 , . . . , Avn .
Thus,
det(AB)/det(A) = det(Av1 , . . . , Avn )/det(A)
Since the properties of the determinant uniquely define it, we need to check that
they hold for det(Av1 , . . . , Avn )/det(A).
1
2
(1) Multilinearity: This follows from linearity of matrix multiplication.
det(Av1 , . . . , A(u + w), . . . , Avn )/det(A)
= det(Av1 , . . . , Au + Aw, . . . , Avn )/det(A)
= (det(Av1 , . . . , Au, . . . , Avn ) + det(Av1 , . . . , Aw, . . . , Avn ))/det(A)
= det(Av1 , . . . , Au, . . . , Avn )/det(A) + det(Av1 , . . . , Aw, . . . , Avn )/det(A);
det(Av1 , . . . , A(cv), . . . , Avn )/det(A) = det(Av1 , . . . , c(Av), . . . , Avn )/det(A)
= c det(Av1 , . . . , Av, . . . , Avn )/det(A).
(2) Alternating:
det(Av1 , . . . , Avi , Avi+1 , . . . , Avn )/det(A) = −det(Av1 , . . . , Avi+1 , Avi , . . . , Avn )/det(A)
(3) Normalization: The columns of A are Ae1 , . . . , Aen , so by definition det(A) =
det(Ae1 , . . . , Aen ) and thus det(Ae1 , . . . , Aen )/det(A) = 1.
Thus, det(AB)/det(A) = det(B) or det(AB) = det(A)det(B)
Corollary. If A is invertible then det(A−1 ) =
1
det(A) .
Proof. By normalization the determinant of the identity matrix is det(I) = 1, so
1 = det(I) = det(AA−1 ) = det(A)det(A−1 )
⇒ det(A−1 ) = 1/det(A)
Proposition. An n × n-matrix A is invertible iff det(A) 6= 0.
Proof. If the matrix is invertible then
det(A) = 1/det(A−1 ) 6= 0
On the other hand, recall that a matrix is invertible iff its columns {v1 , . . . , vn }
are linearly independent. So if A is not invertible so that {v1 , . . . , vn } is linearlly dependent, then by a homework problem, there is an i such that vi ∈
span{v1 , . . . , vi−1 , vi+1 , . . . , vn }. Write
vi = a1 v1 + · · · + ai−1 vi−1 + ai+1 vi+1 + · · · + an vn
By multilinearity we know that determinant is linear in the ith entry so
a1 det(v1 , . . . , vi−1 , v1 , vi+1 , . . . , vn )
..
.
det(v1 , . . . , vn ) =
+ai−1 det(v1 , . . . , vi−1 , vi−1 , vi+1 , . . . , vn )
+ai+1 det(v1 , . . . , vi−1 , vi+1 , vi+1 , . . . , vn )
..
.
+an det(v1 , . . . , vi−1 , vn , vi+1 , . . . , vn )
And since each term is a determinant with repeated entries, each term equals 0 and
thus det(A) = det(v1 , . . . , vn ) = 0.
Definition. The characteristic polynomial of a square matrix is the polynomial
pA (λ) = det(A − λI).
3
Example. A =
1
1
1
0
.
pA (λ) = det
1
1
1
0
−λ
1
0
0
1
1 1
λ 0
1−λ 1
= det
−
= det
= (1−λ)(−λ)−1·1 = λ2 −λ−1
1 0
0 λ
1
−λ
We know that the λ is an eigenvalue for A iff A − λI is not invertible. Thus eigenvalues are roots of the characteristic polynomial. For example A has eigenvalues
which we can find by the quadratic formula.
√
1± 5
λ=
2