2.2 Definition.

The matrix A I M_{m n} (K) whos elements are given by the relation ( ) is called the associated matrix to the linear transformation T with regard to the pair of bases B and B

2.3 Theorem.

If x = has the image y = T (x) = , then

y_i = , i =   (2.2)

Truly,

T(x) = T T (e_j)=

= ,

resulting the relation (2.2)

If we note X = ^t(x₁, x₂, , x_n), Y = ^t(y₁, y₂, , y_m) then relation (2.2) is also written as a matrix equation of the following form:

Y = AX (2.3)

The equation (2.2) or (2.3) is called the equation of the linear transformation T with regard to the considered bases.

Remarks:

1 If L (V_n, W_m) is the set of all linear transformation from V_n with values in W_m and M_{m n} (K) is the set of all matrices of type m n, and B and B are two fixed bases in V_n respectively W_m, then the corespondence

Y : L (V_n, W_m) M_{m n} (K) Y (T) = A,

which associates the matrix A to a linear transformation T relative to the two fixed bases, is a vector space isomorphism. Consequently dim L (V_n, W_m) = m n.

2 This isomorphism has the following properties:

Y (T ₁ T ₂ Y T ₁ Y (T ₂), if T ₁ and T ₂ exist

- T V_n V_n is invertable iff the associated matrix A, with regard to some base from V_n, is invertable.

Let V_n be a K-vector space and TI End(V_n). Considering different bases in V_n, we can associate separate square matrices to the linear transformation T. Naturally, one question rises: when do these two square matrices represent the same endomorphism? The answer is given by the following theorem

Proof. Let B = and B = two bases in V_n and W w_ij) the passing matrix from base B to base B , hence e _j j =

If A = (a_ij) is the associated matrix of T relative to the base B, hence T (e_j) = , j = and A = (a _ij) the associated matrix of T relative to the base B , hence T (e _j , j = , then the images T (e_j) can be written in two ways:

T(e _j respectively

T(e_j) = T T e_i

Out of the two expressions we obtain WA A W

W being a nondegenerated matrix, results the relation A W AW

Remarks:

The similarity relation is an equivalence relation on the set M_n (K). Every equivalence class corresponds to an endomorphism T I End(V_n) and contains all associated matrices of T relative the the bases of the vector space V_n.

2 The matrices A and B have the same determinant

detB = det(C^-1) detA detC = det A

Any two similar matrices have the same rank, number which represents the rank of the endomorphism T . Therefore the rank of an endomorphism does not depend on the chosen base of the vector space V (the rank of an endomorphism is invariant to base change).

3. Eigenvectors and Eigenvalues

Let V be an n-dimensional K-vector space and T I End(V) an endomorphism.

If in the vector space V we consider different bases, then to the same endomorphism T I End(V) several different similar matrices will correspond. Therefore, we are interested in finding the base of V, relative to whom the associated matrix of the endomorphism T has the simplest form, the canonical form. In this case the relations that define the endomorphism T , y_i = , will have the simplest expressions. We will solve this problem with the help of the values and the vectors proper to the endomophism T.

The set of all eigenvalues of T is called the spectrum of the operator T and it is noted with s T

The equation T (x) = l x with x 0 is equivalent to x I Ker(T - lI , where I is the identity endomorphism.

If x is an eigenvector of T , then the vectors kx , k I K are also eigenvectors.

Proof. 1) Let us consider x 0 an eigenvector corresponding to the eigenvalue l I K. Suppose there is another eigenvalue l I K corresponding to the same eigenvector such that T (x) = l x then it would result lx l x l l )x = 0 l l

2) Let x₁, x₂, , x_p be the eigenvectors corresponding to the distinct eigenvalues l l l_p. We show using induction (p) the linear independence of the considered vectors. For p = 1 and x₁ 0, as an eigenvector, the set is linearly independent. Supposing that the property is true for p-1, we show that it is true for p eigenvectors. Applying the endomorphism T to the relation k₁x₁ + k₂x₂ + + k_px_p = 0 we obtain k₁l x + k₂l x + + k_pl_px_p = 0. By substracting the first relation amplified by l_p, we obtain the following from the second relation:

k l l_p)x₁ + + k_p-1(l_p l_p )x_p-1 = 0

The inductive hypothesis prooves k₁ = k₁ = = k_p-1 = 0, and if we use this in the relation k₁x₁ + + k_p-1x_p-1 + k_px_p = 0 we obtain k_px_p = 0 k_p = 0 , that is x₁, x₂, , x_p are linearly independent.

3) For any x, y I S_l and a b I K we have :

T ax by aT (x) + bT (y) = alx bly l ax + by), which means that S_l is a vector subspace of V.

For x I S_l , then T (x) = lx I S_l , followingly T (S_l) S_l

Proof. Let l l I s T l l . Suppose there exists a vector x I , different from 0, for which we can write the relations T (x)= l x si T (x)= l x. We obtain (l l )x = 0 l l which is a contradiction. Therefore = .

The matrix equation AX = lX can be written under the form (A - lI )X = 0 and it is equivalent to the sistem of linear and homogenous equations:

(3.2)

which admits solutions different than the trivial solution if

P l) = det(A - lI ) = (3.3)

If can be proven that the caracteristic polinom can be also written like:

P l) = (-1)ⁿ [lⁿ d l^n-1 + + (-1)ⁿd_n (3.4)

where d_i is the sum of the principal minors of order i belonging to the matrix A.

Remarks

The solutions of the caracteristic equations det(A - lI ) = 0 are the eigenvalues of the matrix A.

If the field K is a closed field, then all roots of the caracteristic equations are in the field K therefore the corresponding eigenvectors are also in the K-vector space M_n (K).

If K is open, i.e. K = R, the caracteristic equation may have also complex roots and the corresponding eigenvectors will be included in the complexified real vector space.

For any real and symmetrical matrix, it can be proven that the eigenvalues are real.

3 Two similar matrices have the same caracteristic polinom.

Truly, if A and A are similar, A = C^-1AC with C nedegenerated, then

P l) = det(A lI ) = det(C^-1AC - lI ) = det[C^-1(A - lI)C] =

= det(C^-1) det(A - lI) detC= det(A - lI) = P(l

If A I M_n(K) and P(x) = a₀xⁿ + a₁x^n-1 + + a_n I K[X] then the polinom P(A) = a₀Aⁿ + a₁A^n-1 + + a_nI is named matrix polinom.

Proof. Let us consider P(l) = det(A - lI) = a₀lⁿ + a₁lⁿ + + a_n.

Likewise the reciproc of the matrix A - lI is given by

(A - lI)* = B_n-1lⁿ + B_n-2lⁿ + + B₁l + B₀, B_i I M_n(K)

and satisfies the relation (A - lI (A - lI)* = P(l I , hence

(A - lI) (B_n-1lⁿ B_n lⁿ + B₁l + B₀) = (a₀lⁿ a lⁿ + a_n)I,

By identrifying the polinoms trough l , we obtain

Now let us consider an n-dimensional K-vector space V_n , and a base B and lets not with A I M_n(K), the associated matrix to the endomorphism T with regard to this base. The equation T x lx is equivalent to (A - lI )X = 0.

The eigenvalues of the endomorphism T, if they exist, are the roots of the polinom P(l)in the field K, and the eigenvectors of T will be the solutions to the matriceal equations (A - lI )X = 0. Because of the invariance of the caracteristic polinom regarding a base change from V_n , P(l) depends only on the endomophism T and does not depend on the matrix representation of T in a given base. Therefore, the naming of the caracteristic polinom / caracteristic equation of the endomorphism T are well justified, respective to the caracteristic polinom P(l) of the matrix A, and for the caracteristic equation (A - lI )X = 0 of the matrix A.

4. An Endomorphisms Canonical Form.

Let us consider the endomorphism T V_n V_n defined on the n-dimensional K-vector space V_n.

If we consider the bases B si B belonging to vector space V_n and we denote the associated matrices of the endomorphism T with respect to these bases with A and A , then we have A W AW, where W represents the passing matrix from base B to base B . Knowing that the associated matrix to an endomorphism depends on the chosen base from the vector space V_n, we will determine that particular base which with regard to the endomorphism has the simplest form, that means that the associated matrix will have a diagonal form.

Proof: If T is diagonalizable then there exists a base B = with respect to whom the associated matrix A = (a_ij) has diagonal form, meaning a_ij = 0, i j. The actoion of T on the elements on base B is given by the relations T (e_i) = a_iie_i , i , followingly e_i , i = are eigenvectors of T .

Reciproc. Let a base in V_n , formed only of eigenvectors, that is T v_i = l_iv_i , i =

From these equations we can construct the associated matrix corresponding to T in the current base:

D = .

where the scalars l_i I K are not necessarily distinct.

In the context of the previous theorem, the matrices from the similarity classes which correspond to the diagonalizable endomorphism T , are called diagonalizable for the different bases referring to the vector space V_n.

An eigenvalue l I K , as the root of the caracteristic equation P(l) = 0, has a multiplicity order which is named algebric multiplicity, and the dimension of the eigensubspace to dimS_l is named the geometric multiplicity of the eigenvalue l

Proof: Let the eigenvalue l I K have the algebric multiplicity m n , and the corresponding eigensubpace have the following dimension : dim = p n , and by considering a base B = in we will obtain the following:

If p = n, then we have n linear independent eigenvalues, therefore a base in V_n reported to whom the associated matrix of the endomorphism T has diagonal form, and its main diagonal having the eigenvalue l . In this case, the caracteristic polinom is written in the following form : P(l) = (-1)ⁿ (l l )ⁿ , resulting p = n.

If p < n , we upgrade the base of the vector subspace to a base = belonging to V_n.

The action of the operator T on the elements of this base is given by :

T (e_i) = l e_i i = and

T (e_j) = , j =

With regard to the base , the endomorphism T has the following matrix:

=

hence the caracteristic polinom T has the form P(l l l)^p Q(l) , meaning that (l l)^p devides P(l) , then p m, q.e.d.

Proof. If T is diagonalizable, then there is a base B V_n , formed only by eigenvectors, with regard to whom the associated matrix has the diagonal form. In these conditions the caracteristic polinom is written in the following form:

P l) = (-1)ⁿ

with l_i I K , the eigenvalues of T, with the multiplicity orders m_i , . Without constraining the generality, we can admit that the first m_i vectors from the base B = are the eigenvectors corresponding to the eigenvalue l , the next m₂ corresponds to l etc. Hence , therefore, m₁ dim . But dim m₁ (theorem 3.12) and we get dim = m₁ , and likewise for the other values.

Reciproc. If l_i I K and dim = m_i, with , then we can consider the set B = , with the convention that the first m₁ vectors form a base in , the following m₂ vectors form a base in , and so on.

Because = and , base B is a base in V_n , with regard to whom the associated matrix of T has the following form:

A =

A being a diagonal matrix, we conclude that the endomorphism T este diagonalizable.

Practically, the required steps for diagonalizing and endomorphis T are the following:

1 We wirte the matrix A , associated to the endomorphism T with regard to a given base of the vector space V_n.

2 We solve the caracteristic equation det(A - lI ) = 0, determing the eigenvalues l l l_p with the corresponding multiplicity orders m₁, m₂, , m_p.

3 We use the result in theorem 3.13 and we have the following cases:

I) If l_i I K i = we determine the dimensions of the subspaces . The dimension of the eigensubspace , representing the vector space of the solutions of the homogenous system (A - l_iI )X = 0, is given by dim = n - rank(A - l_iI ). The dimension of the subspace can be found by identifying the subspace itself.

a) If dim = m_i i = , then T is diagonalizable The associated matrix to T , with regard to the base formed of eigenvectors is a diagonal matrix having on the main diagonal the eigenvalues written as many times its multiplicity order would allow it.

We can check this result by constructing the matrix T = , having as columns the coordonates of the eigenvectors (the diagonalizing matrix) and represents the passing matrix from the base considered initially to the base formed of eigenvectors, where the associated matrix of T with regard to the latter base is the diagonal matrix D, given by

D = T^-1 AT = .

b) if l_i I K such that dim < m_i , then T is not diagonalizable. The following paragraph will analyze this case.

II) If l_i K, then T is not diagonalizable. The problem of T s diagonalization can be considered only if the K-vector space V is considered an extension over the field K

Let us consider V_n a K-vector space and T I End(V_n) an endomorphism defined on V_n.

If A I M_n(K) is the associated matrix of T with regard to a base in V_n , then A can be diagonalized only if the conditions of theorem 3.13 are fulfilled..

If the eigenvalues corresponding to T belong to field K, l_i I K , and the geometric multiplicity is different then the algebric multiplicity dim < m_i , at least for some eigenvalue l_i I K , the endomorphism T is not diagonalizable, but we can still determine a base in the vector space V_n with respect to which the endomorphism T has a more general canonical form, named Jordan form.

For l I K, the matrices of the form :

l), , , , (3.5)

are named Jordan cells attached to the scalar l, of the order 1, 2, 3, ,n .

A Jordan cell of order p attached to the eigenvalue l I K , is a multiple of the order m p , and it corresponds to the linearly independent vectors e₁, e₂, , e_p which satisfy the following relations:

T(e₁) = l e₁

T(e₂) = l e₂ + e₁

T(e₃) = l e₃ + e₂

. . . . . . . . . . . . . . . . . . . .

T(e_p) = l e_p + e_p-1

The vector e₁ is an eigenvector and the vectors e , e₃, , e_p are called main vectors.

Remarks

The diagonal form of a diagonalizable endomorphism is a particular case of a the Jordan canonical form, having all Jordan cells of the first order

The jordan canonical form is not unique. The order in the main diagonal of the Jordan cells depends on the chosen order of the main vectors and the eigenvectors with regard to the given base.

The number of the Jordan cells, equal to the number of the linearly independent eigenvectors, and also their orders are unique.

The following theorem can be prooven :

Practically, for determining the canonical Jordan form of an endomorphism, we will consider the following steps:

We write the matrix A ,associated to the endomorphism T with regard to the given base.

We solve the caracteristic equation det(A - lI ) = 0 having the eigenvalues l l l_p with their multiplicity order m₁, m₂, , m_p determined.

We find the eigensubspaces , for each eigenvalue l_i

We calculate the number of Jordan cells, separately for each eigenvalue l_i , given by dim = n rang(A - lI . That would mean, that for each eigenvalue the number of the linearly independent vectors gives us the number of the corresponding Jordan cells.

The corresponding main vectors of those eigenvalues are determined for which dim < m_i , their number is given by m_i - dim . If v I is some eigenvector from , we will check the compatibility conditions and we will solve the linear equation systems one by one.

(A - l_iI )X₁ = v, , (A - l_iI )X_p = X_s

Keeping in mind the compatibility conditions and the general form of the eigenvectors and those of the main vectors we will determine by giving arbitrary values to the parameters, the linearly independent eigenvectors of and the main vectors associated to each of them.

We write the base of the vector space V_n, uniting the systems of linear independent vectors m_i , i = , formed by eigen- and main vectors.

By using the matrix T having as columns the coordonates of the vectors belonging to the base, which were constructed using the associated eigenvectors and main vectors in this order, we obtain the matrix J = T^-1AT , the Jordan canonical form which contains on the main diagonal the Jordan cells, in the order in which they appear in the constructed base of associated eigen- and main vectors (if they exist). The Jordan cells have the order equal to the number of vectors from the system constructed out of the associated eigen- and main vectors.

5. Linear transformations on euclidean vector spaces.

The properties of the linear transformations defined on an arbitrary vector space, are applicable on the euclidian vector spaces as well. The scalar product that defines the euclidean structure permits us the introduction of particular linear transformation classes.

5.1. Orthogonal transformations

Let V and W two euclidean R-vector spaces. Without being in danger of confusion we can denote the scalar products on the two vector spaces with the same symbol < , > .

Examples.

The identical transformation T: V V, T (x) = x , is an orthogonal transformation.

2 The transformation T : V V, that associates to a vector x I V its inverse T (x) = - x , is an orthogonal transformation.

Proof If T is orthogonal then < T x, T y> = < x, y >, which for x=y becomes

< T x, T y> = < x, x> || T x||² = ||x||² || T x|| = ||x||.

Reciproc. Using the relation <a, b> = , we have

< T x T y > = [ || T x + T y ||² - || T x - T y ||² ] =

= =

= = <x, y> , q.e.d.

Proof. If T is an orthogonal transformation then || T x|| = || x || and in the hypothesis T x = 0 results || x || = 0 x = 0 . Followingly, the kernel Ker T = , meaning T is injective.

Using the consequence 4.3 we find that by using an orthogonal transformation on a linearly independent system we get also a linearly independent system.

Truly, d(T x, T y T x - T y || = || T (x - y)|| = ||x - y|| = d(x, y).

Any orthogonal transformation T, different then the identical transformation is injective, T(0) = 0 is the only fixed point. Conversely, the transformation T is the same with the identical transformation.

If W = V , T ₁ and T ₂ are two orthogonal transformations on V then their composition is also an orthogonal transformation.

If T: V V is also surjective, then T is invertable moreover, T is an orthogonal transformation.

In these conditions the set of all bijective orthogonal transformation of the euclidian vector space V forms a group together with the composing operation (product) of two orthogonal transformations, named the the orthogonal group of the euclidian vector space V denoted GO(V) and which represents a subgroup of the linear group GL(V).

Let us consider in the finitely dimensional euclidian vector spaces V_n and V_m, the orthonormed bases B = , respectively = and the orthogonal linear transformation T : V_n W_m

With regard to the orthonormed bases B and the linear transformation is caracterized by the matrix A I M_{m n}(R), with the following relations :

(T e_j) = , i, j =

The bases B and are orthonormed, meaning that

< e , e₂ > = _ij , i, j = si < f_k, f_h> = _kh , k, h =

Lets evaluate the scalar product of the images of the vectors of B

< T (e_i), T (e_j)> = = =

= =

By using T s orthogonal transformation property < T (e_i), T (e_j)> = _ij we have

, i, j = (5.3)

The relation (5.3) can also be written under the form

^tA A = I_n (5.3)

Reciproc. If T is a linear transformation , caracterized (with regard to the orthonormed bases B and ), by the matrix A I M_{m n}(R) which satisfied the conditions (4.3) , then T is an orthogonal transformation.

Truly, let x = (x₁, x₂, , x_n) and y = (y₁, y₂, , y_n) two vectors in the space V_n. If we calculate the scalar product of the images of these vectors

< T (x), T (y)> = < T , T > =

< T (e_i), T (e_j)> =

= = = = <x, y>,

meaning < T (x), T (y)> = < x, y> is an orthogonal transformation.

We have also proven the following theorem:

By taking into consideration that det ^tA = detA, results that if A is an orthogonal matrix, then detA =

The subset of the orthogonal matrices which have the property detA = 1 form a subgroup denoted with SO(n; R) , of the orthogonal matrix group of the n-th order name the special orthogonal group. An orthogonal transformation caracterized by the matrix A I SO(n; R) is also called rotation.

Remarks

Since an orthogonal transformation is injective, and if B V_n, is a base, then its image trough the orthogonal transformation T: V_n W_m , is a base in Im T W_m. Consequently n m.

Any orthogonal transformation between two euclidian vector spaces of the same dimension is an isomorhpism of euclidian vector spaces.

3 The associated matrix of two orthogonal transformations T , T ₂ : Vn Vn with regard to an orthonormed base, is given by the product of the associated matrices correspondin to the transformations T and T . Thus, the orthogonal group of the euclidian vector space V_n, group GO(V_m), is isomorph with the multiplicative group GO(n; R), of the orthogonal matrices of the n-th order.

5.2 Symmetrical linear transformations

Let V and W two euclidian R-vector spaces and T: V W a linear transformation.

Let V_n a finitely dimensional euclidian R-vector space and an orthonormed base B V_n . If the endomorphism T : V_n V_n is caracterized by the real matrix A I M_n(R), we can easily proove that to any symmetrical (antisymmetrical) endomorphism corresponds a symmetrical (antisymmetrical) matrix, with regard to an orthonormed base.

Proof. Let l an arbitraty root of the caracteristic equation det(A-lI ) = 0 and X = ^t(x₁, x₂, , x_n) and eigenvector that corresponds to the eigenvalue l

Denoting with the conjugated matrix of X and multiplying with ^t the equation (A-l I )X = 0, we obtain ^t AX = l ^t X .

Since A is symmetrical, the first member of the equality is real, because = = ^tX A = ^t( ^tX A ) = ^t A X.

Also, ^tXX 0 is a real number and l is the quotient of two real numbers, hence it is real.

Let the eigenvector v₁ I R correspond to the eigenvalue l I R , and S₁ the subspace generalted by v₁ and V_n-1 V_n , its orthogonal complement, V_n = S₁ V_n-1.

Proof. If v I V_n is an eigenvector corresponding the eigenvalue l , then T v l v . For v I V_n-1 we have satisfied the relation < v₁, v > = 0. Let us proove that T (v) I V_n-1.

< v₁, T v > = < T v , v > = < l, v₁, v > = l < v₁, v > = 0.

Based on this result the next theorem can be easily prooven.

Proof. Let , , the corresponding eigensubspaces to the distinct eigenvalues l_i and l_j which have the algebric multipolicity orders m_i and m_j respectively. For v I and w I , T (v) = l_iv T (w) = l_jw we obtain :

< w, T (v) > = l_i < w, v> si < T (w), v> = l_j < w, v> . Since T is symmetrical we have (l_j l_i) < w, v> = < T (w), v> - < w, T (v)> = 0 .

Thus l_j l_i T < w, v > = 0 . q.e.d.

Truly, the first m_i vectors belong to the eigensubspace and the corresponding caracteristic roots l_i of multiplicity m_i and so on until all eigenvalues are used up m₁ + m₂ + + m_p = n.

With regard to the orthonormed base of these eigenvectors the endomorphism T:V_n V_n has canonical form. The matrix associated to the endomorphism T is this base is diagonal, having on the main diagonal eigenvalues, written as may times as their multiplicirty order allows and are expressed in function of the associated matrix A corresponding to the endomorphism T in the orthonormed base B , trought the relation D = ^tW AW , where W is the othogonal passing matrix form base B to the base formed out of eigenvectors.

5.3 Isometrical Transformations on punctual euclidian spaces.

Let E=(E,V j) a punctual euclidian space, E being the support set, V the director vector space and j the affine structure function.

If E is endowed with a certain geometrical structure, and f satisfies certain conditions referring to this structure, then f will be named geometrical transformation.

By denoting (s_E ) the transformation group of the set E with regard to the composing operation of the functions, and G a subgroup of his, then the pair (E, G ) will be named the geometric space or the space iwth a fundamental group.

A subset of points F E is named a figure of the geometric space (E, G ). Two figures F₁,F₂ E are called congruent if fI G, such that f(F₁) = F₂.

A property or scale referring to the figures of the geometrical space (E, G ) is called geometric if this is invariant to the transforamtions of the group G .

We call the geometry of the space (E, G ) the theory obtained by the study of the notions, propeties and geometric scales of geometrice referring to the figures of the set E .

If E = (E,V j) and E = (E,V j ) are two affine spaces, then an affine transformation t : E E is uniquely determined by the pair of points AIE, and A IE respectively, and by a linear transformation T : V V