In this chapter, every matrix is an element of Mn(ℂ) and x = (x1,…,xn)T ∈ ℂn, for some n ∈ ℕ. We start with a few examples to motivate this chapter.
Whereas, xT Bx = 5 + 10
. Here also the maximum/minimum
displacements occur along the orthogonal lines x + 2y = 0 and 2x - y = 0, where
x + 2y = (x,y)
and 2x - y = (x,y)
.
![]() | = 2a11x1 + 2a12x2 + ![]() | ||
![]() | = ![]() | ||
![]() | = 2an1x1 + 2an2x2 + ![]() |
Therefore, to get the points of extremum, we solve for
We observe the following about the matrices A,B and C that appear in Example 6.1.1.
Thus, we see that given A ∈ Mn(ℂ), the number λ ∈ ℂ and x ∈ ℂn,x≠0 satisfying Ax = λx have certain nice properties. For example, there exists a basis of ℂ2 in which the matrices A,B and C behave like diagonal matrices. To understand the ideas better, we start with the following definitions.
Definition 6.1.2. [Eigenvalues, Eigenvectors and Eigenspace] Let A ∈ Mn(ℂ). Then,
![]() | (6.1.1) |
is called the eigen-condition.
Theorem 6.1.3. Let A ∈ Mn(ℂ) and α ∈ ℂ. Then, the following statements are equivalent.
Proof. We know that α is an eigenvalue of A if any only if the system (A-αIn)x = 0 has a non-trivial solution. By Theorem 2.2.40 this holds if and only if det(A - αI) = 0. _
Definition 6.1.4. [Characteristic Polynomial / Equation, Spectrum and Spectral Radius] Let A ∈ Mn(ℂ). Then,
We thus observe the following.
Remark 6.1.5. Let A ∈ Mn(ℂ).
DRAFT
Almost all books in mathematics differentiate between characteristic value and eigenvalue as the ideas change when one moves from complex numbers to any other scalar field. We give the following example for clarity.
Remark 6.1.6. Let A ∈ M2(F). Then, A induces a map T ∈(F2) defined by T(x) = Ax, for all
x ∈ F2. We use this idea to understand the difference.
Let us look at some more examples.
Definition 6.1.9. Let A ∈(ℂn). Then, a vector y ∈ ℂn\{0} satisfying y*A = λy* is called
a left eigenvector of A for λ.
Theorem 6.1.11. [Principle of bi-orthogonality] Let (λ,x) be a (right) eigenpair and (μ,y) be a left eigenpair of A, where λ≠μ. Then, y is orthogonal to x.
Proof. Verify that μy*x = (y*A)x = y*(λx) = λy*x. Thus, y*x = 0. _
Definition 6.1.13. [Eigenvalues of a linear Operator] Let T ∈(ℂn). Then, α ∈
is called
an eigenvalue of T if there exists v ∈ ℂn with v≠0 such that T(v) = αv.
Proposition 6.1.14. Let T ∈ L(ℂn) and let be an ordered basis in ℂn. Then, (α,v) is an
eigenpair for T if and only if (α,[v]
) is an eigenpair of A = T[
,
].
DRAFT
Proof. Note that, by definition, T(v) = αv if and only if [Tv] = [αv]
. Or equivalently, α ∈ σ(T) if
and only if A[v]
= α[v]
. Thus, the required result follows. _
Remark 6.1.15. [A linear operator on an infinite dimensional space may not have any
eigenvalue] Let V be the space of all real sequences (see Example 3.1.4.8a). Now, define a
linear operator T ∈(V) by
Solution: Let if possible α be an eigenvalue of T with corresponding eigenvector x = (x1,x2,…). Then, the eigen-condition T(x) = αx implies that
Theorem 6.1.16. Let λ1,…,λn, not necessarily distinct, be the A = [aij] ∈ Mn(ℂ). Then,
det(A) = ∏
i=1nλi and tr(A) = ∑
i=1naii = ∑
i=1nλi.
DRAFT
Proof. Since λ1,…,λn are the eigenvalues of A, by definition,
![]() | (6.1.2) |
is an identity in x as polynomials. Therefore, by substituting x = 0 in Equation (6.1.2), we get det(A) = (-1)n(-1)n ∏ i=1nλi = ∏ i=1nλi. Also,
for some a0,a1,…,an-1 ∈ ℂ. Then, an-1, the coefficient of (-1)n-1xn-1, comes from the term
Definition 6.1.18. [Algebraic, Geometric Multiplicity] Let A ∈ Mn(ℂ). Then,
We now state the following observations.
Theorem 6.1.20. Let A and B be two similar matrices. Then,
Proof. Since A and B are similar, there exists an invertible matrix S such that A = SBS-1. So, α ∈ σ(A) if and only if α ∈ σ(B) as
Note that Equation (6.1.5) also implies that Alg.Mulα(A) = Alg.Mulα(B). We will now show that Geo.Mulα(A) = Geo.Mulα(B).So, let Q1 = {v1,…,vk} be a basis of Null(A - αI). Then, B = SAS-1 implies that Q2 = {Sv1,…,Svk}⊆ Null(B -αI). Since Q1 is linearly independent and S is invertible, we get Q2 is linearly independent. So, Geo.Mulα(A) ≤ Geo.Mulα(B). Now, we can start with eigenvectors of B and use similar arguments to get Geo.Mulα(B) ≤ Geo.Mulα(A) and hence the required result follows. _
We will now give a relation between the geometric multiplicity and the algebraic multiplicity.
Proof. Let Geo.Mulα(A) = k. Suppose Q1 = {v1,…,vk} is an orthonormal basis of
Null(A - αI). Extend Q1 to get {v1,…,vk,vk+1,…,vn} as an orthonormal basis of ℂn. Put
P = . Then, P* = P-1 and
![]() ![]() ![]() | (6.1.6) |
So, Alg.Mulα(A) = Alg.Mulα(P*AP) ≥ k = Geo.Mulα(A). _
Remark 6.1.23. Note that in the proof of Theorem 6.1.22, the remaining eigenvalues of A are the eigenvalues of D (see Equation (6.1.6)). This technique is called deflation.
Let A ∈ Mn(ℂ) and let T ∈(ℂn) be defined by T(x) = Ax, for all x ∈ ℂn. In this section, we first
find conditions under which one can obtain a basis
of ℂn such that T[
,
] (see Theorem 4.4.4) is a
diagonal matrix. And, then it is shown that normal matrices satisfy the above conditions. To start
with, we have the following definition.
Definition 6.2.1. [Matrix Diagonalizability] A matrix A is said to be diagonalizable if A is similar to a diagonal matrix. Or equivalently, P-1AP = D ⇔ AP = PD, for some diagonal matrix D and invertible matrix P.
Proof. Let S = . Then, AS = SD gives
Conversely, let {u1,…,un} be n linearly independent eigenvectors of A corresponding
to eigenvalues α1,…,αn. Then, by Corollary 3.3.10, S = is non-singular and
As a direct consequence of Theorem 6.2.3, we obtain the following result.
Corollary 6.2.5. Let A ∈ Mn(ℂ). Then,
Theorem 6.2.6. Let (α1,v1),…,(αk,vk) be k eigen-pairs of A ∈ Mn(ℂ) with αi’s distinct.
Then, {v1,…,vk} is linearly independent.
DRAFT
Proof. Suppose {v1,…,vk} is linearly dependent. Then, there exists a smallest ℓ ∈{1,…,k - 1} and
β≠0 such that vℓ+1 = β1v1 + + βℓvℓ. So,
![]() | (6.2.1) |
and
Now, subtracting Equation (6.2.2) from Equation (6.2.1), we getAn immediate corollary of Theorem 6.2.3 and Theorem 6.2.6 is stated next without proof.
The converse of Theorem 6.2.6 is not true as In has n linearly independent eigenvectors corresponding to the eigenvalue 1, repeated n times.
Corollary 6.2.8. Let α1,…,αk be k distinct eigenvalues A ∈ Mn(ℂ). Also, for 1 ≤ i ≤ k, let dim(Null(A - αiIn)) = ni. Then, A has ∑ i=1kni linearly independent eigenvectors.
Proof. For 1 ≤ i ≤ k, let Si = {ui1,…,uini} be a basis of Null(A-αiIn). Then, we need to prove that
⋃
i=1kSi is linearly independent. To do so, denote pj(A) = ∕
, for
1 ≤ j ≤ k. Then, note that pj(A) is a polynomial in A of degree k - 1 and
![]() | (6.2.3) |
So, to prove that ⋃ i=1kSi is linearly independent, consider the linear system
Corollary 6.2.9. Let A ∈ Mn(ℂ) with distinct eigenvalues α1,…,αk. Then, A is diagonalizable if and only if Geo.Mulαi(A) = Alg.Mulαi(A), for each 1 ≤ i ≤ k.
Proof. Let Alg.Mulαi(A) = mi. Then, ∑ i=1kmi = n. Let Geo.Mulαi(A) = ni, for 1 ≤ i ≤ k. Then, by Corollary 6.2.8 A has ∑ i=1kni linearly independent eigenvectors. Also, by Theorem 6.1.22, ni ≤ mi, for 1 ≤ i ≤ mi.
Now, let A be diagonalizable. Then, by Theorem 6.2.3, A has n linearly independent eigenvectors. So, n = ∑ i=1kni. As ni ≤ mi and ∑ i=1kmi = n, we get ni = mi.
Now, assume that Geo.Mulαi(A) = Alg.Mulαi(A), for 1 ≤ i ≤ k. Then, for each i,1 ≤ i ≤ n, A has ni = mi linearly independent eigenvectors. Thus, A has ∑ i=1kni = ∑ i=1kmi = n linearly independent eigenvectors. Hence by Theorem 6.2.3, A is diagonalizable. _
Example 6.2.10. Let A = . Then,
and
are the only
eigen-pairs. Hence, by Theorem 6.2.3, A is not diagonalizable.
We now prove one of the most important results in diagonalization, called the Schur’s Lemma or the Schur’s unitary triangularization.
DRAFT
Lemma 6.2.12 (Schur’s unitary triangularization (SUT)). Let A ∈ Mn(ℂ). Then, there exists
a unitary matrix U such that A is similar to an upper triangular matrix. Further, if A ∈ Mn(ℝ)
and σ(A) have real entries then U is a real orthogonal matrix.
Proof. We prove the result by induction on n. The result is clearly true for n = 1. So, let n > 1 and assume the result to be true for k < n and prove it for n.
Let (λ1,x1) be an eigen-pair of A with ∥x1∥ = 1. Now, extend it to form an orthonormal basis
{x1,x2,…,un} of ℂn and define X = . Then, X is a unitary matrix and
Further, if A ∈ Mn(ℝ) and σ(A) has real entries then x1 ∈ ℝn with Ax1 = λ1x1. Now, one uses induction once again to get the required result. _
Remark 6.2.13. Let A ∈ Mn(ℂ). Then, by Schur’s Lemma there exists a unitary matrix U such that U*AU = T = [tij], a triangular matrix. Thus,
![]() | (6.2.5) |
Furthermore, we can get the αi’s in the diagonal of T in any prescribed order.
Definition 6.2.14. [Unitary Equivalence] Let A,B ∈ Mn(ℂ). Then, A and B are said to be unitarily equivalent/similar if there exists a unitary matrix U such that A = U*BU.
Remark 6.2.15. We know that if two matrices are unitarily equivalent then they are necessarily similar as U* = U-1, for every unitary matrix U. But, similarity doesn’t imply unitary equivalence (see Exercise 6.2.17.6). In numerical calculations, unitary transformations are preferred as compared to similarity transformations due to the following main reasons:
Example 6.2.16. Consider the two matrices A = and B =
. Then, we show
that they are similar but not unitarily similar.
Solution: Note that σ(A) = σ(B) = {1,2}. As the eigenvalues are distinct, by
Theorem 6.2.7, the matrices A and B are diagonalizable and hence there exists invertible
matrices S and T such that A = SΛS-1, B = TΛT-1, where Λ = . Thus,
A = ST-1B(ST-1)-1. That is, A and B are similar. But, ∑
|aij|2≠∑
|bij|2 and hence by
Exercise 5.4.8.11, they cannot be unitarily similar.
We now use Lemma 6.2.12 to give another proof of Theorem 6.1.16.
Proof. By Schur’s Lemma there exists a unitary matrix U such that U*AU = T = [tij], a triangular matrix. By Remark 6.2.13, σ(A) = σ(T). Hence, det(A) = det(T) = ∏ i=1ntii = ∏ i=1nαi and tr(A) = tr(A(UU*)) = tr(U*(AU)) = tr(T) = ∑ i=1ntii = ∑ i=1nαi. _
We now use Schur’s unitary triangularization Lemma to state the main theorem of this subsection. Also, recall that A is said to be a normal matrix if AA* = A*A.
Theorem 6.2.19 (Spectral Theorem for Normal Matrices). Let A ∈ Mn(ℂ). If A is a normal matrix then there exists a unitary matrix U such that U*AU = diag(α1,…,αn).
Proof. By Schur’s Lemma there exists a unitary matrix U such that U*AU = T = [tij], a triangular matrix. Since A is a normal
Exercise 6.2.20. Let A ∈ Mn(ℂ). If A is either a Hermitian, skew-Hermitian or Unitary matrix then A is a normal matrix.
We re-write Theorem 6.2.19 in another form to indicate that A can be decomposed into linear combination of orthogonal projectors onto eigen-spaces. Thus, it is independent of the choice of eigenvectors.
Remark 6.2.21. Let A ∈ Mn(ℂ) be a normal matrix with eigenvalues α1,…,αn.
We now give the spectral theorem for Hermitian matrices.
Theorem 6.2.22. [Spectral Theorem for Hermitian Matrices] Let A ∈ Mn(ℂ) be a Hermitian matrix. Then,
Proof. The second part is immediate from Theorem 6.2.19 as Hermitian matrices are also normal
matrices. For Part 1, let (α,x) be an eigen-pair. Then, Ax = αx. As A is Hermitian A* = A. Thus,
x*A = x*A* = (Ax)* = (αx)* = αx*. Hence, using x*A = αx*, we get
DRAFT
As an immediate corollary of Theorem 6.2.22 and the second part of Lemma 6.2.12, we give the following result without proof.
Let A ∈ Mn(ℂ). Then, in Theorem 6.1.16, we saw that
![]() | (6.2.6) |
for certain ai ∈ ℂ, 0 ≤ i ≤ n - 1. Also, if α is an eigenvalue of A then PA(α) = 0. So,
xn - an-1xn-1 + an-2xn-2 + + (-1)n-1a1x + (-1)na0 = 0 is satisfied by n complex numbers. It
turns out that the expression
DRAFT
Lemma 6.2.25. Let A1,…,An ∈ Mn(ℂ) be upper triangular matrices such that the (i,i)-th
entry of Ai equals 0, for 1 ≤ i ≤ n. Then, A1A2An = 0.
Proof. We use induction to prove that the first k columns of A1A2Ak is 0, for 1 ≤ k ≤ n. The result
is clearly true for k = 1 as the first column of A1 is 0. For clarity, we show that the first two columns
of A1A2 is 0. Let B = A1A2. Then, by matrix multiplication
Exercise 6.2.26. Let A,B ∈ Mn(ℂ) be upper triangular matrices with the top leading principal submatrix of A of size k being 0. If B[k + 1,k + 1] = 0 then prove that the leading principal submatrix of size k + 1 of AB is 0.
We now prove the Cayley Hamilton Theorem using Schur’s unitary triangularization.
Theorem 6.2.27 (Cayley Hamilton Theorem). Let A ∈ Mn(ℂ). Then, A satisfies its
characteristic equation. That is, if PA(x) = det(A-xIn) = a0-xa1++(-1)n-1an-1xn-1+
(-1)nxn then
Proof. Let σ(A) = {α1,…,αn} then PA(x) = ∏ i=1n(x-αi). And, by Schur’s unitary triangularization there exists a unitary matrix U such that U*AU = T, an upper triangular matrix with tii = αi, for 1 ≤ i ≤ n. Now, observe that if Ai = T - αiI then the Ai’s satisfy the conditions of Lemma 6.2.25. Hence,
We now give some examples and then implications of the Cayley Hamilton Theorem.
Exercise 6.2.30. Miscellaneous Exercises:
DRAFT
The next section deals with quadratic forms which helps us in better understanding of conic sections in analytic geometry.
Definition 6.3.1. [Positive, Semi-positive and Negative definite matrices] Let A ∈ Mn(ℂ). Then,
A is said to be
DRAFT
Lemma 6.3.2. Let A ∈ Mn(ℂ). Then A is Hermitian if and only if at least one of the following statements hold:
Proof. Let S ∈ Mn, (S*AS)* = S*A*S = S*AS. Thus S*AS is Hermitian.
Suppose A = A*. Then, A is clearly normal as AA* = A2 = A*A. Further, if (λ,x) is an eigenpair then λx*x = x*Ax ∈ ℝ implies λ ∈ ℝ.
For the last part, note that x*Ax ∈ ℂ. Thus x*Ax = (x*Ax)* = x*A*x = x*Ax, we get Im(x*Ax) = 0. Thus, x*Ax ∈ ℝ.
If S*AS is Hermitian for all S ∈ Mn then taking S = In gives A is Hermitian.
If A is normal then A = U* diag(λ1,…,λn)U for some unitary matrix U. Since λi ∈ ℝ, A* = (U* diag(λ1,…,λn)U)* = U* diag(λ1,…,λn)U = U* diag(λ1,…,λn)U = A. So, A is Hermitian.
If x*Ax ∈ ℝ for all x ∈ ℂn then aii = ei*Aei ∈ ℝ. Also, aii + ajj + aij + aji = (ei + ej)*A(ei + ej) ∈ ℝ. So, Im(aij) = -Im(aji). Similarly, aii + ajj + iaij - iaji = (ei + iej)*A(ei + iej) ∈ ℝ implies that Re(aij) = Re(aji). Thus, A = A*. _
DRAFT
Remark 6.3.3. Let A ∈ Mn(ℝ). Then the condition x*Ax ∈ ℝ in Definition 6.3.9 is always
true and hence doesn’t put any restriction on the matrix A. So, in Definition 6.3.9, we assume
that AT = A, i.e., A is a symmetric matrix.
Proof. 1 ⇒2: Let A be positive semi-definite. Then, by Lemma 6.3.2 A is Hermitian. If (α,v) is an
eigen-pair of A then α∥v∥2 = v*Av ≥ 0. So, α ≥ 0.
DRAFT
2 ⇒3: Let σ(A) = {α1,…,αn}. Then, by spectral theorem, there exists a unitary matrix U such
that U*AU = D with D = diag(α1,…,αn). As αi ≥ 0, for 1 ≤ i ≤ n, define D = diag(
,…,
).
Then, A = UD
[D
U*] = B*B.
3 ⇒1: Let A = B*B. Then, for x ∈ ℂn, x*Ax = x*B*Bx = ∥Bx∥2 ≥ 0. Thus, the required result follows. _
A similar argument gives the next result and hence the proof is omitted.
Remark 6.3.7. Let A ∈ Mn(ℂ) be a Hermitian matrix with eigenvalues λ1 ≥ λ2 ≥≥ λn. Then,
there exists a unitary matrix U = [u1,u2,…,un] and a diagonal matrix D = diag(λ1,λ2,…,λn) such
that A = UDU*. Now, for 1 ≤ i ≤ n, define αi = max{λi,0} and βi = min{λi,0}. Then
Definition 6.3.8. [Multilinear Function] Let V be a vector space over F. Then,
DRAFT
Definition 6.3.9. [Sesquilinear, Hermitian and Quadratic Forms] Let A = [aij] ∈ Mn(ℂ) be a Hermitian matrix and let x,y ∈ ℂn. Then, a sesquilinear form in x,y ∈ ℂn is defined as H(x,y) = y*Ax. In particular, H(x,x), denoted H(x), is called a Hermitian form. In case A ∈ Mn(ℝ), H(x) is called a quadratic form.
The main idea of this section is to express H(x) as sum or difference of squares. Since H(x) is a quadratic in x, replacing x by cx, for c ∈ ℂ, just gives a multiplication factor by |c|2. Hence, one needs to study only the normalized vectors. Let us consider Example 6.1.1 again. There we see that
Note that both the expressions in Equation (6.3.1) is the difference of two non-negative terms. Whereas, both the expressions in Equation (6.3.2) consists of sum of two non-negative terms. Is this just a coincidence?In general, let A ∈ Mn(ℂ) be a Hermitian matrix. Then, by Theorem 6.2.22, σ(A) = {α1,…,αn}⊆ ℝ and there exists a unitary matrix U such that U*AU = D = diag(α1,…,αn). Let x = Uz. Then, ∥x∥ = 1 and U is unitary implies that ∥z∥ = 1. If z = (z1,…,zn)* then
![]() | (6.3.3) |
where α1,…,αp > 0, αp+1,…,αr < 0 and αr+1,…,αn = 0. Thus, we see that the possible values of H(x) seem to depend only on the eigenvalues of A. Since U is an invertible matrix, the components zi’s of z = U-1x = U*x are commonly known as the linearly independent linear forms. Note that each zi is a linear expression in the components of x. Also, note that in Equation (6.3.3), p corresponds to the number of positive eigenvalues and r - p to the number of negative eigenvalues. For a better understanding, we define the following numbers.
Definition 6.3.12. [Inertia and Signature of a Matrix] Let A ∈ Mn(ℂ) be a Hermitian matrix. The inertia of A, denoted i(A), is the triplet (i+(A),i-(A),i0(A)), where i+(A) is the number of positive eigenvalues of A, i-(A) is the number of negative eigenvalues of A and i0(A) is the nullity of A. The difference i+(A) - i-(A) is called the signature of A.
Exercise 6.3.13. Let A ∈ Mn(ℂ) be a Hermitian matrix. If the signature and the rank of A is known then prove that one can find out the inertia of A.
As a next result, we show that in any expression of H(x) as a sum or difference of n absolute squares of linearly independent linear forms, the number p (respectively, r - p) gives the number of positive (respectively, negative) eigenvalues of A. This is popularly known as the ‘Sylvester’s law of inertia’.
Lemma 6.3.14. [Sylvester’s Law of Inertia] Let A ∈ Mn(ℂ) be a Hermitian matrix and let x ∈ ℂn. Then, every Hermitian form H(x) = x*Ax, in n variables can be written as
Proof. Equation (6.3.3) implies that H(x) has the required form. We only need to show that p and r are uniquely determined by A. Hence, let us assume on the contrary that there exist p,q,r,s ∈ ℕ with p > q such that
where y = Now, this can hold only if = 0, a contradiction to
being a non-trivial solution.
Hence p = q. Similarly, the case r > s can be resolved. This completes the proof of the
lemma. __
Remark 6.3.15. Since A is Hermitian, Rank(A) equals the number of nonzero eigenvalues.
Hence, Rank(A) = r. The number r is called the rank and the number r - 2p is called the
inertial degree of the Hermitian form H(x).
DRAFT
We now look at another form of the Sylvester’s law of inertia. We start with the following definition.
Definition 6.3.16. [Star Congruence] Let A,B ∈ Mn(ℂ). Then, A is said to be *-congruent (read star-congruent) to B if there exists an invertible matrix S such that A = S*BS.
Theorem 6.3.17. [Second Version: Sylvester’s Law of Inertia] Let A,B ∈ Mn(ℂ) be Hermitian. Then, A is *-congruent to B if and only if i(A) = i(B).
Proof. By spectral theorem U*AU = ΛA and V *BV = ΛB, for some unitary matrices U,V and
diagonal matrices ΛA,ΛB of the form diag(+,,+,-,
,-,0,
,0). Thus, there exist invertible
matrices S,T such that S*AS = DA and T*BT = DB, where DA,DB are diagonal matrices of the
form diag(1,
,1,-1,
,-1,0,
,0).
If i(A) = i(B), then it follows that DA = DB, i.e., S*AS = T*BT and hence A = (TS-1)*B(TS-1).
Conversely, suppose that A = P*BP, for some invertible matrix P, and i(B) = (k,l,m). As T*BT = DB, we have, A = P*(T*)-1DBT-1P = (T-1P)*DB(T-1P). Now, let X = (T-1P)-1. Then, A = (X-1)*DBX-1 and we have the following observations.
Thus, it now follows that i(A) = (k,l,m). _
We now obtain conditions on the eigenvalues of A, corresponding to the associated quadratic form, to characterize conic sections in ℝ2, with respect to the standard inner product.
Definition 6.3.18. [Associated Quadratic Form] Let f(x,y) = ax2+2hxy+by2+2fx+2gy+c be a general quadratic in x and y, with coefficients from ℝ. Then,
Proposition 6.3.19. Consider the general quadratic f(x,y), for a,b,c,g,f,h ∈ ℝ. Then, f(x,y) = 0 represents
Proof. As A is symmetric, by Corollary 6.2.23, A = U diag(α1,α2)UT , where U = is an
orthogonal matrix, with (α1,u1) and (α2,u2) as eigen-pairs of A. Let
= xT U. As u1 and u2 are
orthogonal, u and v represent orthogonal lines passing through origin in the (x,y)-plane. In most
cases, these lines form the principal axes of the conic.
We also have xT Ax = α1u2 + α2v2 and hence f(x,y) = 0 reduces to
![]() | (6.3.6) |
for some g1,f1 ∈ ℝ. Now, we consider different cases depending of the values of α1,α2:
Let H(x) = x2 + 4y2 + 4xy be the associated quadratic form for a class of curves. Then,
A = , α1 = 0,α2 = 5 and v = x + 2y. Now, let d1 = -3 and vary d2 and d3 to get
different curves (see Figure 6.2 drawn using the package “MATHEMATICA”).
![]() | (6.3.7) |
whose understanding requires the following subcases:
DRAFT
Let H(x) = 10x2 - 5y2 + 20xy be the associated quadratic form for a class of curves. Then,
A = , α1 = 15,α2 = -10 and
u = 2x + y,
v = x - 2y. Now, let
d1 =
,d2 = -
to get 3(2x + y + 1)2 - 2(x - 2y - 1)2 = d3. Now vary d3 to get
different curves (see Figure 6.3 drawn using the package ”MATHEMATICA”).
![]() | (6.3.8) |
We consider the following three subcases to understand this.
Let H(x) = 6x2 + 9y2 + 4xy be the associated quadratic form for a class of curves. Then,
A = , α1 = 10,α2 = 5 and
u = x + 2y,
v = 2x-y. Now, let d1 =
,d2 = -
to
get 2(x + 2y + 1)2 + (2x - y - 1)2 = d3. Now vary d3 to get different curves (see Figure 6.4
drawn using the package “MATHEMATICA”).
Thus, we have considered all the possible cases and the required result follows. _
Remark 6.3.20. Observe that the condition =
u1u2
implies that the principal axes
of the conic are functions of the eigenvectors u1 and u2.
Exercise 6.3.21. Sketch the graph of the following surfaces:
As a last application, we consider a quadratic in 3 variables, namely x,y and z. To do so, let
A = , x =
, b =
and y =
with
![]() | (6.3.11) |
Example 6.3.22. Determine the following quadrics f(x,y,z) = 0, where
Solution: Part 1 Here, A = , b =
and q = 2. So, the orthogonal matrices
P =
and PT AP =
. Hence, f(x,y,z) = 0 reduces to
Part 2 Here f(x,y,z) = 0 reduces to -
-
= 1 which is the equation of a
hyperboloid consisting of two sheets with center 0 and the axes x, y and z as the principal
axes.
Part 3 Here f(x,y,z) = 0 reduces to -
+
= 1 which is the equation of a
hyperboloid consisting of one sheet with center 0 and the axes x, y and z as the principal
axes.
Part 4 Here f(x,y,z) = 0 reduces to z = y2 - 3x2 + 10 which is the equation of a hyperbolic paraboloid.
The different curves are given in Figure 6.5. These curves have been drawn using the package “MATHEMATICA”.