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A. Alaca

Linear Algebra II

Vector Spaces

MATH 2107 LINEAR ALGEBRA II LECTURE NOTES c se Alaca Ay¸

Lecture # 1: Thursday, September 9, 2021

Section 4.1 VECTOR SPACES and SUBSPACES

Text book: Linear Algebra and its applications, 6E, By David C. Lay, Steven R. Lay, Judi J. McDonald

Please do not repost these notes elsewhere. These notes may be incomplete or contain errors/typos.

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A. Alaca

Linear Algebra II

Vector Spaces

2

VECTOR SPACES Definition: Let V be a set on which two operations ⊕ (addition) and ⊙ (multiplication) have been defined. If the following axioms hold for all u, v, w ∈ V and for all c, d ∈ R, then V is called a real vector space and its elements are called vectors. It is denoted by (V, ⊕, ⊙). 1. u ⊕ v ∈ V

(Closure under addition)

2. u ⊕ v = v ⊕ u

(Commutativity)

3. (u ⊕ v) ⊕ w = u ⊕ (v ⊕ w)

(Associativity)

4. ∃ (unique) 0 ∈ V such that u ⊕ 0 = u 5. For each u ∈ V , ∃ (unique) −u ∈ V such that u ⊕ (−u) = 0. 6. c ⊙ u ∈ V

(Closure under scalar multiplication)

7. c ⊙ (u ⊕ v) = c ⊙ u ⊕ c ⊙ v

(Distributivity)

8. (c + d) ⊙ u = (c ⊙ u) ⊕ (d ⊙ u)

(Distributivity)

9. (c d) ⊙ u = c ⊙ (d ⊙ u) 10. 1 ⊙ u = u Remarks: An element of a vector space is called a “vector”. Instead of R, if we use C, then (V, ⊕, ⊙) is called a complex vector space.

A. Alaca

Linear Algebra II

Vector Spaces

Theorem: Let V be a vector space, u be a vector in V and c be a scalar. Then, (a) 0 u = 0. (b) c 0 = 0. (c) (−1)u = −u. (d) If c u = 0, then either c = 0 or u = 0.

Proof. (a) 0u = (0 + 0)u = 0u + 0u, 0u + (−0u) = (0u + 0u) + (−0u) = 0u + (0u + (−0u)) 0 = 0u + 0 = 0u

(by axiom (8)) (adding −0u on both sides) (by axiom (3)) (by axiom (5)) (by axiom (4)).

(b) c 0 = c (0 + 0) = c 0 + c 0, c 0 + (−c 0) = (c 0 + c 0) + (−c 0), = c 0 + (c 0 + (−c 0)) 0 = c0+0 = c0

(by axiom (7)) (adding −c 0 on both sides) (by axiom (3)) (by axiom (5)) (by axiom (4)).

(c) 0

= =

0u (1 + (−1))u

(by (a))

= =

1 u + (−1) u u + (−1) u

(by axiom (8)) (by axiom (10))

=⇒ (−1)u = −u

(by axiom (5)).

(d) Suppose cu = 0 but c 6= 0 (If c = 0, there is nothing to prove). Then, u = 1u =

1  1 1 c u = (c u) = 0 = 0 c c c

(by (b)).

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Linear Algebra II

Vector Spaces

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Example 1: Let V := {(1, y) | y ∈ R} with (1, x) ⊕ (1, y) = (1, x + y),

c ⊙ (1, x) = (1, cx).

Show that (V, ⊕, ⊙) is a vector space. Solution: Geometricaly, the set V is the x = 1 line in R2 . y

(1, 2) ⊕ (1, 3) = (1, 5)

(1, 3) (1, 2) (1, 1) (1, 0)

x

−2 ⊙ (1, 1) = (1, −2)

l:x=1 We have to show that ⊕ and ⊙ satisfy the ten conditions in the definition of a vector space. 1. (1, x) ⊕ (1, y) = (1, x + y) ∈ V . 2. (1, x) ⊕ (1, y) = (1, x + y) = (1, y + x) = (1, y) ⊕ (1, x). 3. 

 (1, x) ⊕ (1, y) ⊕ (1, z) = (1, x + y) ⊕ (1, z )     = 1, (x + y) + z = 1, x + (y + z) = (1, x) ⊕ (1, y + z )   = (1, x) ⊕ (1, y) ⊕ (1, z) .

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Linear Algebra II

Vector Spaces

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4. (1, 0) ∈ V and for all (1, x) ∈ V we have (1, 0) ⊕ (1, x) = (1, 0 + x) = (1, x). Hence “zero vector” of V is (1, 0). 5. If (1, x) ∈ V , then −1 ⊙ (1, x) = (1, −x) ∈ V and we have (1, x) ⊕ (1, −x) = (1, x − x) = (1, 0). So , the additive inverse of (1, x) is (1, −x). 6. c ⊙ (1, x) = (1, cx) ∈ V . 7.     c ⊙ (1, x) ⊕ (1, y ) = c ⊙ (1, x + y) = 1, c(x + y)

= (1, cx + cy) = (1, cx) ⊕ (1, cy )     = c ⊙ (1, x) ⊕ c ⊙ (1, y) .

8. (c + d) ⊙ (1, x) =



 1, (c + d)x = (1, cx + dx)

= (1, cx) ⊕ (1, dx)     = c ⊙ (1, x) ⊕ d ⊙ (1, x) . 9. 

c ⊙ d ⊙ (1, x)



= c ⊙ (1, dx) = (1, c(dx)) = (1, (cd)x) = (cd) ⊙ (1, x).

10. 1 ⊙ (1, x) = (1, 1x) = (1, x). Warning: The set V := {(1, y) | y ∈ R} is not a vector space with the usual addition and the scalar multiplication that we know in R2 .

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Linear Algebra II

Vector Spaces

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Example 2: Set R∗ = R \ {0}. Let V := {(x, y) | x, y ∈ R∗ } with (x, y) ⊕ (z, w) = (xz, yw),

c ⊙ (x, y) = (cx, cy ).

Is (V, ⊕, ⊙) a vector space? Solution: 1. (x, y) ⊕ (z, w) = (xz, yw) ∈ V . 2. (x, y) ⊕ (z, w) = (xz, yw) = (zx, wy ) = (z, w) ⊕ (x, y). 3. 

 (a, b) ⊕ (c, d) ⊕ (e, f ) = (ac, bd) ⊕ (e, f )     = (ac)e, (bd)f = a(ce), b(df ) = (a, b) ⊕ (ce, df )   = (a, b) ⊕ (c, d) ⊕ (e, f ) .

4. The zero vector of V is (1, 1) ∈ V as (1, 1) ⊕ (a, b) = (1a, 1b) = (a, b). Remark: We can find the zero vector of V as follows: Take any (a, b) ∈ V . Then a 6= 0 and b 6= 0. We need to find (x, y) ∈ V such that (a, b) ⊕ (x, y) = (a, b). Now (a, b) ⊕ (x, y) = (ax, by) = (a, b) ⇐⇒ ax = a and by = b ⇐⇒ x = 1 and y = 1 (since a, b 6= 0) ⇐⇒ (x, y ) = (1, 1). 1 1 ∈ R∗ , ∈ R∗ , and we have b a 1 1   1 1 (a, b) ⊕ , = a· , b· = (1, 1). b a b a  1 1 , . Thus the negative of (a, b) is a b

5. Let (a, b) ∈ V . Then a ∈ R∗ , b ∈ R∗ ,

6. c ⊙ (x, y) = (cx, cy) ∈ V if c 6= 0. So, the axiom 6 is not true for all c ! Hence (V, ⊕, ⊙) is not a vector space.

A. Alaca

Linear Algebra II

Vector Spaces

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Example 3: For any n ≥ 1, Rn is a vector space. For n = 1, we have R1 = R, and ⊕ and ⊙ coincides with the usual + and ·, respectively, over R. It is easy to verify all the axioms in the definition of vector space for (R, +, ·). For n = 2:           v1 u1 + v1 cu1 u1 u1 ⊕ = , c⊕u= c⊕ = . u⊕v = u2 v2 u2 cu2 u2 + v2       0 u1 −u1 2 0= (zero vector), If u = ∈ R =⇒ −u = ∈ R2 and −u2 0 u2         u1 −u1 u1 − u1 0 u ⊕ (−u) = ⊕ = = = 0. u2 −u2 u2 − u2 0 Example 4: For any positive integers m and n, the set of all m × n matrices form a vector space with the usual matrix addition and scalar multiplication. This vector space is denoted by Mm×n or Mmn . Example 5: For n ≥ 0, set Pn = {all polynomials of degree ≤ n with real coefficients}. If p(x), q(x) ∈ Pn . Then, p(x) = a0 + a1 x + · · · + an xn , q (x) = b0 + b1 x + · · · + bn xn , p(x) ⊕ q(x) = (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn , c ⊙ p(x) = ca0 + ca1 x + · · · + can xn . (Pn , ⊕, ⊙) is a real vector space. The zero vector (zero polynomial) of Pn is 0 = 0 + 0x + · · · + 0xn , whose degree is not defined. The negative of p ∈ Pn is (−p)(x) = −p(x) = −a0 − a1 x − · · · − an xn . Example 6: Let F = { All real-valued functions}. Then (F, ⊕, ⊙) is a vector space, where (f ⊕ g)(x) = f (x) + g (x),

(c ⊙ f )(x) = cf (x).

The zero vector (zero function) is f0 (x) = 0 for all x. −f is defined by (−f )(x) = −f (x).

A. Alaca

Linear Algebra II

Vector Spaces

8

Example 7: Let Z be the set of all integers. Is (Z, +, ·) a real vector space? 1 Solution: Take z = 2 ∈ Z and c = ∈ R. Then 4 cz =

1 1 2 = 6∈ Z =⇒ Z is not closed under “ · ”. 2 4

Thus (Z, +, ·) is not a vector space over R.  Example 8: Let Nn = {A det(A) = 0} ⊂ Mn×n . Is Nn a vector space? Soution: Take n = 2. Let     2 0 0 0 A= , B= =⇒ det A = 0, det B = 0 =⇒ A, B ∈ N2 . 0 0 0 5   2 0 A+B = =⇒ det(A + B) = 10 6= 0 =⇒ A + B 6∈ N2 . 0 5 Thus N2 is not closed under matrix addition, and so N2 is not a vector space.  Example 9: Let Kn = {A A is invertible} ⊂ Mn×n . Is Kn a vector space? Soution: Take     −1 0 1 0 , B= . n = 2, A = 0 1 0 −1 Then, A, B are invertible (A−1 = A, B −1 = B) =⇒ A, B ∈ K2 .   0 0 =⇒ A + B is not invertible =⇒ A + B 6∈ K2 . A+B = 0 0 Hence K2 is not closed under matrix addition, and so K2 is not a vector space.  Example 10: Let V = {f f is differentiable and f ′ (0) = 1}. Is V a vector space? Solution: Let f, g ∈ V . Since f and g are differentiable, f + g is also differentiable. Since f ′ (0) = 1, g ′ (0) = 1, (f ′ + g ′ )(0) = f ′ (0) + g ′ (0) = 1 + 1 = 2 6= 1 =⇒ f + g 6∈ V. V is not a vector space since it is not closed under addition.

A. Alaca

Linear Algebra II

Vector Spaces

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SUBSPACES Definition: A subset H of a vector space V is called a subspace of V if H is itself a vector space with the same scalars, addition, and scalar multiplication as V . Theorem: Let V be a vector space and let H be a subset of V . (i) The zero vector of V is in H , Then H is a subspace of V ⇐⇒ (ii) If u, v ∈ H =⇒ u + v ∈ H, (iii) If u ∈ H and c is a scalar =⇒ c u ∈ H. Notes: 1. V and {0} are two trivial subspaces of V .     x  2.  x ∈ R is a subspace of R2 . 0

3. None of the following subsets is a subspace of R2 : Why?         x  x  2 2 (i)  x ≥ 0, y ≥ 0 , (ii) x + y = 1 , y y         x  x  (iii)  x ≥ 0, y ≥ 0 ∪  x ≤ 0, y ≤ 0 . y y       2 −6 −4 Note that + = . 3 −1 2 4. Each of the following subsets is a subspace of R3 : Why?         2x   0   x      0   x, y ∈ R . (i)  0  , (ii)  y   x, y ∈ R , (iii)        0 0 −3y 5. None of the following subsets is a subspace of R3 :        x    1   x, y ∈ R , (ii) (i)    y

Why?  0  0  .  1

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Linear Algebra II

Vector Spaces

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Theorem: If v1 , v2 , . . . , vp are in a vector space V , then Span{v1 , v2 , . . . , vp } is a subspace of V . Example 11: Let  a + 2b − c    b + 2c W =   3a − 4c    a+b

     2 1            1 0   a, b, c ∈ R = a   + b    0    3      1 1 



  −1      2     a, b, c ∈ R .  + c   −4     0 

Then W is a subspace of R4 as       1 2 −1     0   1   2    ,  ,   . W = Span   3   0   −4      1 1 0     a b  Example 12: Let H = a, b, c ∈ R ⊂ M2×2 . c 0 Is H a subspace of M2×2 ? Solution: We will show that H is a subspace in two different ways. (1) By using the above theorem: Since         1 0 a b 0 1 0 0 =a +b . +c c 0 1 0 0 0 0 0       1 0 0 1 0 0 H is a spanning set, that is, H = Span , , . 0 0 0 0 1 0 (2) By using the theorem on page 9:   0 0 (i) ∈ H. 0 0       a 1 b1 a 2 b2 a 1 + a 2 b1 + b2 (ii) + = ∈ H. c1 0 c2 0 c1 + c2 0     ka kb a b = ∈ H. (iii) k c 0 kc 0 Hence H is a subspace of M2×2 .

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Linear Algebra II

Vector Spaces

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Notation: Let F = The set of all real-valued functions defined on R = {f : R → R}, C = The set of all continuous real-valued functions defined on R  = {f ∈ F f is continuous},

D = The set of all differentiable real-valued functions defined on R  = {f ∈ C f is differentiable},  P = The set of all polynomials = {f : R → R f polynomial}.

Recall that F is a vector space with (see Example 6 on page 7) (f + g)(x) = f (x) + g(x),

(cf )(x) = cf (x)

( for f, g ∈ F and scalars c).

We have    P is a subspace of D, D is a subspace of C, P ⊂ D ⊂ C ⊂ F and   C is a subspace of F.

Definition: The trace of an n × n square matrix A is the sum of its main diagonal entries, and it is denoted by trA. That is, trA =

n X

aii .

i=1

Properties of the trace: trA = trAT tr(A + B) = trA + trB tr(AB ) = tr(BA) tr(cA) = c trA

Homework 4.1  1. Let H = {f ∈ D f ′ (1) = 0}. Is H a subspace of D?    2. Let S = A ∈ M2×2  trA = 0 . Is S a subspace of M2×2 ?

A. Alaca

MATH 2107

(Nul A, Col A, Row A, Bases)

LINEAR ALGEBRA II LECTURE NOTES  c Ay¸se Alaca

Sections 4.2 and 4.3 NULL SPACES, COLUMN SPACES, ROW SPACES & LINEAR TRANSFORMATIONS LINEARLY INDEPENDENT SETS & BASES

Lecture # 2 Lecture # 3

Pages: 2–12 Pages: 13–23

Text book: Linear Algebra and its applications, 6E, By David C. Lay, Steven R. Lay, Judi J. McDonald

Please do not repost these notes elsewhere. These lecture notes may be incomplete or contain errors/typos.

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MATH 2107

(Nul A, Col A, Row A, Bases)

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Some notations: • A matrix with only one column is called a column a vector.        −1 1 a a+b  5  2  ,  b  ,  2a − 3c  ,   3 3 c a−b+c 4

vector or simply 

 . 

• Sometimes  for  convenience (also to save space) we may write a column 1 vector  2  as (1, 2, 3). In this case, the parentheses amd comma 3   distinguish the vector (1, 2, 3) from the 1 × 2 row matrix 1 2 3 . With other words,   1    2  6= 1 2 3 3 Although these two matrices have the same entries, their sizes are different. The first one is a 3 × 1 matrix and the second one is a 1 × 3 matrix.

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MATH 2107

(Nul A, Col A, Row A, Bases)

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Nul spaces, column spaces and row spaces Definition: Let A be an m × n matrix. • The null space of A, denoted by Nul A, is the set of all solutions of the homogenous equation Ax = 0. In set notation, Nul A = {x ∈ Rn | Ax = 0} • The column space of A, denoted by Col A, is the set of all linear combinations of the columns of A. Col A = {b ∈ Rm | Ax = b for some x in Rn } • The set of all linear combinations of the row vectors is called the row space of A, denoted by Row A. Since each row of A has n entries, Row A is a subspace of Rn . Since the rows of A are the columns of AT , Row A = Col AT .

Remarks: (a) If A is an n × n square invertible matrix, then (i) Nul A = {the zero vector of Rn }. (ii) Col A = Row A = Rn . (b) Let A be an m × n matrix. Then, (i) Col A is a subspace of Rm .  Ax = b is consistent. (ii) If b ∈ Col A, then b is a linear combination of the columns of A. (iii) Col A = Rm ⇐⇒

the equation Ax = b has a solution for each b ∈ Rm .

(iv) Both Nul A and Row A are subspace of Rn . (c) Suppose that A ∼ B. Then Row A = Row B .

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MATH 2107

(Nul A, Col A, Row A, Bases)

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Example 1: Let         1 0 1 4 1 2 0 0 A= ,B = ,C = ,D = . 0 1 1 5 0 0 0 0 Then, 2

Col A = Row A = R , Nul A =





0 0



,



0 , Col B = Row B = R2 , Nul B = 0   1 = x − axis, Col C = Span 0   1 = the line through the origin and the point (1, 2), Row C = Span 2   −2 Nul C = Span = the line through the origin and the point (−2, 1). 1   0 Col D = Row D = , Nul D = R2 . 0     1 2 3 1 2 3 Example 2: Let A =  4 5 6  and B =  4 5 6 . 2 8 9 2 1 0 Then we have the following:  Row A = Col A = R3       1 2 3      A ∼ 0 1 2 =⇒  0    Nul A =  0  . 0 0 1      0

 1 2 3 B ∼  0 1 2  =⇒   0 0 0              

     2   1  4  , 5    2 1     0   1 Row B = Span  2  ,  1    3 2   1   Nul B =  −2  .   1

     Col B = Span           

A. Alaca

MATH 2107

(Nul A, Col A, Row A, Bases)

 1 −3 −4  7 6 , u =  Example 3: A =  −3 

5

 5  3  3 , v =  −5 .

−4 6 −2 −1 0 (a) Is u in the null space of A? (b) Find the null space of A. (c) Is v in the column space of A? (d) If v is in Col A, then find a vector x ∈ R3 such that Ax = v . (e) Find the column space of A. (f ) Find the row space of A. Solution: (a)        1 −3 −4 5 0 5 7 6   3  =  0  =⇒ u =  3  ∈ Nul A. Au =  −3 −4 6 −2 −1 0 −1

(b) We need to find all the solutions of the equation Ax = 0.       1 −3 −4 0 1 0 5 0 −5  −3 7 6 0  ∼  0 1 3 0  =⇒ x = t  −3  , t ∈ R. −4 6 −2 0 0 0 0 0 1    −5  Nul A = Span  −3  .   1     1 −3 −4 3 1 0 5 −3 7 6 −5  ∼  0 1 3 −2  =⇒ v ∈ Col A. (c) [A | v] =  −3 −4 6 −2 0 0 0 0 0 

 −3 − 5t (d) Ax = v ⇐⇒ x =  −2 − 3t  , t ∈ R. t     −3 2 For x1 =  −2  (t = 0) and x2 =  1  (t = −1), Ax1 = Ax2 = v. 0 −1           1 −3 −4 1 0 5 −4 1 −3 7 6  ∼  0 1 3  = B =⇒  6  = 5  −3  +3  7  . (e) A =  −3 −4 6 −2 0 0 0 −2 −4 6 | {z } | {z } u1

Col A = Span {u1 , u2 }.

(f ) Since A is a symmetric matrix (AT = A), Row A = Col A = Span {u1 , u2 }.

u2

A. Alaca

MATH 2107

(Nul A, Col A, Row A, Bases)

6

Remarks: (1) Elementary row operations change the column space! That is, in general Col A 6= Col B, where B is an echelon form of A. Note that, in the above example, every vector in Col B has its third component equal to 0, which is not true for Col A. Hence the pivot columns of the matrix B do not form a basis for Col A.           0  1 0   1  1          0 , 1 0 , 1  . Col A 6= Span since −3 6∈ Span     0 0 0 −4 0

(2) Another way to find a basis for Col A is to find all b ∈ R3 such that Ax = b is consistent:       1 −3 a 1 −3 a 1 −3 a  −3  7 b  ∼  0 −2 3a + b  ∼  0 −2 3a + b −4 6 c 0 −6 4a + c 0 0 −5a − 3b + c The system is consistent if c = 5a + 3b. Hence    ) ( ) ( a  a      a, b ∈ R b Col A = b =  b   a, b, c ∈ R =    c 5a + 3b      ( ) 1 0   = a  0  +b  1   a, b ∈ R .  5 3 | {z } | {z } w1

w2

Hence {w1 , w2 } is also a basis for Col A.

Remark: Let us show that Span {u1 , u2 } = Span {w1 , w2 }:     1 0 1 −3 1 0 1 −3 7  ∼  0 1 −3 7 . [u1 u2 | w1 w2 ] =  0 1 −3 5 3 −4 6 0 0 0 0

So, we have (

u1 = w1 − 3w2 u2 = −3w1 + 7w2

=⇒

(

w1 = − 27 u1 − 23 u2 w2 = − 23 u1 − 21 u2 .

Thus, Col A = Span {u1 , u2 } = Span {w1 , w2 }.

A. Alaca

MATH 2107

(Nul A, Col A, Row A, Bases)

 1 2 −3 4 0 5  1 2 −2 7 6 7  Example 4: Let A =  2 4 −5 11 6 12 0 0 3 9 19 7

Find Col A, Nul A and Row A. Solution: Set    1 2  1   2   c1 =   2  , c2 =  4 0 0



 . 

  −3 4   −2   7  , c3 =     −5  , c4 =  11  3 9 

7





  0 5   6   7  , c5 =     6  , c6 =  12  19 7 



 . 

Then, we know by the definition of Col A that Col A = Span {c1 , c2 , c3 , c4 , c5 , c6 }. We find row echelon form of A:     1 2 −3 4 0 5 1 2 −3 4 0 5  1 2 −2 7 6 7  0 0 1 3 6 2     = B. A=  2 4 −5 11 6 12  ∼  0 0 0 0 1 1  0 0 3 9 19 7 0 0 0 0 0 0

In the matrix B, the first, third and fifth columns have pivot. So, the first, third and fifth columns of A span Col A. Hence

Col A = Span {c1 , c2 , c3 , c4 , c5 , c6 } = Span {c1 , c3 , c5 }. To find Nul A we need to find all the solutions of the equation Ax = 0, which is the same as Bx = 0. x5 = −x6 , x3 = −3x4 − 6x5 − 2x6 = −3x4 + 6x6 − 2x6 = −3x4 + 4x6 , x1 = −2x2 + 3x3 − 4x4 − 5x6 = −2x2 + 3(−3x4 + 4x6 ) − 4x4 − 5x6 = −2x2 − 13x4 + 7x6 . So, if x ∈ Nul A, then    x1 −2x2 − 13x4 + 7x6  x2   x2     x3   −3x4 + 4x6   x=   x4  =  x4     x5   −x6 x6 x6 Nul A = Span {u, v, w}.





       = x2        |

−2 1 0 0 0 0 {z u





}

|

       +x4       

−13 0 −3 1 0 0 {z v





}

|