1. VEKTORINĖS ERDVĖS.

Apibrėžimas. Tegu K− kūnas. Aibė V , kurioje apibrėžta sudėties ir daugybos iš kūno K elemento operacijos, vadinama vektorine erdve, jeigu su visais u, v, w ∈ V ir a, b ∈ K išpildytos šios sa̧lygos:

1. u + v ∈ V;

2. (u + v) + w = u + (v + w) ;

3. u + v = v + u;

4. ∃o : v + o = v;

5. ∃ (−v) : v + (−v) = o;

6. a (u + v) = au + av;

7. (a + b) u = au + ab; 8.(ab) u = a (bu) ;

9. 1_{K} · u = u. Vektorinės erdvės elementai vadinami vektoriais. Pavyzdžiai.1 . Su kiekvienu n K^{n} = {(a_{1}, ..., a_{n}) : a_{i} ∈ K} - eilučiu̧ su n kom-

ponenčiu̧ vektorinė erdvė. Ši erdvė dar vadinama aritmetine n− erdve. Eilučiu̧ komponentės vadinamos vektoriaus koordinatėmis.

2. Visu̧ realiu̧ju̧ tolydžiu̧ intervale [0, 1] funkciju̧ vektorinė erdvė C [0, 1] virš kūno R.

3. Su visais k ir n visu̧ k × n matricu̧ vektorinė erdvė M_{k,n} (K) virš kūno K.

4. Kompleksiniu̧ skaiciu̧ aibė C - vektorinė erdvė virš kūno R. Apibrėžimas. Išraiška̧ a_{1}v_{1}+a_{2}v_{2}+⋯+a_{n}v_{n} vadiname vektoriu̧ v_{1}, v_{2}, ..., v_{n}

tiesine kombinacija; čia a_{1}, a_{2}, ..., a_{n} ∈ K.

Apibrėžimas. Vektoriu̧ sistema̧ v_{1}, ..., v_{n} vadinsime tiesiškai priklausoma, jei tarp šiu̧ vektoriu̧ yra toks v_{i} , kuris yra likusiu̧ sistemos vektoriu̧ tiesinė kombinacija: v_{i} = a_{1}v_{1} + ⋯ + a_{i−1}v_{i−1} + a_{i+1}v_{i+1} + ⋯ + a_{n}v_{n}. Jeigu vektoriu̧ sistemoje v_{1}, ..., v_{n} nėra vektoriaus, tiesiškai priklausomo nuo likusiu̧, tai šia̧ sistema̧ vadina teisiškai nepriklausoma.

Susitarta, kad tuščios vektoriu̧ sistemos tiesinė kombinacija lygi noliniam vektoriui: \underbrace{v + ⋯ + u}_{0 dėmenu̧} = 0 .

Examples.

1. Every two vectors in the line R are linearly dependent (one vector is proportional to the other one).

2. Every three vectors a, b, c on a plane R^{2} are linearly dependent. Indeed, if a and b are parallel then one of them is a multiple of another one, and so a and b are linearly dependent which implies that all three vectors are linearly dependent. If a and b are not parallel then we know that every vector on the plane, including the vector c, is a linear combination of aand b. Thus a, b, c are linearly dependent.

3. If a subset S of R^{n} consists of more than n vectors then S is linearly dependent. ( Prove it!)

4. The set of polynomials x + 1, x^{2} + x + 1, x^{2} − 2x − 2, x^{2} − 3x + 1 is linearly dependent. To prove that we need to find numbersa, b, c, d not all equal to 0 such that a(x + 1) + b(x^{2} + x + 1) + c(x^{2} − 2x − 2) + d(x^{2} − 3x + 1) = 0. This leads to a homogeneous system of linear equations with 4 unknowns and 3 equations. Such a system must have a non-trivial solution by the theorem about homogeneous systems of linear equations.

5. Let f_{1}, f_{2}, ..., f_{n} be functions in C[0, 1] each of which has firstn − 1 derivatives. The determinant of the following matrix

( \begin{matrix}f_{1}(x) f_{2}(x) ... .f_{n}(x) \\ f′_{1}(x) f′_{2}(x) .... f′_{n}(x) \\ f′′_{1}(x) f′′_{2}(x) .... f′′_{n}(x) \\ ⋯ \\ f_{1}^{(n−1)}(x) f_{2}^{(n−1)}(x) ....f_{n}^{(n−1)}(x) \end{matrix} )

is called the Wronskian of this set of functions.

Theorem. Letf_{1}, f_{2}, ..., f_{n} be functions in C[0, 1] each of which has first n − 1derivatives. If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent

Apibrėžimas. Jeigu A ⊂ {v_{1}, ..., v_{n}} ⊂ B , tai A vadinamas sistemos v_{1}, ..., v_{n} posistemiu, o B− viršsistemiu.

Vektoriu̧ sistemu̧ savybės.

1. Sistema iš vieno vektoriaus {v} yra tiesiškai priklausoma kai v = 0 ir tiesiškai nepriklausoma kai v ≠ 0.

2. Tiesiškai priklausomos sistemos v_{1}, ..., v_{n} viršsistemis yra tiesiškai priklausomas.

3. Tiesiškai nepriklausomos sistemos v_{1}, ..., v_{n} posistemis yra tiesiškai nepriklausomas.

4. Vektoriu̧ sistema, kurioje yra nulinis vektorius, yra tiesiškai priklausoma.

5. Vektoriu̧ sistema, kurioje yra du sutampantys vektoriai, yra tiesiškai priklausoma.

6. Vektoriu̧ sistema v_{1}, ..., v_{n} yra tiesiškai priklausoma tada ir tik tada, kai egzistuoja tokie a_{1}, ..., a_{n} ∈ K ir a_{i} ≠ 0 , 1 ≤ i ≤ n, kad a_{1}v_{1} + ⋯ + a_{n}v_{n} = 0.

I̧rodymas. (⇒)Tegu v_{1}, ..., v_{n}− tiesiškai priklausoma sistema. Tada egzistuoja v_{i} = a_{1}v_{1} + ⋯ + a_{i−1}v_{i−1} + a_{i+1}v_{i+1} + ⋯ + a_{n}v_{n} ir todėl a_{1}v_{1} + ⋯ + a_{i−1}v_{i−1} + (−1) v_{i} + a_{i+1}v_{i+1} + ⋯ + a_{n}v_{n} = 0.

(⇐) Tegu a_{1}v_{1} + ⋯ + a_{i−1}v_{i−1} + a_{i}v_{i} + a_{i+1}v_{i+1} + ⋯ + a_{n}v_{n} = 0 ir a_{i} ≠ 0 , tada v_{i} = −\frac{a_{1}}{a_{i}} v_{1} − ⋯ − \frac{a_{i−1}}{a_{i}} v_{i−1} − \frac{a_{i+1}}{a_{i}} v_{i+1} − ⋯ − \frac{a_{n}}{a_{i}} v_{n}.

I̧rodyta.

7. Vektoriu̧ sistema v_{1}, ..., v_{n} yra tiesiškai nepriklausoma tada ir tik tada, kai iš a_{1}v_{1} + ⋯ + a_{n}v_{n} = 0 turime a_{1} = ⋯ = a_{n} = 0.

I̧rodymas. Akivaizdu.

Apibrėžimas. Vektorinės erdvės V virš kūno K netuščias poaibis U vadinamas poerdviu, jeigu

1) u_{1} + u_{2} ∈ U su visais u_{1}, u_{2} ∈ U;

2) au ∈ U su visais a ∈ K ir su visais u ∈ U.

Vektorinės erdvės V poerdvis U yra pats vektorinė erdvė tu̧ pačiu̧ operaciju̧ kaip ir V atžvilgiu.

Positive Examples.

1. The whole space R^{n} is a subspace of itself. And the set consisting of one vector, 0, is a subspace of any space.

2. In R^{2}, consider the set W of all vectors which are parallel to a given line L. It is clear that the sum of two vectors which are parallel to L is itself parallel to L, and a scalar multiple of a vector which is parallel to L is itself parallel to L. Thus W is a subspace.

3. A similar argument shows that in R^{3}, the set W of all vectors which are parallel to a given plane (line) is a subspace.

4. The set of all polynomials is a subspace of the space of continuous functions on [0, 1],C[0, 1]. The set of all polynomials whose degrees do not exceed a given number, is a subspace of the vector space of polynomials, and a subspace of C[0, 1].

5. The set of differentiable functions is also a subspace of C[0, 1].

Negative Examples.

1. In R^{2}, the set of all vectors which are parallel to one of two fixed nonparallel lines, is not a subspace. Indeed, if we take a non-zero vector parallel to one of the lines and add a non-zero vector parallel to another line, we get a vector which is parallel to neither of these lines.

2. The set of polynomials of degree 2 is not a subspace of C[0, 1]. Indeed, the sum of x^{2} + x and −x^{2} is a polynomial of degree 1.

Apibrėžimas. Tegu v_{1}, ..., v_{n} ∈ V, − vektoriu̧ sistema. Vektorinės erdvės V poaibis {a_{1}v_{1} + ⋯ + a_{n}v_{n}|a_{1}, ..., a_{n} ∈ K} = [v_{1}, ..., v_{n}] vadinamas vektoriu̧ sistemos v_{1}, ..., v_{n} tiesiniu apvalkalu. Vektoriu̧ sistema v_{1}, ..., v_{n} vadinama generuojančia ši̧ tiesini̧ apvalkala̧ sistema.

Aišku, kad vektoriu̧ sistemos v_{1}, ..., v_{n} tiesinis apvalkalas [v_{1}, ..., v_{n}] yra poerdvis, o bet koks generuojančios sistemos virš- sistemis yra generuojanti sistema.

Examples.

1. Let V = R^{3}. Let S consist of one non-zero vector A. Then spanA consists of all vectors of the form xA. In other words, spanA consists of all vectors which are proportional to A, or span(A) is the set of vector parallel to A.

2. Let V = R^{3}, S = (1, 0, 0), (0, 1, 0). Then span(S) consists of all vectors of the form x(1, 0, 0) + y(0, 1, 0) = (x, y, 0), that is span(S) consists of all vectors parallel to the (x, y)-plane. More generally, if we take any two non-parallel vectors A and B in R^{3} then spanA, B is the subspace of all vectors which are parallel to the plane containing A and B.

3. Let V = R^{3}, S = (1, 0, 0), (0, 1, 0), (0, 0, 1). Then span(S) coincides with the whole R^{3}. More generally if V = R^{n} thenV is spanned by the set of basic vectors(1, 0, ..., 0), (0, 1, ..., 0), ..., (0, 0, ..., 1).

Theorem. Two subsets S_{1} and S_{2} of a vector space V span the same subspace if and only if every vector of S_{1} is a linear combination of vectors of S_{2} and every vector of S_{2} is a linear combination of vectors of S_{1}.

The proof is left as an exercise.

Examples.

1. Vectors a = (1, 2, 3), b = (0, 1, 2),and c = (0, 0, 1) span R^{3}. Indeed, we know that R^{3} is spanned by the vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1).Thus we need to show that the sets a, b, c and i, j, k span the same subspace. By the previous theorem, we need to show that every vector from the first subset is a linear combination of vectors from the second subset and conversely every vector from the second subset is a linear combination of vectors of the first subset. It is clear that a, b, c are linear combinations of i, j, k (as any vector in R^{3}). So we need to show only that i, j, k are linear combinations of a, b, c. This is easy to check :i = a − 2b + c; j = b − 2c; k = c.

2. Polynomials f_{1} = x^{2} + 2x + 1, f_{2} = x + 1and f_{3} = x + 2 in the space of all polynomials P span the subspace P_{2} of all polynomials of degree not exceeding 2. Indeed, it is clear that P_{2} is spanned by the polynomials p_{1} = 1, p_{2} = x and p_{3} = x^{2}. So we need to show that the setsf_{1}, f_{2}, f_{3} and p_{1}, p_{2}, p_{3} span the same subspace. It is clear that f_{1}, f_{2}, f_{3} are linear combinations of p_{1}, p_{2}, p_{3}. Conversely, it is easy to check that p_{1} = f_{3} − f_{2}; p_{2} = 2f_{2} − f_{3}; p_{3} = f_{1} − 3f_{2} + f_{3}.

3. Leta = (1, 2, 3, 4), b = (3, −1, 5, 2), c = (−1, 2, 0, 1), d = (2, 1, 4, 5). Determine whether spana, b=spanc, d. We need to check if a and bare linear combinations of vectors candd, and whetherc and d are linear combinations of vectors a and b. By definition of a linear combination, the vector a is a linear combination of c and d if there exist x and y such that a = xc + yd This gives the following system of linear equations:

1 = −x + 2y

2 = 2x + y

3 = 0x + 4y

4 = x + 5y.

This system does not have a solution, so a is not a linear combination of c and d, so spana, b is not equal to spanc, d.

Apibrėžimas. Generuojančia̧ sistema̧ vadiname minimalia, jeigu bet koks jos posistemis nėra generuojanti sistema.

Tiesiškai nepriklausoma sistema vadinama maksimalia, jeigu bet koks jos viršsistemis yra tiesiškai priklausoma sistema.

Dabar pateiksime svarbias tiesiniu̧ apvalkalu̧ savybes.

Teiginys 1. Jeigu vektoriu̧ sistemos v_{1}, ..., v_{n} vektoriai yra vektoriu̧ sistemos u_{1}, ..., u_{m} vektoriu̧ tiesinės kombinacijos, t.y. v_{1}, ..., v_{n} ∈ [u_{1}, ..., u_{m}] , ir vektorius v ∈ [v_{1}, ..., v_{n}] , tai v ∈ [u_{1}, ..., u_{m}] .

I̧rodymas.Paliekame i̧rodyti studentams.

Teiginys 2. Tegu v_{1}, ..., v_{n}− tiesiškai nepriklausoma sistema, o v, v_{1}, ..., v_{n}− tiesiškai priklausoma sistema. Tada vektorius v ∈ [v_{1}, ..., v_{n}] .

I̧rodymas.Paliekame i̧rodyti studentams.

Teorema. Tegu v_{1}, ..., v_{n}− vektorinės erdvės V vektoriu̧ sistema. Šie trys teiginiai yra ekvivalentūs:

1. v_{1}, ..., v_{n}− tiesiškai nepriklausoma ir vektorinȩ erdvȩ V generuojanti sistema.

2. v_{1}, ..., v_{n}− maksimali tiesiškai nepriklausoma sistema.

3. v_{1}, ..., v_{n}− minimali vektorinȩ erdvȩ V generuojanti sistema.

I̧rodymas. 1⇒2. Kadangi v_{1}, ..., v_{n}− generuojanti erdvȩ V sistema, tai kiekvienas vektorius v ∈ V yra tiesinė vektoriu̧ sistemos v_{1}, ..., v_{n} kombinacija, t.y. v ∈ [v_{1}, ..., v_{n}] ⟹ sistema v, v_{1}, ..., v_{n}− tiesiškai priklausoma ⟹ v_{1}, ..., v_{n}− maksimali tiesiškai nepriklausoma vektoriu̧ sistema.

2⇒3. Su kiekvienu v ∈ V sistema v, v_{1}, ..., v_{n} yra tiesiškai priklausoma sistema. pagal teigini̧ 1 v ∈ [v_{1}, ..., v_{n}] ⟹ v_{1}, ..., v_{n}− generuojanti vektoriu̧ sistema. \underline{Sakykime}, vektoriu̧ sistema v_{2}, ..., v_{n} irgi yra vektorinȩ erdvȩ V generuojanti sistema⟹ kiekvienas vektorius v ∈ V ( tame tarpe ir v_{1}, ..., v_{n} ) yra vektoriu̧ v_{2}, ..., v_{n} tiesinė kombinacija⟹ v_{1}, ..., v_{n} ∈ [v_{2}, ..., v_{n}] ⟹ v_{1} = a_{2}v_{2} + ⋯ + a_{n}v_{n} ⟹ sistema v_{1}, ..., v_{n}− tiesiškai priklausoma, prieštaravimas prielaidai.

3⇒ 1. \underline{Sakykime}, v_{1} = a_{2}v_{2} + ⋯ + a_{n}v_{n} ⟹ v_{1} ∈ [v_{2}, ..., v_{n}] . Iš teiginio 2 turime, kad kiekvienas vektorinės erdvės V vektorius v ∈ [v_{2}, ..., v_{n}] , prieštaravimas prielaidai.

I̧rodyta.

Apibrėžimas. Vektorinės erdvės V vektoriu̧ sistema, tenkinanti viena̧ iš teoremos sa̧lygu̧, vadinama vektorinės erdvės V baze. Bazėje esančiu̧ vektoriu̧ skaičius vadinamas vektorinės erdvės V dimensija, žymima dim_{K} V arba dim V.

Positive examples.

1.The set of vectors V_{1} = (1, 0, ..., 0), V_{2} = (0, 1, ..., 0), ..., V_{2} = (0, ..., 1) is a basis of R^{n}.

2.The set of vectors(1, 2), (2, 3) is a basis of R^{2}. Indeed, these vectors are linearly independent because they are not proportional. In order to check that R^{2} is spanned by these vectors, it is enough to check that (1, 0) and (0, 1) are linear combinations of them (theorem about spans):

(1, 0) = −3 · (1, 2) + 2 · (2, 3);

(0, 1) = 2 · (1, 2) − (2, 3).

In fact, every two non-parallel vectors in the plane R^{2} form a basis of R^{2}.

3. The set of polynomials 1, x, x^{2} is a basis of the space of polynomials of degree at most 2.

4.The set of functionsx, e^{x}, e^{2x} is a basis of the subspaceV of C[0, 1] spanned by these functions. Indeed, these functions are linearly independent and V is spanned by these functions by the definition of V.

5.The set of matrices:( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} )

is a basis of the space of all 2 by 2 matrices. First we need to prove that these matrices are linearly independent. Indeed, if we take a linear combination of these

matrices with coefficients a, b, c, d, we get the matrix ( \begin{matrix}a & b \\ c & d \end{matrix} ) . This matrix is

equal to the zero matrix only if a = b = c = d = 0. Second, we need to show that these 4 matrices span the space of all 2 by 2 matrices. Indeed, every 2 by 2 matrix

( \begin{matrix}a & b \\ c & d \end{matrix} ) is the linear combination of our four matrices with coefficientsa, b, c, d.

Negative examples

1.The set of two vectors u = (1, 2, 3) and v = (2, 3, 4) is not a basis of R^{3}. Indeed, although these vectors are linearly independent, they do not span R^{3}. For example, the vector (1, 0, 0) is not equal to a linear combination of u andv.

2.The set of four vectors u = (1, 2, 3), i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) is not a basis of R^{3}. Indeed, although these vectors span R^{3} (even i, j, k span R^{3}), these vectors are not linearly independent because u = i + 2j + 3k.

3.The set of functions u = sin(x), v = cos(x), w = x is not a basis of C[0, 1] because the function e^{x} does not belong to the spanu, v, w ( prove it using the Wronskian).

In fact C[0,1] does not have a finite basis at all. Vektorinės erdvės dimensijos apibrėžimas yra korektiškas. Tai rodo tokia teo-

rema.

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.

Pastaba. Matricos A rangas - tai matricos eilučiu̧ ( stulpeliu̧) tiesinio apvalkavo dimensija, žymima rankA . Tai maksimalios tiesiškai nepriklausomos eilučiu̧ (stulpeliu̧) sistemos elementu̧ skaičius.

Svarbiu vektorinio poerdvio pavyzdžiu yra homogeninės tiesiniu̧ lygčiu̧ sistemos sprendiniu̧ aibė.

Teorema. Homogeninės tiesiniu̧ lygčiu̧ sistemos

\{ \begin{matrix}a_{11}x_{1} + ⋯ + a_{1n}x_{n} = 0 \\ ⋯ \\ a_{m1}x_{1} + ⋯ + a_{mn}x_{n} = 0 \end{matrix} arba AX = O

sprendiniu̧ aibė yra stulpeliu̧ aritmetinės erdvės K_{n} poerdvis, kurio dimensija lygi n−rankA.

I̧rodymas. Homogeninȩ tiesiniu̧ lygčiu̧ sistema̧ užrašykime taip x_{1}u_{1} + ⋯ + x_{n}u_{n} = 0, čia u_{1}, ..., u_{n}− matricos A stulpeliai.

Tegu u = ( \begin{matrix}α_{1} \\ ⋯ \\ α_{n} \end{matrix} ) ir v = ( \begin{matrix}β_{1} \\ ⋯ \\ β_{n} \end{matrix} ) − sistemos sprendiniai, t.y.

\begin{matrix}α_{1}u_{1} + ⋯ + α_{n}u_{n} ≡ 0 \\ β_{1}u_{1} + ⋯ + β_{n}u_{n} ≡ 0 \end{matrix} , todėl (aα_{1} + bβ_{1}) u_{1} + ⋯ (aα_{n} + bβ_{n}) u_{n} ≡ 0 ir vek-

torius au+bv yra sistemos sprendinys. Gavome, kad sprendiniu̧ aibė yra stulpeliu̧ aritmetinės erdvės K_{n} poerdvis.

Tarp tiesinės lygčiu̧ sistemos stulpeliu̧ yra r = rank A tiesiškai neprikausomu̧ stulpeliu̧. Tegu tai stulpeliai u_{1}, ..., u_{r}. Tada likusius n − r stulpelius užrašykime ju̧ tiesinėmis kombinacijomis:

\begin{matrix}u_{r+1} = b_{r+1,1}u_{1} + ⋯ + b_{r+1,r}u_{r} \\ ⋯ \\ u_{n} = b_{n1}u_{1} + ⋯ + b_{nr}u_{r} \end{matrix} .

Stulpeliai z_{r+1} = ( \begin{matrix}b_{r+1,1} \\ ⋯ \\ b_{r+1,r} \\ −1 \\ 0 \\ ⋯ \\ 0 \end{matrix} ) , ..., z_{n} = ( \begin{matrix}b_{n1} \\ ⋯ \\ b_{nr} \\ 0 \\ ⋯ \\ 0 \\ −1 \end{matrix} ) yra homogeninȩ tiesiniu̧ lygčiu̧

sistemos sprendinys. Šie stulpeliai yra tiesiškai nepriklausomi.

Tegu x = ( \begin{matrix}x_{1}^{0} \\ ⋯ \\ x_{r}^{0} \\ x_{r+1}^{0} \\ ⋯ \\ x_{n}^{0} \end{matrix} ) − koks nors sistemos sprendinys.

Tada y = x+x_{r+1}^{0}z_{r+1} +⋯+x_{n}^{0}z_{n} irgi yra sistemos sprendinys. Šio sprendinio komponentės, pradedant (r + 1) − a̧ja̧, lygios 0. Lygiomis nuliui bus ir likusios r komponenčiu̧, nes vektoriu̧ sistema u_{1}, ..., u_{r}− tiesiškai nepriklausoma.

Taigi, y = 0 ir x = −x_{r+1}^{0}z_{r+1} − ⋯ − x_{n}^{0}z_{n} ,t.y. stulpeliai z_{r+1}, ..., z_{n}− tiesiškai neprikausoma generuojanti sistemos sprendiniu̧ vektorinio poerdvio sistema. Ji vadinama fundamentalia̧ja sprendiniu̧ sistema.

I̧rodyta.

Ištirkime taip pat nehomogeninės tiesiniu̧ lygčiu̧ sistemos sprendiniu̧ aibės struktūra̧.

Teorema. Nehomogeninės tiesiniu̧ lygčiu̧ sistemos

\{ \begin{matrix}a_{11}x_{1} + ⋯ + a_{1n}x_{n} = b_{1} \\ ⋯ \\ a_{m1}x_{1} + ⋯ + a_{mn}x_{n} = b_{m} \end{matrix} arba AX = b

sprendiniu̧ aibė yra v_{0}+U = {v_{0} + u : u ∈ U}, čia v_{0}− atskiras nehomogeninės sistemos sprendinys, o U− atitinkamos homogeninės sistemos AX = O sprendiniu̧ poerdvis.

I̧rodymas. Tegu v_{0}− koks nors atskiras sistemos AX = b sprendinys, o u− bet koks atitinkamos homogeninės sistemos AX = O sprendinys, u ∈ U. Tada A (v_{0} + u) = Av_{0} + Au = b + O = b ir v_{0} + u− sistemos AX = b sprendinys.

Priešingai.Tegu v_{0}− atskiras sistemos AX = b sprendinys, o v− bet koks šios sistemos sprendinys. Tada A (v_{0} − v) = Av_{0} − Av = b − b = O ir v_{0} − v ∈ U, t.y. v_{0} − v = u ir v = v_{0} + u ∈ v_{0} + U.

I̧rodyta.

Apibrėžimas. Tegu U yra vektorinės erdvės V poerdvis, o v ∈ V. Aibė v + U = {v + u : u ∈ U} vadinama tiesine daugdara arba plokštuma erdvėje V. Jeigu dim U = 1, tai tiesinė daugdara vadinama tiese.

Taigi, nehomogeninės tiesiniu̧ lygčiu̧ sistemos sprendiniu̧ aibė yra tiesinė daugdara stulpeliu̧ aritmetinėje erdvėje K_{n}.

Pavyzdys. Nagrinėkime homogeninȩ lygti̧

ax + by = 0.

Šios homogeninės lygties sprendiniu̧ aibė yra stulpelio ( \begin{matrix}−b \\ a \end{matrix} ) tiesinis apval-

kalas T = [( \begin{matrix}−b \\ a \end{matrix} )] .Geometriškai - tai tiesė l dvimatėje plokštumoje, einati per

koordinačiu̧ pradžia̧ vektoriaus ( \begin{matrix}−b \\ a \end{matrix} ) kryptimi.

Nagrinėkime dabar nehomogeninȩ lygti̧

ax + by = c, c ≠ 0

Turime a (−b)+b (a + \frac{c}{b}) = c ir todėl aibė ( 0_{\frac{c}{b}} )+T = ( 0_{\frac{c}{b}} )+[( \begin{matrix}−b \\ a \end{matrix} )] yra

lygties sprendiniu̧ aibė. Geometriškai - tai tiesė dvimatėje plokštumoje lygiagreti tiesei l ir paslinktai vektoriaus ( 0_{\frac{c}{b}} ) kryptimi.

Figure 1.1: 13