Geometria Lingotto.
LeLing12: More on determinants.
Contents: ¯ • Laplace expansion of the determinant. • Cross product and generalisations. • Rank and determinant: minors. • The characteristic polynomial.
Recommended exercises: Geoling 14. ¯
Laplace expansion of the determinant
The expansion of Laplace allows to reduce the computation of an n × n determinant to that of n (n − 1) × (n − 1) determinants. The formula, expanded with respect to the ith row (where A = (aij)), is:
i+1 i+n det(A) = (−1) ai1det(Ai1) + ··· + (−1) aindet(Ain)
where Aij is the (n − 1) × (n − 1) matrix obtained by erasing the row i and the column j from A. With respect to the j th column it is:
j+1 j+n det(A) = (−1) a1jdet(A1j) + ··· + (−1) anjdet(Anj) Example 0.1. We do it with respect to the first row below.
1 2 1 4 1 3 1 3 4 3 4 1 = 1 − 2 + 1 = (4 − 6) − 2(3 − 5) + (3.6 − 5.4) = 0 6 1 5 1 5 6 5 6 1 The proof of the expansion along the first row is as follows. The determinant’s linearity, proved in the previous set of notes, implies Ej n A X 2 det(A) = a1j det . j=1 . An
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where Ej is the canonical basis of the rows, i.e. Ej is zero except at position j where there is 1. Thus we have to calculate the determinants
0 0 ··· 0 1 0 0 ··· 0
a a ··· a a a ······ a 21 22 2(j−1) 2j 2(j+1) 2n ...... ··· . . ··· . ··· .
an1 an2 ··· an(j−1) anj an(j+1) ······ ann
The whole column j does not intervene in the computation
0 0 ··· 0 1 0 0 ··· 0
a a ··· a 0 a ······ a 21 22 2(j−1) 2(j+1) 2n ...... ··· . . ··· . ··· .
an1 an2 ··· an(j−1) 0 an(j+1) ······ ann Since swapping two rows changes the sign of the determinant, Ej 1 0 0 ··· 0 0 0 ··· 0
A 0 a a ··· a a ······ a 2 j−1 21 22 2(j−1) 2(j+1) 2n 1+j det . = (−1) . . . . . = (−1) det(A1j) . . . ··· . ··· . ··· .
An 0 an1 an2 ··· an(j−1) an(j+1) ······ ann
Altogether, we have the expansion along the first row: det(A) = a11det(Ai1) ± · · · ± a1ndet(A1n).
a b Note that the determinant = ad − bc is a special case of Laplace’s formula. c d
Cross product a x Given a plane vector 2 3 −→v = we wish to find an orthogonal vector −→w = . R b y
x y The determinant vanishes when x = a and y = a, as confirmed by the formula a b x y x b −→ b −→ a = xb − ya = · . Hence w = is orthogonal to v = , a b y −a −a b a b i.e. · = 0. b −a
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This argument generalises to space, and puts us in the position of solving the problem a1 −→ −→ −→ −→ of finding a vector w orthogonal to two given vectors v1 , v2 . In fact, let v1 = b1 , c1 a2 x −→ −→ v2 = b2 and w = y. Consider the determinant: c2 z
x y z b1 c1 a1 c1 a1 b1 a1 b1 c1 = x − y + z b2 c2 a2 c2 a2 b2 a2 b2 c2 where the equality is a consequence of Laplace expanded along the first row. We may interpret the latter equality as a dot product
b1 c1
b2 c2 x y z x b1 c1 a1 c1 a1 b1 a1 c1 a1 b1 c1 = x − y + z = y · − , b2 c2 a2 c2 a2 b2 a2 c2 a2 b2 c2 z a1 b1
a2 b2 b1 c1
b2 c2 −→ a1 c1 −→ −→ so w = − is orthogonal to both v1 and v2 . a2 c2 a1 b1
a2 b2 −→ 1 −→ −→ −→ −→ The vector w is called the cross product of v1 and v2 , and denoted v1 × v2 .
−→ Similarly, using an n × n determinant and Laplace we can find a vector w ∈ Rn perpendicular to n − 1 given vectors.
The cross product seen geometrically −→ −→ −→ −→ The cross product v1 × v2 is orthogonal to v1 and v2 . But as the vectors orthogonal to −→ −→ −→ −→ v1 and v2 form a straight line, to determine v1 × v2 it suffices to know its length and orientation. 1Also known as vector product, or wedge product because of the symbol ∧.
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−→ −→ −→ −→ Proposition 0.2. The length of v1 × v2 is |v1 ||v2 | sin(θ), where 0 ≤ θ < π is the angle −→ −→ formed by v1 and v2 . −→ −→ v1 × v2 −→ 2 Proof. Recall that det( v1 ) is the volume of the parallelepiped spanned by −→ v2 −→ −→ −→ −→ the three vectors v1 × v2 , v1 , v2 : −→ −→ v1 × v2 −→ −→ −→ −→ −→ −→ −→ 2 V olume(v1 × v2 , v1 , v2 ) = det( v1 ) = |v1 × v2 | . −→ v2
Notice: −→ −→ v1×v2 −→ −→ −→ −→ −→ −→ |v1−→×v2| −→ −→ |v1 ||v2 | sin(θ) = Area(v1 , v2 ) = det( v1 ) = |v1 × v2 | −→ v2 2 −→ −→ v1×v2 −→ −→ |v1−→×v2| Since det( v1 ) is positive, we conclude that the −→ v2 −→ −→ right-hand-rule is suited to tell the orientation of v1 × v2 . −→ −→ −→ −→ Hence v1 × v2 is the unique vector orthogonal to v1 , v2 −→ −→ with length equal to the area Area(v1 , v2 ) of the paral- −→ −→ lelogram spanned by v1 and v2 , and whose orientation is found with the right-hand-rule.
The cross product −→v × −→w can be interpreted also by a 90◦ -degree rotation. Namely, assuming that −→v ⊥ −→w , then −→v × −→w is the vector obtained by rotating by 90◦ in the counter-clockwise direction the vector −→w on the plane orthogonal to −→v .
2because the determinant stems from the need to compute areas and volumes formed by vectors.
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Rank and determinant: minors
Recaal that the determinant makes sense for square matrices exclusively. If not square, we can compute certain quantities called minors or mini-determinants. A mini-determinant of order k for the matrix A is the determinant of a k × k matrix obtained from A by selecting k rows and k columns. For instance, any number aij is a mini-determinant of order 1. 1 8 2
1 2 0 6 −3 Example 0.3. The determinant = −2 is a minor of order 2 in , 3 4 3 −1 4 1 1 1 corresponding to choosing the first and third rows and columns. Also the determinant
1 8 2 6 −3 = 21 is a mini-determinant for A. Here’s a 3rd order minor: 0 6 −3 = −1 4 3 −1 4 −87 c1 c 2 A column C = . is non-zero iff at least one minor is not zero. In fact, the minors . cn are the numbers c1, ··· , cn . This is a general fact, it holds for arbitrary matrices n × m; A is non-zero iff at least one mini-determinant is non-zero. Furthermore, a matrix has rank zero iff all mini-determinants are zero. This generalises, too, and is expressed by Kronecker’s theorem 3.
Theorem 0.4. An n × m matrix A has rank ρ(A) = k if and only if there exists a non-zero minor of order k and all minors of order > k are zero.
Proof . Let Mk be a mini-determinant of order k. If there were a non-trivial linear combination among the k rows (columns) involved in building Mk , such combination would also be a non-trivial linear combination of k rows in the mini-determinant Mk . That would imply Mk = 0, so Mk = 0 if k > ρ(A). To prove that if k = ρ(A) then there is a non-zero minor of order k, we may assume A has k columns, i.e. we merely pay attention k LI columns. Since the rank is the dimension of the column space, there are k LI rows as well. Hence the k × k minor thus built cannot be trivial, for the k rows a re LI. 2 3Leopold Kronecker (1823 - 1891), german mathematician.
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The choice of free parameters There are linear systems where some variables are free. In the Gauß-Jordan elimination method explained at the beginning of the course the free variables were the last ones: if x1, x2, ··· , xn were the unknowns, then ··· , xn−1, xn are those we tried to leave free, so to speak. What if we want to leave the first ones x1, x2 free? More generally, if we want that a certain subset xi1 , xi2 , ··· , xil is free. In practice the easiest way is simply to change the order, i.e. we write the columns of xi1 , xi2 , ··· , xil at the end of the matrix, and proceed as before. x1 + 2x2 + 7x3 + 5x4 = 1 Example 0.5. To solve , using x1, x2 as parameters, 3x1 + 4x2 + 8x3 + 3x4 = 0 we use the matrix 7 5 1 2 1 8 3 3 4 0
The first column corresponds to x3 , the second to x4 , the third to x1 and the last one to x2 . Now, Gauß-Jordan:
5 1 2 1 5 1 2 1 7 5 1 2 1 1 7 7 7 7 1 7 7 7 7 → → −19 13 12 −8 → 8 3 3 4 0 8 3 3 4 0 0 7 7 7 7
5 1 2 1 12 14 −3 1 7 7 7 7 1 0 19 19 19 −13 −12 8 → −13 −12 8 0 1 19 19 19 0 1 19 19 19 12 14 −3 x4 + 19 x1 + 19 x2 = 19 The system is equivalent to: −13 −12 8 The solution, with free param- x3 + 19 x1 + 19 x2 = 19 eters x , x , is: 1 2 x1 0 1 0 x2 0 0 1 = 8 + x1 13 + x2 12 x3 19 19 19 −3 −12 −14 x4 19 19 19
In the next example, instead, x1, x2 cannot be taken as fre parameters . x = 1 Example 0.6. 1 3x1 + 4x2 + 8x3 + 3x4 = 0
Hence the question is clearly: when can we choose a set of unknowns xi1 , xi2 , ··· , xil as free parameters? Here’s the answer.
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Theorem 0.7. Let (A|B) be a consistent system in the n unknowns x1, x2, ··· , xn .
The unknowns xi1 , xi2 , ··· , xil can be chosen as free parameters iff the rank of A equals the rank of the matrix obtained by erasing the columns of the variables xi1 , xi2 , ··· , xil . Moreover, this is possible iff there exists a non-zero mini-determinant of order l in the matrix obtained erasing the columns relative to xi1 , xi2 , ··· , xil . The proof is an easy corollary of the theory developed so far.
The previous theorem is a linear version of the Implicit function theorem of Calculus. The latter allows to define a (vector) function using a system of equations that are not necessarily linear.
The characteristic polynomial
To a square matrix A is associated a very important polynomial found using determi- nants. It is called characteristic polynomial:
a11 − x a12 ··· a1n
a a − x ··· a 21 22 2n χA(x) = det(A − xId) = ......
an1 an2 ··· ann − x In other words, one subtracts x on the diagonal of A and calculates the determinant. 1 2 Example 0.8. Here is the characteristic polynomial of A = : 3 4
1 − x 2 2 χA(x) = = (1 − x)(4 − x) − 6 = x − 5x − 2. 3 4 − x
Substituting 0 to x we have det(A), so the constant term of the characteristic poly- nomial is the determinant of A. Proposition 0.9. The characteristic polynomial of an n × n matrix has degree n.
Proof. Easy using the formula of Laplace. 2 a b The characteristic polynomial of A = is c d
a − x b 2 χA(x) = = x − (a + d)x + (ad − bc) c d − x
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It has to be noticed that the characteristic polynomial of the n × n zero matrix is (−1)nxn , whereas for the n × n identity matrix we have (1 − x)n .
An important result.
Theorem 0.10. Let A be a matrix and P an invertible matrix. The characteristic polynomial of A coincides with that of P AP −1 .
Proof. Binet’s formula has
−1 −1 det(P (A − xId)P ) = det(P )det(A − xId)det(P ) = det(A − xId) = χA(x),
−1 −1 but det(P (A − xId)P ) = det(P AP − xId) = χP AP −1 (x) 2
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