Linear Algebra 242

Thm 9 (5.5): Let A be a real 2x2 matrix with a complex eigenvalue λ= a-bi (b≠0) and an associated eigenvector v in C²

A=PCP⁻¹, where P=[real v imaginary v] & C=|a -b| |b a| The transformation x→C(x) may be viewed as the composition of a rotation through the angle ω and a scaling by |λ| C= r |a/r -b/r| |b/r a/r| =|r 0| |cosω -sinω| |0 r| |sinω cosω| where ω is the angle between the positive x asix and the ray from (0,0) to (a,b) the angle ω is called the argument of λ=a+bi

definition of an eigenvector of any general linear operator T: V→V

An eigenvector v of T with eigenvalue λ is a nonzero vector so that T(v) = λv.

Theorem about subspaces and vector spaces:

Any subspace H of Rn is a subspace of the vector space Rn

The Invertible Matrix Theorem

Either all of these statements are true, or all of them are false: a. A is an invertible matrix b. A is row equivalent to the nxn identity matrix c. A has n pivot positions d. The equation Ax= 0 has only the trivial solution. e. The columns of A form a linearly independent set. f. The linear transformation x→ Ax is one to one g. The equation Ax=b has at least one solution for each b in Rⁿ h. The columns of A span Rⁿ i. The linear transformation x→Ax maps Rn onto Rn j. There is an nxn matrix C such that CA= I k. There is an nxn matrix D such that AD= I l. A^T is an invertible matrix

Other rules for Vector Space that have proofs

If V is any vector space, then for any v∈V (called vectors), and any c ∈R 1 0v = 0 2 c0 = 0 3 (−1)v = −v.

Theorem 4 5.2

If nxn matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues with the same multiplicities

Properties of Determinants 5.2

Let A and B be nxn matrices 1. A is invertible if an only if det A≠0 2. det AB=detA × detB 3. det A^T= det A 4. If A is triangular, then det A is the product of the entries on the main diagonal of A 5. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor.

Properties of row operations to find determinants by reducing to echelon form

Let A be a square matrix. a. if a multiple of one row of A is added to another row to produce a matrix B, then det A=det B b. if two rows of A are interchanged to produce B, then det B= (-1) det A c. if one row of A is multiplied by a constant k to produce B, then det B= (k) det A

Theorem 8 page 221

Let B= {b1...bn} be a basis for a vector space V. Then the coordinate mapping x→ [x]sub is a one to one linear transformation from V onto Rn

Unique Representation Theorem: Page 218

Let B= {b1...bn} be a basis for a vector space V. Then, for each vector x in V, there exists a unique set of scalars c1...cn such that x= c1b1+...+cnbn

definition of a basis

Let H be a subspace of Rn. Then, a linearly independent set {v1,...,vk} of vectors in H which spans H is called a basis of H. -Given a basis {v1,...,vk} of a subspace H of Rn, then any element w∈H can be written uniquely as a linear combination of {v1,...,vk} w = x1v1 +···+xkvk.

1.9 pg 72: Theorem 10

Let T: Rn→Rn be a lin trans. Then there exists a unique matrix A such that T(x)= Ax for all x in Rn. A is given by the mxn matrix whose jth column is the vector T(ej) where ej is the jth column of the identity matrix in Rn. A= [T(e1)....T(en)]

Given an m×n matrix A, there are 3 subspaces that it defines

Nul(A) Col(A) Row(A)

def: An orthogonal basis for a subspace W is

a basis for W that is also an orthogonal set (i.e. if each pair of distinct vectors from the set is orthogonal, if <ui,uj>= 0 whenever i≠j)

Each column of AB is

a linear combination of the columns of A using weights from the corresponding column of B

definition of an orthogonal matrix

a square invertible matrix U such that U⁻¹=U^T. By Thm 6, an orthogonal matrix has orthonormal columns ANY square matrix with orthonormal columns is an orthogonal matrix; -this type of matrix will also have orthonormal rows→ U^T is an orthogonal matrix

Definition of a Vector Subspace

a vector subspace of a vector space V is a nonempty subset H ⊂V so that, for each v,w∈H, and c ∈R, 1. v+w∈H 2. cv∈H Another way to say this is that a subspace H ⊂V is any subset of V which, with the same rules, is a vector space in its own right. This must include the fact that H is closed under the operations of addition and scalar multiplication, and, since the rules above all still hold for H (since they hold for V), that is all you need to check separately if you already know that V is a vector space.

Thm 1: 6.1: Let u, v and w be vectors in Rn and let c be a scalar

a. udotv =vdotu b. (u+v)dotw= (udotw) + (vdotw) c. (cu)dotv=c(udotv)=udot(cv) d. udotu≥0 and udotu =0 if and only if u =0 Combination of B&C rule: (c1u1+...+cpup)dotw= c1(u1dotw)+...+cp(updotw)

The standard basis {e₁....en} for Rn is

an orthonormal set, as is any nonempty subset of {e₁....en}

The matrix product u^Tv is a 1x1 matrix, which we write as a single real number (scalar) without brackets u1v1+u2v2+...+unvn

inner product/dot product definiton

A set {u₁...up} is an orthonormal set if

it is an orthogonal set of unit vectors

if A^T =A (A is symmetric)

know eigenvalues are real and eigenvectors are perpendicular (are a basis)

why det A^T A =0 for a 2 by 4 matrix A

rank(AB) ≤ min{rank(A),rank(B)}), so rank A^T A is at most 2, but the matrix is 2x2

to find the change of basis matrix P(C←B)

row reduce the partitioned matrix [c1....cn | b1...bn] to [I | P(C←B)]

ways to verify that a matrix A is diagonalizable

show that S⁻¹AS=D or A=SDS⁻¹ or AS=SD *A is diagonalizable if A is similar to a diagonal matrix D. D is the matrix of A with respect to the basis S →Suppose A is a square matrix of size n. Then A is diagonalizable if and only if there exists a linearly independent set S that contains n eigenvectors of A.

Now think of composition. Let B be an n×p matrix — a linear map from Rp to Rn, and A an m×n matrix. Then, the composition of linear maps A(Bx) = y

takes the x∈Rp to z∈Rn to y∈Rm. How? It maps Rp to Rm, so should be a p×m matrix. The notation even allows for a mnemonic device: AB should be (m×n)(n×p) and you cancel the adjacent n's to get m×p.

A linear transformation T with the matrix A is one-to-one if

the equation T(x)=0 has only the trivial solution, meaning the matrix Ax=0 will only have the trivial solution, so the columns of A will be LI -each b in Rm is the image of at most one x in Rn

The multiplicity of the eigenvalue λ(sub i) is

the largest positive integer mi so that (λ−λ(sub i))^m(sub i) is a factor of p(λ), the characteristic polynomial of A.

The point identified with yhat is the closest point of L (aka the subspace spanned by u) to y because

the line segment between yhat and y is perpendicular to L (prove with geometry) and so the distance between y and L = ||y-yhat||

if det A ≠ 0 for a square matrix:

the matrix is invertible and the set of vectors making up the columns is linearly independent and the matrix spans Rn

Nul A^T is

the orthogonal complement of ColA, aka the set of all vectors orthogonal to ColA * given any x is in NulA, x is orthogonal to each of row of A, so it is orthogonal to RowA because RowA spans the rowspace of A. Conversely, if x is perp to Row A, then x is certainly perp to each row of A, hence Ax=0. Since this statement is true for any matrix, it is true for A^T -so, the orthogonal complement of the Row A^T is is Nul A^T. We know Row A^T= Col A, so Nul A^T= (Col A)perp

NulA is

the orthogonal complement of RowA, aka the set of all vectors orthogonal to RowA

The number of elements of a basis of a subspace H of Rⁿ

the same for each basis. That number is called the dimension of H, and dim(H) ≤n, dim(H) = n only when H = Rn.

book definition of a vector space

the set of all possible inputs (functions) or a nonempty set of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers)

Null Space of an m×n matrix A = Nul(A)

the set of all vectors x∈Rn so that Ax = 0 is a subspace of Rn The reason that the nullspace of A is a subspace is that, for any two vectors v and w in the nullspace, A(v+w) = Av+Aw = 0+0 = 0, and A(cv) = cAv = c0 = 0, so the nullspace is closed under addition and scalar multiplication. It is also nonempty since, no matter what A is, 0∈Nul(A).

The orthogonal complement of W is

the set of all vectors z orthogonal to every vector in W

Column Space of an m×n matrix A= Col(A)

the span of the columns of A. Since it is a span, it is a subspace, of Rm. Col(A) is also the range of A, the set of all vectors y∈Rm so that y = Ax for some x∈Rn. *side note: The column space of A is not the same as the column space of the reduced row-echelon form of A. The rowspace and nullspace are the same for A as for the reduced form, but the columns change.

Row Space of an m×n matrix A Row(A)

the span of the rows of A. Since it is a span, it is a subspace, of Rn this time. You can think of this as a space of row vectors, or you can take the transpose of the rows and consider it in the usual way.

the orthogonal projection of y onto u is determined by

the subspace L spanned by u (the line through u and 0). -so if c is any nonzero scalar and if u is replaced by cu in the definition of yhat, then the orthogonal projection of y onto cu is exactly the same as the orthogonal projection of y onto u

Thm5: 6.2: Let {u1...up} be an orthogonal basis for a subspace W of Rn. For each y in W,

the weights in the linear combination: y= c1u1+...+cpup are given by cj= (y dot uj) / (uj dot uj) for j= 1...p

if A matrix a is similar to a matrix B such that S⁻¹AS=B,

then B is the matrix of A with respect to the basis S, and A and B have the same characteristic polynomial, and in turn the same eigenvalues, with the same dimensional eigenspaces corresponding to those eigenvalues (not necessarily with the same eigenvectors)

Thm 4: 6.2: If S={u1...up} is is an orthogonal set of nonzero vectors in Rn,

then S is linearly independent and hence is a basis for the subspace spanned by S

if W is the subspace spanned by an orthonormal set {u₁...up},

then {u₁...up} is an orthonormal basis for W, since the set is automatically LI by Thm 4

if u and v are nonzero vectors in either R2 or R3,

u dot v= |u| |v| cosθ where θ is the angle between u and v

unit vector

vector whose length is 1, found by dividing a nonzero vector by its length, also called normalizing a vector. when a vector is normalized, the unit vector created has the same direction

Fact: A vector x is in Wperp if and only if

x is orthogonal to every vector in a set that spans W

how to express a given vector y in W in terms of an orthogonal basis {u1...up} for W

y= (<y1, u1>/ <u1,u1>) u1+....+ (<yp, up>/ <up, up>) up

H={[ s t 0 ]: s and t are real} is it a subspace of R3?

yes, although R2 is not even a subset of R3, this subset H in R3 contains the origin(like another example of a subspace in R3, the x₁x₂-plane) and H is closed under vector addition and scalar multiplication because these operations on vectors in H always produce vectors whose third entries are zero (and so belong to H). Side note: a plane in R3 NOT through the origin is not a subspace of R3 because the plane does not contain the zero vector of R3

yhat=projL (y)= [(ydot u)/(udotu)] u

yhat is the orthogonal projection of y onto u, or the component of the vector y that is a multiple of u. -The other component of y is z, some vector orthogonal to u * y= yhat +z; and given that z=y- yhat, the set {yhat, z} is an orthogonal set

For a vector z to be orthogonal to a subspace W of Rn

z must be orthogonal to every vector in W

Pythagorean Thm: two vectors u and v are orthogonal if and only if

|u+v|²= |u|²+|v|²

definition of the distance between 2 vectors

|u-v| length of the vector u-v

the law of cosines

|u-v|²=|u|²+|v|²- 2|u| |v| cosθ

Tr(AB) =Tr(BA)

∑n i=1 ∑ n k=1 aikbki =∑n i=1 ∑n k=1 bikaki

To find an eigenvector v for an eigenvalue λ

find a vector v so that Av = λIv, or (A−λI)(v) = 0. Fact Equivalently find Nul(A−λI)

Properties of Matrix Multiplication

(AB)C= A(BC) A(B+C) = AB+AC (B+C)A= BA + CA r(AB)= (rA)B= A(rB) if an invertible matrix A is multiplied by an invertible matrix B, then the resultant matrix is also invertible

Properties of the Inverses of Matrices

(AB)^-1 = B^-1 A^-1 AT−1 =A−1T

Properties of the Transposes of Matrices

(AB)^T = B^T A^T (A+B)^T = A^T + B^T (A^T)^T = A for any scalar r, (rA)^T = rA^T The transpose of a matrix has the same rank as its original matrix

To find an eigenvalue λ of A

, you need to solve Av = λIv, or (A−λI)(v) = 0. Equivalently, find all λ so that det(A−λI) = 0. det(A−λI) = p(λ) is a degree-n polynomial in the variable λ, called the characteristic polynomial of A. Roots of that polynomial are eigenvalues of A.

the rank of an mxn matrix

1 The number of leading ones (or pivot columns) 2 The number of nonzero rows in the reduced row-echelon form 3 The dimension of the rowspace. 4 The dimension of the column space.

5.2 The Characteristic Equation

A scalar λ is an eigenvalue of an nxn matrix A if and only if λ satisfies the characteristic equation det(A-λI)=0

Definition of an orthogonal set of vectors

A set of vectors {u1...up} in Rn is said to be orthogonal if each pair of distinct vectors from the set is orthogonal, if <ui, uj>= 0 whenever i≠j

Fact: W perp is

A subspace of Rn (so if W is a plane, it must be through the origin)

6.2: Thm 7: Let U be an mxn matrix with orthonormal columns and let x and y be in Rn. Then:

1. |Ux|=|x| 2. (Ux) dot (Uy)= x dot y 3. (Ux) dot (Uy)= 0 if and only if x dot y=0 properties 1 and 3 say that the mapping x→Ux preserves lengths and orthogonality

det (B⁻¹)

1/ (detB)

det AB

= (detA)(detB)

rank (A) [or dim(Col(A)) or dim(Row(A))] + dim(Nul(A))

=n

the length of a vector |v|

=√(vdotv)=√[(v1)²+(v2)²+...+(vn)²] and |v|²= vdotv and for any scalar c |cv|= |c| |v|

Definition of A Real Vector Space We finally move beyond Rn. Instead of only thinking about Rn as vectors, or subspaces of Rn, we now expand the idea of vectors to any set, or vector space, in which the notions of addition and scalar multiplication make sense, and fit the same rules as on Rn

A (real) vector space is a set V, so that, for any u,v,w∈V (called vectors), and any c,d ∈R (called scalars), the operations (u,v) →u+v and (c,v) →cv are defined (as elements of V) and: 1 u+v = v+u 2 (u+v) +w = u+ (v+w) 3 There is a vector 0∈V so that v+0 = v for all v∈V 4 For each u∈V there is a vector −u∈V so that u+ (−u) = 0. 5 c (u+v) = cu+cv 6 (c +d)v = cv+dv 7 (cd)v = c (dv) 8 1v = v. 9 There is a scalar multiple of u by c, denoted by cu, that is in V 10 The sum of u and v, denoted by u+v, is in V

Representing Linear Transformations as Matrices with respect to different bases

For a linear transformation of the form T:V→V, we know that representations relative to different bases are similar matrices. We also know that similar matrices have equal characteristic polynomials. -eigenvalues of a linear transformation T are precisely the eigenvalues of any matrix representation of T. -Since the choice of a different matrix representation leads to a similar matrix, there will be no "new" eigenvalues obtained from this second representation. Similarly, the change-of-basis matrix can be used to show that eigenvectors obtained from one matrix representation will be precisely those obtained from any other representation. So we can determine the eigenvalues and eigenvectors of a linear transformation by forming one matrix representation, using any basis we please

definition of an eigenvector of a matrix transformation

For an n×n matrix A, an eigenvector v∈Rn of A, with eigenvalue λ, is a nonzero vector v so that Av = λv So, the matrix, as a linear transformation, stretches by a factor of λ in that direction. Also, this generalizes to any linear operator T : V →V . Definition

Theorem about upper triangular matrices and eigenvalues

If A is an n×n matrix which is upper-triangular (that is, aij = 0 if i > j), then the eigenvalues of A are the entries on the diagonal.

Theorem about eigenvalues and non-defectiveness

If A is an n×n matrix with n distinct eigenvalues, then A is non-defective (diagonalizable)

theorem about the dimension of an eigenspace for λ(sub i)

If A is an n×n matrix, with eigenvalue λ(sub i) of multiplicity m(sub i), then Eλ(sub i) is a vector space of dimension 1≤dim(Eλ(sub i))≤m(sub i)

definition of eigenspace

If A is an n×n matrix, with eigenvalue λ(sub i), the eigenspace Eλ(sub i) is defined by Eλ(sub i) := {v ∈Rn| Av = λ(sub i)v}

theorem about eigenvectors and LI

If A is an n×n matrix, with eigenvalues λ1 and λ2, λ1 6= λ2, then any eigenvectors v1 of λ1 and v2 of λ2 are linearly independent

fact about inverses of orthogonal matrices

If U ∈O(n), then U−1 ∈O(n). Proof: Since U−1 = U^T, and U^T has columns that are orthonormal (since the rows of U are orthonormal), then U−1 is also an orthogonal

fact about matrix products of orthogonal matrices

If U,V ∈O(n), then UV ∈O(n). Proof. Since (UV)−1 = V−1U−1 = VTUT = (UV)T , UV must be orthogonal

An open, dense set of all A∈M(n,n,R) are diagonalizable matrices.

Remark A set S ⊂RN is open if every point in the set is an interior point, and a set S ⊂RN is dense if for every point x∈RN, a ball of any positive radius centered at x will meet S. Proof. [Sketch] This depends on the fact that an open dense set of polynomials p(λ) have distinct roots (including complex roots). But all that this means is that polynomials with repeated roots are "special cases". The characteristic map taking A7→pA(λ) is continuous, so the inverse image of an open set will be open. In this case, it is also true that the inverse image of a dense set will be dense, and the set of matrices then with distinct eigenvalues is an open dense set — all of them being diagonalizable

For any m×n coefficient matrix A, and any b∈Rm, the set H of solutions to the system Ax = b is a subspace if and only if b = 0.

Remark: Solutions of homogeneous systems are subspaces, solutions of inhomogeneous systems are not subspaces. Proof: If b6= 0, then 0 / ∈H, so H is not a subspace. If b = 0, then it is easy to see that the set of all solutions to Ax = 0 is indeed a subspace.

Examples of vector spaces

Rⁿ 0 M(m,n, R):= {m×n real matrices A} R[x] = {p(x) = a0 +···+anxn}, the set of all polynomials with real coefficients Ck ([a,b]), the set of k-times differentiable functions f(x), with f(k)(x) still continuous, on the interval [a,b]. C0 ([a,b]) would be the set of continuous functions on [a,b]. R+, the set of positive real numbers, with the addition defined to be (x,y) →xy := x ⊕y, and for any real number c, the scalar multiplication defined by (c,x) →x∧c := c⊗x.

proof for (-1)v= -v

Since v = 1v = (1+0)v = v+0v, then adding −v to both sides gives 0 = 0v. For (2): 0 = 00 by the first line. Then, c0 = c (00) = (c0)0 = 00 = 0. For (3), since 0 = (1−1)v = 1v+ (−1)v = v+ (−1)v. Add −v to both sides, and you get −v = (−1)v,

Definition of Similar Matrices

Suppose A and B are two square matrices of size n. Then A and B are similar if there exists a nonsingular matrix of size n, S, such that A=S⁻¹BS. We will say "A is similar to B via S" when we want to emphasize the role of S in the relationship between A and B. Also, it does not matter if we say Ais similar to B, or B is similar to A. If one statement is true then so is the other, as can be seen by using S⁻¹ in place of S (see Theorem SER for the careful proof). Finally, we will refer to S⁻¹BS as a similarity transformation when we want to emphasize the way S changes B.

definition of the coordinates of a vector relative to a basis B

Suppose B= {b1...bn} is a basis for V and x is in V. The coordinates of x relative to the basis B (or the B coordinates of x) are the weights c1...cn such that c1b1+...+cnbn= x * in Rn: [x]sub B = [c1 ....cn] (picture as a column vector) Ex: consider a basis B= {b1 b2} for R2, where b1= [1 0] and b2= [1 2] (picture both as column vectors). given a vector x in R2 with the B-coordinate vector [x]sub B = [-2 3]. find x. → the B coordinates of x tell how to build x from the vectors in B. so, x= (-2)b1 + (3)b2 =(-2) [1 0] + (3)[1 2] = [1 6]. [1 6] are the coordinates of x relative to the standard basis {e1 e2} for R2.

A transformation is linear if

T (x + y)= Tx + Ty T(cx)= c T(x)

Theorem from Notes on Orthogonal Matrices

U ∈O(n) if and only if (U∧T)U = I, that is U⁻¹ = U∧T. also, because U^T will also be an orthogonal matrix with U as its transpose,U (U^T)=I Proof. The condition UTU = I, taken component by component, says that, if U = [u1,...,un], then (ui^T)uj = δij (δij =(1 for i = j; 0 for i ≠ j , the so-called Kronecker delta), which says that the ith column is orthogonal to the jth column (i ≠ j), since ui ·uj := uT i uj = δij, and the length of the ith column is 1, so the columns are orthonormal, so U ∈O(n).

Corollary to Theorem on OMs from Notes

U ∈O(n) if and only if the rows of U are orthonormal. Proof. If U is orthogonal, U⁻¹= U^T, so not only is U^T U = I, which says that the columns are orthonormal, but also U U^T = I, which term for term says that the rows are orthonormal.

An mxn matrix U has orthonormal columns if and only if

U^T U =I

the matrix AB, if A is m×n and B is n×p, is an m×p matrix C, with C = AB, given by entries

c(sub ij) = ∑ (k=1 to n) a(sub ik) b(sub kj)

the matrix of a basis B in Rn, [b1 b2 .... bn]

changes the B-coordinates of a vector x into the coordinates for x with respect to the standard basis for Rn, [e1... en] x= c1b1 + c2b2 + ... + cnbn is equivalent to x= [b1 b2.... bn] [x]subB

how to show a set is orthonormal

compute the dot product of each vector in the set with each of the other vectors in the set. if the set is orthonormal, all of the dot products will equal 0 and the length of each vector in the set will equal 1

Properties of conjugates for complex numbers carry over to matrix algebra

conjugate of (rx)= conjugate of r * conjugate of x conjugate of (Bx)= conjugate of B * conjugate of x conjugate of (BC)= conjugate of B * conjugate of C conjugate of (rB)= conjugate of r * conjugate of B

A mapping T : Rn to Rm is said to be onto Rm if

each b in Rm is the image of at least one x in Rn -this is an existence question; the mapping T is NOT onto when there is some b in Rm for which the equation T(x) =b has no solution; this would mean that for the corresponding matrix A, there exists a vector b in Rm for which the system Ax=b is incosistent, so A would NOT SPAN Rm

Finding determinants using Cofactor Expansion

expanding the first row: det A = a₁₁C₁₁+a₁₂C₁₂+...+a₁ⁿC₁ⁿ where C₁₂=(-1)¹⁺²det A₁₂ where A₁₂ is the submatrix formed by deleting the 1st row and 2nd column of the original matrix

theorem about complex eigenvalues

if A is an n×n matrix with real entries, and if λ is a complex eigenvalue of A, then so is the conjugate of λ. proof: The proof of this is simply that, if v is a (complex) eigenvector for λ, then Av = λv. But, then Av = λv, by taking the conjugate of everything, and since A = A, because A has real entries, then λ is an eigenvalue with eigenvector v.

To prove that a set of vectors, S, does not span Rn

if S spanned all of Rn, the rank of the matrix A formed by the columns of S would be n, (rankA = dimColA = dimRowA). If rankA= (n-1), the span of the vectors making up S is n-1 dimensional, so it cannot be all of Rn

Theorem about spans being subspaces

if V₁....Vp are in a vector space V, then Span {V₁,...Vp} is a subspace of V

Properties of Similar Matrices

if there exists a nonsingular matrix of size n, S, such that A=S⁻¹BS, then the characteristic polynomials of A and B are equal→ A and B have equal eigenvalues, and the multiplicities of these eigenvalues will be the same →det(A-λI)=det(B-λI) →pA(λ)=pB(λ) When a matrix A is similar to a diagonal matrix D: The eigenvalues of A are the entries on the main diagonal of D, and A^(any ridiculously high power)= S D^(that same ridiculously high power) S⁻¹

An n×n matrix A has a complete set of eigenvectors

if there is a basis of Rn (or Cn) consisting of eigenvectors of A (not all with the same eigenvalue, usually). If A does not have a complete set of eigenvectors, then A is defective. If it does have a complete set of eigenvectors, A is non-defective.