Linear Algebra 2270
1
Existence of the determinant. Expansion according to a row.
We define the determinant for 1 × 1 matrices as
det([a]) = a
(1)
It is an easy check that it satisfies D1)-D3).
For any other n we define the determinant as follows.
Assuming the determinant is defined already for (n − 1) × (n − 1) matrices, for any A ∈ Rn×n define
det(A) using an expansion according to the i-th row:
n
det(A) = ∑ (−1)i+j ai j det(Ai j )
(2)
j=1
where Ai j is an (n − 1) × (n − 1) matrix that is matrix A without the i-th row and j-th column. For
example (using an example from the textbook at page 165), if
Notice that the formula for the determinant of a 2 × 2 matrix, which was presented in the lecture:
a
a
det(A) = det ([ 1 1 1 1 ]) = a1 1 a2 2 − a1 2 a2 1
a2 1 a2 2
(3)
is compatible with expansion according to the first row:
(−1)1+1 a1 1 det(A1 1 ) + (−1)1+2 a1 2 det(A1,2 ) = a1 1 det([a2 2 ]) − a2 1 det([a1,2 ]) = a1 1 a2 2 − a1 2 a2 1
(4)
where at the last step we used the definition of the determinant for 1 × 1 matrices. This definition is also
compatible with expansion according to the second row:
(−1)2+1 a2 1 det(A2 1 ) + (−1)2+2 a2 2 det(A2,2 ) = −a2 1 det([a1 2 ]) + a2 2 det([a1,1 ]) = a1 1 a2 2 − a1 2 a2 1
(5)
In the lecture we have proven that formula (3) satisfies D1)-D3), so it gives a valid definition of the
determinant for 2 × 2 matrices.
Our calculation shows that one can use expansion according to a row to calculate the determinant.
To understand how expansion according to a row works, let’s use an example from the textbook for
a 3 × 3 matrix and an expansion according to the first row:
1
Let us now prove the definition of the determinant using the expansion according to a row is valid.
So assume that the determinant is defined for (n − 1) × (n − 1) matrices and for those matrices it satisfies
D1)–D3) (and as a consequence DP1)–DP5)). We will prove that no matter which row is used, expansion
(2) satisfies D1)–D3).
D1): linearity wrt any column.
Consider k-th column. Consider a term in the sum (2):
(−1)i+j ai j det(Ai j )
For j ≠ k, then ai j does not depend on the k-th column and det(Ai j ) depends linearly on the k-th column
(as the determinant for (n − 1) × (n − 1) matrices satisfies D1)). If j = k then ai j depends linearly on the
k-th column and det(Ai j ) does not depend on the k-th column. In any case our term depends linearly
on the k-th column. Since det(A) is a sum of such terms, it depends linearly on the k-th column.
D2): If two adjacent columns are equal, then the determinant is equal to 0.
Suppose two adjacent columns of A are equal, namely a ⋅ k = a ⋅ k+1 . Again consider a term in the sum
(2):
(−1)i+j ai j det(Ai j )
If j is not equal to k and k + 1, then the matrix Ai j has two adjacent columns equal, and hence its
determinant is equal to 0. Thus the term corresponding to an index j not equal to k and k + 1 gives a
zero contribution to det(A). The other two terms can be written as
(−1)i+k ai k det(Ai k ) + (−1)i+k+1 ai k+1 det(Ai k+1 )
The two matrices Ai k and Ai k+1 are equal because of our assumption that the k-th column of A is
equal to the (k + 1)-th column. Similarly ai k = ai k+1 . Hence these two terms cancel as they have the
opposite signs.
D3): det(I) = 1
Let A be the identity matrix. Then ai j = 0 unless i = j, in which case ai i = 1. Thus the only term is
the sum which gives a non-zero contribution is
(−1)i+i ai i det(Ai i ) = det(Ai i )
As Ai i is the (n − 1) × (n − 1) identity matrix, its determinant is equal to 1, as the determinant for an
(n − 1) × (n − 1) matrices satisfies D3).
Thus one can use expansion to any row to define the determinant of an n × n matrix and such a
determinant satisfies D1)–D3). As there is only one function satisfying D1)–D3) (proven in the lecture)
those expansions have to give the same results.
We have proven that if the determinant is defined for (n − 1) × (n − 1) matrices, then we may define
it for n × n matrices.
Formally it is a definition using a mathematical induction, which may be explained as follows. As the
determinant is defined for 1 × 1 matrices, define it for 2 × 2 matrices using the expansion according to a
row, then define it for 3 × 3 matrices using the expansion according to a row and its definition for 2 × 2
matrices. Then define it for 4 × 4 matrices using expansion according to a row and its definition for 3 × 3
matrices. Then define it for 5 × 5 matrices . . .
2
2
Expansion according to a column. Determinant of AT .
We have proven that the determinant satisfying D1) – D5) exists and we have also proven (in the lecture)
that it is unique. We have two algorithms to find the determinant. One using an expansion according
to a row and another using column operations (or row operations on AT ). We will prove now that one
could also use expansion according to a column.
Let A ∈ Rn×n . Expansion according to the j-th column is defined as
n
detcj (A) = ∑(−1)i+j ai j det(Ai j )
(6)
detcj (A) = det A
(7)
i=1
Theorem 2.1 Let A ∈ Rn×n
Proof For n = 1 there is no expansion. For n = 2, the fact that detc1 (A) = detc2 (A) = det A is easy to
show.
For a proof by induction, assume that if A ∈ R(n−1)×(n−1) , then (7) is satisfied. We will prove that
then (7) is satisfied for any j and for any A ∈ Rn×n .
To prove it, we will show that for any j, the function detcj acting on n × n matrices satisfies D1), D2),
D3) and that implies that it has to be equal to the determinant.
Proofs that D1) and D3) are satisfied by detcj are similar to those for expansion by a row, so we will
skip them.
Let us focus on D2) which says that if two adjacent columns are the same, then detcj (A) = 0.
Let a ⋅ k = a ⋅ k+1 . For an expansion according the j-th columns, where j is not equal to k or k + 1, each
term det(Ai j ) has two adjacent columns that are equal, so det(Ai j ) = 0. Thus the sum is equal to zero.
Consider j = k.
Using the induction assumption, as Ai k ∈ R(n−1)×(n−1) , we have detck (Ai k ) = det(Ai k ). In fact we are
performing another expansion according to the k-th column of Ai k , which corresponds to the (k + 1)-th
column of A:
n
n
i=1
i=1
n
detck (A) = ∑(−1)i+k ai k det(Ai k ) = ∑(−1)i+k ai k ∑ (−1)pl +k al k+1 det((Ai j )pl k )
l=1,l≠i
= ∑ ∑ (−1)
i+k
(−1)
pl +k
(8)
ai k al k+1 det((Ai j )pl k )
i=1 l=1,l≠i
where pl = l if l < i and pl = l − 1 if l > i. Each matrix (Ai j )pl k is the matrix A without two rows and
without two columns k, k + 1. Consider two rows i1 , i2 where i1 < i2 . Let A(i1 ,i2 ) (k,k+1) denote matrix A
without rows i1 , i2 and without columns k, k + 1. Expression det(A(i1 ,i2 ) (k,k+1) ) appears in the sum (8)
exactly twice, as det((Ai1 k )i2 −1 k ) and as det((Ai2 k )i1 k ). As a result once it is multiplied by
(−1)i1 +k (−1)i2 −1+k ai1 k ai2 k+1 = (−1)i1 +i2 −1 ai1 k ai2 k
and the second time it is multiplied by
(−1)i2 +k (−1)i1 +k ai2 k ai1 k+1 = −(−1)i1 +i2 −1 ai1 k ai2 k
As a result those two terms sum up to 0. This implies that the whole sum in (8) is equal to 0.
The proof for the case of expansion according to the j = k + 1 column is similar.
This implies that D2) is satisfied for n. Together with D1),D3) (not shown here) it means that if (7)
is true for any A ∈ R(n−1)×(n−1) , then D1), D2), D3) satisfied by function detcj for n × n matrices. As there
is only one function satisfying D1),D2),D3), it has to be that for A ∈ Rn×n (7) is satisfied.
Using mathematical induction, we have proven that (7) for any square matrix A ∈ Rn×n .
This allows us to consider another tool for finding the determinant, but also allows us to prove that:
Theorem 2.2 Let A ∈ Rn×n , then
det(AT ) = det(A)
3
(9)
Proof Proof is by induction. For A ∈ R1×1 it is a trivial result, as then
T
AT = [a] = [a] = A
Let us assume that (9) is satisfied for any (n − 1) × (n − 1) matrix. We shall prove that it must be
satisfied for any n × n matrix.
Let
⎡ a1 1 a1 2 . . . a1 n ⎤
⎥
⎢
⎥
⎢a
⎢ 2 1 a2 2 . . . a2 n ⎥
⎥
A=⎢
⎢ ⋮
⋱
⋮ ⎥⎥
⎢
⎢an 1 an 2 . . . an n ⎥
⎦
⎣
Define B as
⎡ b1 1 b1 2 . . . b1 n ⎤
⎡ a1 1 a2 1 . . . an 1 ⎤
⎥
⎥
⎢
⎢
⎢b
⎥
⎥
⎢a
⎢ 2 1 b2 2 . . . b2 n ⎥
⎢ 1 2 a2 2 . . . a n n ⎥
T
⎥=A =⎢
⎥
B=⎢
⎢ ⋮
⎢ ⋮
⋱
⋮ ⎥⎥
⋱
⋮ ⎥⎥
⎢
⎢
⎢bn 1 bn 2 . . . bn n ⎥
⎢a1 n a2 n . . . an n ⎥
⎦
⎦
⎣
⎣
Let us perform an expansion according to the first row of B = AT :
n
det(AT ) = det(B) = ∑ (−1)1+j b1 j det(B1 j )
(10)
j=1
Notice that
B1 j = (AT )1 j = (Aj 1 )T
And as (Aj 1 ) is an (n − 1) × (n − 1) matrix, we assumed that (9) is satisfied for it, so
det((Aj 1 )T ) = det(Aj 1 )
Additionally b1 j = aj 1 and thus (10) is equal to
n
∑ (−1)
j+1
aj 1 det(Aj 1 )
j=1
which is the expansion according to the 1st column for A. Theorem 2.1 says it has to be equal to det(A).
As a result as the rows of A are columns of A, we obtain that D1) − D2) and DP1)–DP5) are true
for columns as well. Thus one can do row operations to find the determinant.
3
Determinant of the product of two matrices
Theorem 3.1 Let A, B ∈ Rn×n , then
det(A B) = det(A) det(B)
Proof We will use, what was proven in class, that for any matrix A ∈ Rn×n , we have
⎡
⎤
⎡
⎛⎢⎢ a1 1 a1 2 . . . a1 n ⎥⎥⎞
∣
∣ ⎤⎥⎞
⎛⎢⎢ ∣
⎜⎢ a2 1 a2 2 . . . a2 n ⎥⎟
⎥
⎥⎟ = det ⎜⎢a ⋅ 1 a ⋅ 2 . . . a ⋅ n ⎥⎟
det(A) = det ⎜⎢
⎜⎢ ⋮
⎥⎟
⎢
⎥
⋱
⋮
⎝⎢ ∣
∣
∣ ⎥⎦⎠
⎝⎢⎢an 1 an 2 . . . an n ⎥⎥⎠
⎣
⎣
⎦
⎡
∣
∣ ⎤⎥⎞
n
n
n
⎛⎢⎢ ∣
⎥
= ∑ ∑ . . . ∑ ai1 1 ai2 2 . . . ain n det ⎜⎢ei1 ei2 . . . ein ⎥⎟
⎢
⎥
i1 =1 i2 =1
in =1
⎝⎢ ∣
∣
∣ ⎥⎦⎠
⎣
Let C = A B
4
(11)
Let us perform some calculations for n = 3 first.
⎤
⎡
∣
∣
∣
⎥
⎢
⎥
⎢
C = ⎢a ⋅ 1 b1 1 + a ⋅ 2 b2 1 + a ⋅ 3 b3 1 a ⋅ 1 b1 2 + a ⋅ 2 b2 2 + a ⋅ 3 b3 2 a ⋅ 1 b1 3 + a ⋅ 2 b2 3 + a ⋅ 3 b3 3 ⎥
⎥
⎢
⎢
⎥
∣
∣
∣
⎦
⎣
as a result, using linearity of the determinant with respect to the each column, we obtain:
⎤
⎡
∣
∣
⎥⎞
n
⎛⎢⎢ ∣
⎥
det(C) = ∑ bi1 1 det ⎜⎢a ⋅ i1 a ⋅ 1 b1 2 + a ⋅ 2 b2 2 + a ⋅ 3 b3 2 a ⋅ 1 b1 3 + a ⋅ 2 b2 3 + a ⋅ 3 b3 3 ⎥⎟
⎥⎠
⎢
i1 =1
⎝⎢ ∣
⎥
∣
∣
⎣
⎦
⎤
⎡ ∣
∣
∣
⎥⎞
n
n
⎛⎢⎢
⎥
= ∑ bi1 1 ∑ bi2 ,2 det ⎜⎢a ⋅ i1 a ⋅ i2 a ⋅ 1 b1 3 + a ⋅ 2 b2 3 + a ⋅ 3 b3 3 ⎥⎟
⎢
⎥⎠
i1 =1
i2 =1
⎝⎢ ∣
⎥
∣
∣
⎦
⎣
⎡ ∣
∣
∣ ⎤⎥⎞
n
n
n
⎛⎢⎢
⎥
= ∑ bi1 1 ∑ bi2 ,2 ∑ bi3 ,3 det ⎜⎢a ⋅ i1 a ⋅ i2 a ⋅ i3 ⎥⎟
⎥
⎢
i1 =1
i2 =1
i3 =1
⎝⎢ ∣
∣
∣ ⎥⎦⎠
⎣
⎡
∣
∣ ⎤⎥⎞
n
n
n
⎛⎢⎢ ∣
⎥
= ∑ ∑ ∑ bi1 1 bi2 ,2 bi3 ,3 det ⎜⎢a ⋅ i1 a ⋅ i2 a ⋅ i3 ⎥⎟
⎥
⎢
i1 =1 i2 =1 i3 =1
⎝⎢ ∣
∣
∣ ⎥⎦⎠
⎣
A similar calculation for n × n matrices shows that
n
n
n
det(C) = ∑ ∑ . . . ∑ bi1 1 bi2 ,2 . . . bin ,n
i1 =1 i2 =1
i3 =1
⎡
⎛⎢⎢ ∣
det ⎜⎢a ⋅ i1
⎝⎢⎢ ∣
⎣
∣
a ⋅ i2
∣
∣ ⎤⎥⎞
⎥
. . . a ⋅ in ⎥⎟
⎥
∣ ⎥⎦⎠
(12)
⎡ ∣
∣
∣ ⎤⎥
⎢
⎥
⎢
Now notice that if any two columns of ⎢a ⋅ i1 a ⋅ i2 . . . a ⋅ in ⎥ are equal the determinant of this matrix is
⎥
⎢
⎢ ∣
∣
∣ ⎥⎦
⎣
zero. So the only non-zero contribution comes from those terms, where this matrix has the same columns
as matrix, but in a different order. By performing column interchanges, one can transform this matrix
to matrix A. As a result
⎡
⎡
∣
∣ ⎤⎥⎞
∣
∣ ⎤⎥⎞
⎛⎢⎢ ∣
⎛⎢⎢ ∣
⎥
⎥
det ⎜⎢a ⋅ i1 a ⋅ i2 . . . a ⋅ in ⎥⎟ = (−1)k det ⎜⎢a ⋅ 1 a ⋅ 2 . . . a ⋅ n ⎥⎟ = (−1)k det(A)
(13)
⎥
⎢
⎢
⎥⎠
⎝⎢ ∣
⎝⎢ ∣
⎥⎠
⎥
∣
∣
∣
∣
⎦
⎣
⎣
⎦
where k is the number of columns interchanges needed. Notice that the same columns interchanges will
transform matrix
⎡ ∣
∣
∣ ⎤⎥
⎢
⎢
⎥
⎢ei1 ei2 . . . ein ⎥
⎢
⎥
⎢ ∣
∣
∣ ⎥⎦
⎣
to matrix
⎡∣
∣
∣ ⎤⎥
⎢
⎢
⎥
⎢e1 e2 . . . en ⎥ = I
⎢
⎥
⎢∣
∣
∣ ⎥⎦
⎣
Thus we obtain:
⎡
⎛⎢⎢ ∣
det ⎜⎢ei1
⎝⎢⎢ ∣
⎣
∣ ⎤⎥⎞
⎥
ei2 . . . ein ⎥⎟ = (−1)k det(I) = (−1)k ⋅ 1 = (−1)k
⎥
∣
∣ ⎥⎦⎠
Using the above, (13) one can rewrite (12) as
n
∣
n
n
det(C) = ∑ ∑ . . . ∑ bi1 1 bi2 ,2 . . . bin ,n
i1 =1 i2 =1
i3 =1
n
n
n
⎡
⎛⎢⎢ ∣
det ⎜⎢ei1
⎝⎢⎢ ∣
⎣
= det(A) ∑ ∑ . . . ∑ bi1 1 bi2 ,2 . . . bin ,n
i1 =1 i2 =1
i3 =1
5
∣ ⎤⎥⎞
⎥
ei2 . . . ein ⎥⎟ det(A)
⎥
∣
∣ ⎥⎦⎠
⎡
∣
∣ ⎤⎥⎞
⎛⎢⎢ ∣
⎥
det ⎜⎢ei1 ei2 . . . ein ⎥⎟
⎢
⎥
⎝⎢ ∣
∣
∣ ⎥⎦⎠
⎣
∣
which, using (11) for matrix B, yields
det(C) = det A det B
6