7 paskaita Vektorinės erdvės dimensija.

Apibrėžimas. Vektorinės erdvės bazėje esančiu̧ vektoriu̧ skaičius vadinamas vektorinės erdvės V dimensija, žymima dim_{K} V arba dim V.

Apibrėžimas. Matricos A rangu vadinama matricos eilučiu̧ (stulpeliu̧) tiesinio apvalkalo dimensija ir žymima rankA. Tai maksimalios tiesiškai nepriklausomos eilučiu̧ (stulpeliu̧) sistemos elementu̧ skaičius.

Vektorinės erdvės dimensijos apibrėžimas yra korektiškas. Tai rodo tokia teorema.

Teorema( tiesiniu̧ kombinaciju̧ tiesinis priklausomumas). Tegu v_{1}, ..., v_{m} ∈ [u_{1}, ..., u_{n}] , t.y.

v_{1} = a_{11}u_{1} + a_{12}u_{2} + ⋯ + a_{1n}u_{n}
v_{2} = a_{12}u_{1} + a_{22}u_{2} + ⋯ + a_{2n}u_{n}
⋯
v_{m} = a_{1m}u_{1} + a_{2m}u_{2} + ⋯ + a_{mn}u_{n}

ir m > n. Tada v_{1}, ..., v_{m}− tiesiškai priklausoma sistema.

I̧rodymas. Indukcija pagal n.

1. Jeigu n = 0, ir m > n tai v_{1} = ⋯ = v_{m} = O , o sistema, kurioje yra nulinis vektorius - tiesiškai priklausoma.

2. Tegu n > 0 . Galimi keli atvejai.

i) a_{11} = a_{12} = ⋯ = a_{1m} = 0. Tada v_{1}, ..., v_{m} ∈ [u_{2}, ..., u_{n}] ir pagal indukcijos prielaida̧ v_{1}, ..., v_{m}− tiesiškai priklausoma sistema.

ii) Sakykime ne visi a_{i1} lygūs 0. Tegu a_{11} ≠ 0. Sudarykime naujus vektorius w_{2} = v_{2} − \frac{a_{21}}{a_{11}} v_{1} = γ_{22}u_{2} + ⋯ γ_{2n}u_{n} ⋯ w_{m} = v_{m} − \frac{a_{m1}}{a_{11}} v_{1} = γ_{m2}u_{2} + ⋯ γ_{mn}u_{n}, čia γ_{ij} = a_{ij} − \frac{a_{i1}}{a_{11}} . Turime, kad m − 1 vektorius w_{2}, ..., w_{m} reiškiamas n − 1 vektoriumi u_{2}, ..., u_{n}.

Pagal indukcijos prielaida̧ vektoriai w_{2}, ..., w_{m}− tiesiškai priklausomi: egzistuoja ne visi lygūs nuliui tokie b_{2}, ..., b_{m} (pvz. b_{i} ≠ 0) , kad b_{2}w_{2} + ⋯ + b_{m}w_{m} = 0 ,t.y.

(−b_{2} \frac{a_{21}}{a_{11}} − ⋯ − b_{m} \frac{a_{m1}}{a_{11}} ) v_{1} + b_{2}v_{2} + ⋯ + b_{m}v_{m} = 0.

Kairėje lygybės pusėje esančioje tiesinėje kombinacijoje ne visi koeficientai lygūs nuliui (pvz. b_{i} ≠ 0). Taigi, vektoriu̧ sistema v_{1}, ..., v_{m}− tiesiškai priklausoma sistema.

I̧rodyta.

Išvada. Vektorinės erdvės bazėje esančiu̧ vektoriu̧ skaičius yra pastovus dydis.

I̧rodymas.[....]

Teorema. 1.S is linearly independent. If S is a linearly dependent set in an n-dimensional

space V and V=span(S) then by removing some elements of S we can get a basis of V.

2. If S is a linearly independent subset of V which is not a basis of V then we can get a basis of V by adding some elements to S.

Examples.

1. Consider the following 5 vectors in R^{4}: (1, 2, 3, 4), (1, 1, 0, 0), (1, 2, 1, 0), (0, 1, 2, 3), (1, 0, 0, 0). It can be shown (check!) that these vectors span R^{4}. Since R^{4} is 4-dimensional

(it has the standard basis with 4 vectors), these 5 vectors must be linearly dependent by the theorem about bases. By the theorem about dimension we can through away one of these vectors and get a basis of R^{4}. By the theorem about throwing away extra elements from a spanning set, we can through away a vector which is a linear combination of other vectors in the set. Let us check that the vector (1, 2, 3, 4) is such a vector. In order to find the linear combination which is equal to this vector, we need to solve the system of linear equation:

(1, 2, 3, 4) = (1, 1, 0, 0) · x_{1} + (1, 2, 1, 0) · x_{2} + (0, 1, 2, 3) · x_{3} + (1, 0, 0, 0) · x_{4} . This system of equations has the following augmented matrix:

( \begin{matrix}1 & 1 & 0 & 1| & 1 \\ 1 & 2 & 1 & 0| & 2 \\ 0 & 1 & 2 & 0| & 3 \\ 0 & 0 & 3 & 0| & 4 \end{matrix} )

Using the Gauss-Jordan procedure, we get the following matrix:

( \begin{matrix}1 & 0 & 0 & 0| & 0 \\ 0 & 1 & 0 & 0| & \frac{1}{3} \\ 0 & 0 & 1 & 0| & \frac{4}{3} \\ 0 & 0 & 0 & 1| & \frac{2}{3} \end{matrix} )

Thus x_{1} = 0, x_{2} = \frac{1}{3}, x_{3} = \frac{4}{3}, x_{4} = \frac{2}{3}. So the vector (1, 2, 3, 4) can be thrown away. The other vectors, (1, 1, 0, 0), (1, 2, 1, 0), (0, 1, 2, 3), (1, 0, 0, 0), form a basis of R^{4}. Indeed, they span R^{4} by the theorem about throwing away extra elements, and by the theorem about dimension, every four vectors in a 4-dimensional vector space which span the vector space, form a basis of this vector space.

2. Take two vectors (1, 2, 3, 4), (2, 1, 1, 1) in R^{4}. These vectors are linearly independent because they are not proportional (see the theorem about linearly dependent sets). Thus by the theorem about dimension we can add two vectors and get a basis of R^{4}. Let us add (1, 0, 0, 0) and (0, 1, 0, 0). Notice that when we add vectors we need to make sure that the added vectors are not linear combinations of the previous vectors. In order to check that the four vectors (1, 2, 3, 4), (2, 1, 1, 1), (1, 0, 0, 0), (0, 1, 0, 0) form a basis of R^{4}, we need to check only that they are linearly independent, that is the system of equations:

(0, 0, 0, 0) = (1, 2, 3, 4) · x_{1} + (2, 1, 1, 1) · x_{2} + (1, 0, 0, 0) · x_{3} + (0, 1, 0, 0) · x_{4} has only one, trivial, solution (see the theorem about dimension). This is an

homogeneous system with 4 equations and 4 unknowns. We know that this system has only one solution if and only if the matrix of coefficients is invertible (see the second theorem about inverses). And we know that a square matrix is invertible if and only if its determinant is not zero (see the third theorem about determinants). Thus we need to check that the determinant of the matrix of coefficients of our system is not zero. Maple says that

det ( \begin{matrix}1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 3 & 1 & 0 & 0 \\ 4 & 1 & 0 & 0 \end{matrix} ) = −1

Thus, our four vectors form a basis of R^{4}.

3. Let us prove that the space of functions C[0, 1]is not finite dimensional. By contradiction, suppose that C[0, 1] has a finite dimension n. Consider the

set of n + 1 functions 1, x, x^{2}, ..., x^{n}. It is easy to check that the Wronskian of this set of functions is non-zero (the matrix of derivatives is upper triangular). Thus by the theorem about Wronskian, this set of functions is linearly independent. This contradicts statement theorem about bases: if a vector space is n-dimensional then every set of more than n vectors in this vector space is linearly dependent.

Apibendrinti veiksmai su matricomis.

Teorema. Tegu M, N, P, Q, R, S - šešios Abelio grupės, tarp kuriu̧ apibrėžtos operacijos: M ×N → Q, N ×P → R, Q×P → S, M ×R → S , t.y. ∀m ∈ M, n ∈ N, p ∈ P apibrėžta mn ∈ Q, (mn) p ∈ S, m (np) ∈ S .Tegu visos sandaugos yra distributyvios abieju̧ argumentu̧ atžvilgiu ir ∀m, n, p teisinga (mn) p = m (np) . Jeigu A_{1}, A_{2}, B, C yra matricos virš atitinkamai M, M, N, P ,tai (AB) C = A (BC) ir (A_{1} + A_{2}) B = A_{1}B + A_{2}B .

Be i̧rodymo.

Baziu̧ keitimo matrica, vektoriaus koordinatės.

Lema. Tegu v_{1}, ..., v_{n} tiesiškai nepriklausoma vektorinės erdvės (V, k) sistema, o A ir B matricos virš k , turinčios m stulpeliu̧ ir n eilučiu̧. Jeigu (v_{1}, ..., v_{n}) A = (v_{1}, ..., v_{n}) B , tai A = B.

I̧rodymas. Tegu C = A − B ir (v_{1}, ..., v_{n}) C = 0. Tada

(v_{1}, ..., v_{n}) C = (v_{1}, ..., v_{n}) ( \begin{matrix}c_{11} & ⋯ & c_{1n} \\ & ⋯ \\ c_{n1} & ⋯ & c_{nn} \end{matrix} ) =

= (∑_{s=1}^{n} v_{s}c_{s1}, ..., ∑_{s=1}^{n} v_{s}c_{sn}) = (0, ..., 0) ,

t.y. ∑_{s=1}^{n} v_{s}c_{sj} = 0 su visais j = 1, ..., n. Sistema v_{1}, ..., v_{n} tiesiškai nepriklausoma

, todėl c_{sj} = 0 su visais 1 ≤ s, j ≤ n .

I̧rodyta.

Tegu (V, k) - vektorinė erdvė ir dim V = n , o v_{1}, ..., v_{n} - V bazė. Jeigu v ∈ V ,

tai v = α_{1}v_{1}+⋯+α_{n}v_{n} = (v_{1}, ..., v_{n}) ( \begin{matrix}α_{1} \\ ⋯ \\ α_{n} \end{matrix} ) ir ( \begin{matrix}α_{1} \\ ⋯ \\ α_{n} \end{matrix} ) vadinamas koordinatiniu

stulpeliu.

Tegu v′_{1}, ..., v′_{n} kita V bazė. Tada v′_{1} = c_{11}v_{1} + ⋯ + c_{n1}v_{n}

⋯

v′_{n} = c_{1n}v_{1} + ⋯ + c_{nn}v_{n}

(v′_{1}, ..., v′_{n}) = (v_{1}, ..., v_{n})·C , čia C = ( \begin{matrix}c_{11} & ⋯ & c_{1n} \\ ⋯ & ... & ⋯ \\ c_{n1} & ⋯ & c_{nn} \end{matrix} ) vadinama vienos bazės

keitimo kita baze keitimo matrica.

Iš lemos turime, kad taip apibrėžta matrica C yra vienintėlė. Kaip keičiasi vektoriaus v koordinatinis stulpelis keičiantis bazei ?

v = (v_{1}, ..., v_{n}) ( \begin{matrix}α_{1} \\ ⋯ \\ α_{n} \end{matrix} ) = (v′_{1}, ..., v′_{n}) ( \begin{matrix}α′_{1} \\ ⋯ \\ α′_{n} \end{matrix} ) = [(v_{1}, ..., v_{n}) · C] ( \begin{matrix}α′_{1} \\ ⋯ \\ α′_{n} \end{matrix} ) =

(v_{1}, ..., v_{n}) · [C ( \begin{matrix}α′_{1} \\ ⋯ \\ α′_{n} \end{matrix} )].

Iš lemos turime, kad

( \begin{matrix}α_{1} \\ ⋯ \\ α_{n} \end{matrix} ) = C· ( \begin{matrix}α′_{1} \\ ⋯ \\ α′_{n} \end{matrix} ) .

Teiginys. Keitimo matrica pasižymi šiomis sa̧vybėmis:

1. Bazės keitimo ta pačia baze keitimo matrica yra vienetinė.

2. Jeigu pirmosios bazės keitimo antra̧ja baze keitimo matrica yra C , o antrosios bazės keitimo trečia̧ja baze yra D , tai pirmosios bazės keitimo trečia̧ja baze matrica yra CD.

3. Keitimo matrica yra neišsigimusi matrica.

I̧rodymas.1. (v_{1}, ..., v_{n}) C = (v_{1}, ..., v_{n}) I ⟹ C = I.

2. Tegu bazės v_{1}, ..., v_{n} keitimo baze v′_{1}, ..., v′_{n} matrica yra C, o bazės v′_{1}, ..., v′_{n} keitimo baze v1^{′′}, ..., vn^{′′} matrica yra D. Tada

(v′′_{1}, ..., v′′_{n}) = (v′_{1}, ..., v′_{1}) D = ((v_{1}, ..., v_{n}) C) D = (v_{1}, ..., v_{n}) CD.

3. Tada, kai v′′_{1} = v_{1}, ..., v′′_{n} = v_{n} iš 2. ir 1. teiginiu̧ turime CD = I ⟹ C = D^{−1}, taigi, matrica C− neišsigimusi.