Đề thi số 1 môn Đại số tuyến tính | Đại học Sư Phạm Hà Nội

λ f λ λ Pf (x) R3  x0  x0  1 = x1 − 2x2 + x3  1 = 3x1 + 2x2 + x3 x02 = x1 − 2x2 + x3 x02 = 2x1 + 2x2 + x3  x0  3 = x1 − 2x2 + x3 x03 = x1 + x2 + x3 A det A 6= 0 A−1 A = (aij) ∈ M at(n, R)  b...i 6= j aij = a...i = j E f ∈ End(E), Rankf = 1 , Dim E = n λ ∈ K f 2 = λf λ 6= 1 IdE − f B A B Det (A + B) = Det A f V α1, α2, ..., αm λ1, λ2, ..., λm α1, α2, ..., αm f V f 2 = f f R3  x0  x0  1 = x1 + x2 + x3  1 = x1 + 2x3 − x3 a) x02 = x1 − x2 + x3 b) x02 = x1 + 2x2 − x3  x0  3 = 3x1 − x2 + 3x3 x03 = 2x1 + 4x2 + x3 A = (aij) ∈ M at(n, R) aij = min(i, j) V K f ∈ End(V ) Dim V = n ≥ 1, Rank f = 1 λ ∈ K f 2 = λf λ 6= 1 Id − f V f ∈ End(V ) n f n = Id V f A p A p + 1
 3x1 + 4x2 + x3 + 2x4 + 2 = 0    3x1 + 5x2 + 3x3 + 2x4 = 0 6x x x  1 + 8 2 + x3 + 5 4 + 7 = 0 
 3x1 + 5x2 + 3x3 + 7x4 + 5 = 0 n  1 1 1 ... 1   1 2 3 ... n   0 1 1 ... 1   0 1 2 ... n − 1       0
0 1 ... 1  X =  0 0 1 ... n − 2       . . . ... .   . . . ... .  0 0 0 ... 1 0 0 0 ... 1  1 0 0  A = C  0 0 4  1 −1 4 C−1AC V K f ∈ End(V ) f 2 = 0 h ∈ End(V ) hf + f h = Id Ker f = Im f f : V → V V Dim V = Dim (Kerf ) + Dim (Imf ) f K V f 2 = f f V K, Dim V = n f ∈ End(V ) g ∈ End(V ) f gf = f  2x1 − 5x2 + x3 + 2x4 = 0  
 5x1 − 9x2 + 2x3 + 7x4 = 0 3x1 − 7x2 + x3 − 4x  4 = 0 
 4x1 + 6x2 + x3 − λx4 = 0 A n ab 1 0 n Det A = P akbn−k i=0 a = 1, b = −1 n (n > 0) A n B n ABA = A B A V E, F V Dim E = Dim F V E F a λ  λx + y + x + t = a3    x + λy + z + t = a2 x + y + λz + t = a2    x + y + z + λt = 1
A ∈ M at(3, R), A 6= 0, A2 = 0
V = M ∈ M at(3, R) : AN + M A = 0 V n n R M at(n) M at(n) S(n) A = (aij) ∈ M at(n) A S(n) R M at(n) S(n) S(n) M at(n) V R n End(V ) R V M at(n) End(V ) f : R3 → R3
f (x1, x2, x3) = (4x1 − 5x2 + 2x3, 5x1 − 7x2 + 3x3, 6x1 − 9x2 + 4x3) f f
 2x1 − 3x2 + 5x3 + 7x4 = 8    11x1 + 7x2 + 2x3 + x4 = 16
−16x1 − 23x2 + 11x3 + 19x4 = 18  
 −x1 − 22x2 + 23x3 + λx4 = 40 λ = 3 K Pn n Pn R R Pn d : Pn → Pn R f : R3 R → 3
f (x1, x2, x3) = (x1 + 2x2 − x3, 4x1 + 3x2 − 5x3, 2x1 − 2x2 + 7x3) f  x0  1 = x1 + x2 + x3 x02 = −x3  x03 = x2 λ  λx + y + z = 1  x + λy + z = λ  x + y + λz = λ2 R3 (− → e1 , − → e2 , − → e3 )  x0  1 = x1 + 2x2 + 3x3 f : R3 → R3 x02 = 2x1 + 2x2 + 4x3  x03 = x1 + x2 + 2x3 Rank f Ker f Im f  0 1 1 ... 1   1 0 1 ... 1  n A   n =  1 1 0 ... 1     . . . ... .  1 1 1 ... 0 Det An An  λx − y − z = 1  −x + λy − z = λ  −x − y + λz = λ2 z + 1 zm = 2 cos α zm + 1 zm = 2 cos mα  x − y + 2z + 2u + v = 3     2x + y + 5z + 2u + 2v = 6
 −x + 4y − 6u + v = −3 
 −2x − 4y − 4z − u + v = −3  
 2x + 4y + 4z + 7u − v = 9 P 2 K(R, C) 1, x, x2) P 2 : f : P 2 → P 2
f (a + bx + cx2) = a + (a + b)x + (2a − 3b)x2 f Ker f, I m f f R4 H(α) = xy − 2yt α = (x, y, z, t) H H H H f K V − → α1, − → α2, ..., −→ αm f λ − → 1, λ2, ..., λm α1, − → α2, ..., −→ αm R4 − → α1 = (2, 1, 3, −1); − → α2 = (2, 2, 6, −2); − → α3 = (6, 3, −9, 3); − → α4 = (1, 1, 1, 1); − → α5 = (2, 1, 5, 1)  1 4 6  R3 f  −3 −7 −7  4 8 7 − → α1 = (1, −1, 1); − → α2 = (1, 2, 0); − → α3 = (0, 0, 1) − → α = (− → α1, − → α2, − → α3) R3 f α f A n A2 − 3A + In = 0 A A−1 A