Linear Algebra [1]
4.1 Vectors and Lines
• Definition – scalar : magnitude vector : magnitude and direction
Geometrically, a vector v can be represented by an arrow. We denote the length of v by kvk.
– zero vector 0 : k0k = 0
– Given v, we have the negative −v.
– v = w if the same length and the same direction
Kyu-Hwan Lee Linear Algebra [2] – sum v + w v + w
v
w w v + w
v
Kyu-Hwan Lee Linear Algebra [3] – scalar multiplication av (a ∈ R)
v
1 2v 2v −2v – subtraction u − v = u + (−v)
Kyu-Hwan Lee Linear Algebra [4]
Thm. u, v, w : vectors, k, p ∈ R 1. u + v = v + u, 2. u + (v + w) = (u + v) + w 3. ∃ 0 s.t. 0 + u = u for each u. 4. For each u, ∃ −u s.t. u + (−u) = 0. 5. k(u + v) = ku + kv , (k + p)u = ku + pu 6. (kp)u = k(pu), 7. 1 · u = u
Kyu-Hwan Lee Linear Algebra [5]
Thm. A, B, C : matrices of the same size, k, p ∈ F 1. A + B = B + A, 2. A + (B + C) = (A + B) + C 3. ∃ O s.t. O + A = A for each A. 4. For each A, ∃ −A s.t. A + (−A) = O. 5. k(A + B) = kA + kB , (k + p)A = kA + pA 6. (kp)A = k(pA), 7. 1 · A = A
Kyu-Hwan Lee Linear Algebra [6]
Thm. f, g, h : continuous functions on D, k, p ∈ F 1. f + g = g + f, 2. f + (g + h) = (f + g) + h 3. ∃ 0 s.t. 0 + f = f for each f. 4. For each f, ∃ −f s.t. f + (−f) = 0. 5. k(f + g) = kf + kg , (k + p)f = kf + pf 6. (kp)f = k(pf), 7. 1 · f = f
Kyu-Hwan Lee Linear Algebra [7]
The notion of vector space! 1. The set of matrices of the same size
2. The set of vectors in R3
3. The set of continuous functions on D
4. ...
5. ... and so on.
Kyu-Hwan Lee Linear Algebra [8]
The theorem says that we can manipulate vectors as if they are variables w.r.t. addition and scalar multiplication.
Eg.
5(u − 2v) + 6(5u + 2v) = 5u − 10v + 30u + 12v = 35u + 2v.
Kyu-Hwan Lee Linear Algebra [9] • Coordinates Consider a point P = (x, y, z). Then we obtain a vector −−→ p = OP : the position vector. Conversely, a vector p determines a unique point P . Thus we identify each point with the corresponding position vector.
P (x, y, z)
p O
Kyu-Hwan Lee Linear Algebra [10]
Given u = (x, y, z) and u1 = (x1, y1, z1), we have
u + u1 = (x + x1, y + y1, z + z1),
au = (ax, ay, az), u − u1 = (x − x1, y − y1, z − z1).
Kyu-Hwan Lee Linear Algebra [11]
• Lines P0 d P
p0 p
O
Assume that p0 and d are given. Then p is the position vector of a point P on the line if and only if
p = p0 + td (t ∈ R).
Kyu-Hwan Lee Linear Algebra [12]
If p = (x, y, z), d = (a, b, c), p0 = (x0, y0, z0), then we have
x = x0 + ta, y = y0 + tb, (t ∈ R). z = z0 + tc, This is the equation of the line through p0 parallel to d.
• Planes Later ... we need the notion of inner product and cross product of vectors.
Kyu-Hwan Lee Linear Algebra [13] 5.1 Subspaces and Dimension
• Subspaces of Fn
vector = point in R3 ↔ (x, y, z) coordinates
????? ↔ (a1, a2, · · · , an)
n R = {(a1, a2, · · · , an)|ai ∈ R}
a1 a2 ∼ i R = . a ∈ . an
Kyu-Hwan Lee Linear Algebra [14]
n C = {(a1, a2, · · · , an)|ai ∈ C}
a1 a2 ∼ i C = . a ∈ . an Fn = Rn or Cn The n-tuples in Fn will be called vectors.
Kyu-Hwan Lee Linear Algebra [15]
• Subspaces A subset U of Fn is called a subspace if it satisfies the following conditions. 1. If X, Y ∈ U, then X + Y ∈ U. 2. If X ∈ U, then rX ∈ U for r ∈ F.
Eg. 1. Fn
2. {0} : the zero subspace
Kyu-Hwan Lee Linear Algebra [16] 3. a line through the origin in Rn : {td}
If t1d and t2d on the line, then t1d + t2d = (t1 + t2)d and r(t1d) = (rt1)d.
4. Let A be an m × n matrix. We define
n nullA = kerA = {X ∈ F |AX = O} and
m n imA = {Y ∈ F |Y = AX for some X ∈ F }.
If X1, X2 ∈ kerA, then A(X1 +X2) = AX1 +AX2 = O and A(rX1) = r(AX1) = O. If Y1, Y2 ∈ imA, then ∃ X1, X2 s.t. AX1 = Y1 and AX2 = Y2. Now A(X1 + X2) = Y1 + Y2 and A(rX1) = rY1.
Kyu-Hwan Lee Linear Algebra [17] 5. U = {(x, y) ∈ R2|x2 + y2 = 1}.
We have (1, 0), (0, 1) ∈ U, but (1, 0) + (0, 1) = (1, 1) ∈/ U. Thus U is not a subspace of R2.
Kyu-Hwan Lee Linear Algebra [18]
• Spanning sets
n Def. Assume that X1, X2, · · · , Xk ∈ F . An expression
a1X1 + a2X2 + · · · + akXk
is called a linear combination of X1, X2, · · · , Xk (ai ∈ F).
The span of X1, X2, · · · , Xk is the set of all linear combinations of X1, X2, · · · , Xk.
span{X1, X2, · · · , Xk} = {a1X1+a2X2+· · ·+akXk|ai ∈ F}
Kyu-Hwan Lee Linear Algebra [19]
n Thm. Assume that X1, X2, · · · , Xk ∈ F . n 1. The span{X1, X2, · · · , Xk} is a subspace of F .
2. If W is a subspace containing X1, X2, · · · , Xk, then
span{X1, X2, · · · , Xk} ⊂ W .
Proof. 1. Let U = span{X1, X2, · · · , Xk}. If
Y = s1X1 + · · · + skXk, Z = t1X1 + · · · + tkXk ∈ U,
then Y + Z = (s1 + t1)X1 + · · · + (sk + tk)Xk ∈ U and rY = rs1X1 + · · · + rskXk ∈ U. 2. Clear! 2
Kyu-Hwan Lee Linear Algebra [20]
The span{X1, X2, · · · , Xk} is the smallest subspace containing X1, · · · , Xk.
If U = span{X1, X2, · · · , Xk}, then {X1, X2, · · · , Xk} is a spanning set of U, and U is spanned by the Xi’s.
Eg. Recall
Thm. Given AX = O, every solution is a linear combination of the basic solutions.
Equivalently, the kerA is the span of the basic solutions.
Kyu-Hwan Lee Linear Algebra [21]
Assume A = C1 C2 · · · Cn : m × n matrix. Then imA = span{C1, C2, · · · , Cn}.
Proof. For X ∈ Fn,
x1 x2 AX = C1 C2 · · · Cn . = x1C1+x2C2+· · ·+xnCn . xn
n imA = {AX|X ∈ F } = {x1C1 + x2C2 + · · · + xnCn}
= span{C1, C2, · · · , Cn}
2
Kyu-Hwan Lee Linear Algebra [22]
• Independence
Def. {X1, X2, · · · , Xk} : linearly independent if t1X1 + t2X2 + · · · + tkXk = 0 implies t1 = t2 = · · · = tk = 0.
Thm. If {X1, X2, · · · , Xk} is linearly independent, X ∈ span{X1, X2, · · · , Xk} has a unique representation as a linear combination of the Xi’s.
Proof.
r1X1 + · · · + rkXk = s1X1 + · · · + skXk
(r1 − s1)X1 + · · · + (rk − sk)Xk = 0 Thus we have ri = si for all i. 2
Kyu-Hwan Lee Linear Algebra [23]
Eg. X1, X2, X1 + X2
2X1 + 2X2 = 2(X1 + X2)
Eg.
1 0 1 1 0 1 0 0 1 1 0 1 1 0 , , r + s + t = 1 1 1 1 1 1 0 2 2 3 2 2 3 0
1 0 1 0 1 0 0 r 0 1 1 0 0 1 0 s = ⇒ , r = s = t = 0 1 1 1 0 0 0 1 t 2 2 3 0 0 0 0
Kyu-Hwan Lee Linear Algebra [24]
Eg. 1 1 −3 2 2 , −2 , 2 , 0 −1 1 1 0 1 1 −3 2 0 r1 2 + r2 −2 + r3 2 + r4 0 = 0 −1 1 −1 0 0 r1 1 1 −3 2 0 r2 2 −2 2 0 = 0 r3 −1 1 −1 0 0 r4
Kyu-Hwan Lee Linear Algebra [25]
1 1 −3 2 , −2 , 2 −1 1 −1 1 1 −3 1 0 −1 2 −2 2 ⇒ 0 1 −2 −1 1 −1 0 0 0
Eg. {X, Y } : indep. ⇒ {2X + 3Y, X − 5Y }: indep.
r(2X + 3Y ) + s(X − 5Y ) = O
(2r + s)X + (3r − 5s)Y = O 2r + s = 0, 3r − 5s = 0 r = s = 0
Kyu-Hwan Lee Linear Algebra [26]
Eg. lin. dep. lin. indep.
lin. dep.
lin. indep.
Kyu-Hwan Lee Linear Algebra [27]
Thm. TFAE 1. A is invertible.
2. The columns of A are linearly independent.
3. The columns of A span Fn.
4. The rows of A are linearly independent.
5. The rows of A span Fn.
6. imA = Fn.
7. kerA = O.
Kyu-Hwan Lee Linear Algebra [28]
Proof.
x1 x2 AX = C1 C2 · · · Cn . = x1C1+x2C2+· · ·+xnCn . xn
AX = O ⇔ x1C1 + x2C2 + · · · + xnCn = O 2 ⇔ AX = O has only the trivial solution. ⇔ kerA = O
AX = B ⇔ x1C1 + x2C2 + · · · + xnCn = B
3 ⇔ AX = B has a solution for every B ∈ Fn ⇔ imA = Fn A is invertible ⇔ AT is invertible 2
Kyu-Hwan Lee Linear Algebra [29]
Eg. 1 5 2 1 0 −3 8 3 , , , −2 7 −1 0 4 0 6 2
1 5 2 1 0 −3 8 3 A = , det A = 36 6= 0 −2 7 −1 0 4 0 6 2
Kyu-Hwan Lee Linear Algebra [30]
• Dimension
Def. U ∈ Fn : a subspace A set {X1, X2, · · · , Xk} is a basis of U, if
1. {X1, X2, · · · , Xk} is linearly independent,
2. U = span{X1, X2, · · · , Xk}.
Thm. If {X1, X2, · · · , Xk} and {Y1, Y2, · · · , Ym} are two bases of U, then k = m.
Def.
the number of vectors in a basis of U = the dimension of U = dim U
Kyu-Hwan Lee Linear Algebra [31]
Eg. For Fn,
0 1 0 1 . 0 . E1 = . , E2 = 0 , · · · , En = ; . . 0 . 0 1 0
the standard basis of Fn
n Eg. If {X1, X2, · · · , Xn} is a basis of F and A is invertible, n then {AX1, AX2, · · · , AXn} is also a basis of F .
Kyu-Hwan Lee Linear Algebra [32]
Eg. Consider AX = O. Recall that kerA=the span of the basic solutions. In fact, the basic solutions are linearly independent. The basic solutions form a basis for kerA.
Eg. Subspaces of R3. 1. If dim U = 3, then U = R3.
2. If dim U = 2, then U is a plane through O.
3. If dim U = 1, then U is a line through O.
4. If dim U = 0, then U = {O}.
Kyu-Hwan Lee Linear Algebra [33]
Thm. Assume that dim U = m = |B|. Then
B is linearly independent ⇔ B spans U;
in either case, B is a basis of U.
Kyu-Hwan Lee