# basis and the dimension

1) (10) Discuss the rank of A varies with t 1 1 0 A = 3 t 1 2 t −1 1 2 − 3 0 1 1 For an 3×5 matrix A = 2 1 − 4 − 1 3 − 2 − 1 1 − 2 3 2) (20) a) Find a basis and the dimension of the row space and the column space of A b) Find an orthonormal basis and the dimension of the null space of A c) d) Find an orthonormal basis and the dimension of the left null space of A Conclusion: Rank, nullity of A? Hint: Follow the example 5 of 4.8 3) (10) Given the set W = (3t − s, t + 2s,2t + s) a) Determine whether W a subspace in R 3 . (You must show the set satisfies a sub-space properties) b) 4) (10) 5) (10) Find the basis and the dimension. Describe this subspace geometrically. Find a basis for orthonormal complement W ⊥ for a set W, a subspace in R 4 x1 − x2 − x3 + x4 = 0 W= 2 x1 + x2 + x3 + 2 x4 = 0 Find the coordinate vector of p( x) in P2 , relative to the basis S = p1 , p2 , p3 , where p( x) = 3 + 6 x − 10 x 2 , p1 ( x) = 2 − 4 x, p 2 ( x) = x + 3x 2 , p3 ( x) = 4 + 6 x 2 6) (10) x x x 2 a) (5) Find the standard coordinate basis = for the = and y y S y B1 − 1 3 − 4 B1 = . 2 1 x − 1 4 ? b) (5) What is the coordinate basis of in the basis B2 = y 2 0 7) (10) Are the following matrices linearly independent? Basis for M 22 ? 3 6 3 − 6 − 8 0 − 1 0 1 0 − 1 0 − 12 − 4 and − 1 2 8) (10) Find the solution set for the equation AX = b where 0 − 1 0 2 x 0 a) A= X = ;b = − 1 1 0 1 y 1 b) Deduce from this, find that the null space of A c) What is the left null space A? 9) (5) Given that the set {1, 1 + x + x 2 , p( x)} is a basis for AX = P2 , which of the following is a possible value for p (x ) ? (A) 0 (B) 1 + x (C) −1 (D) 2 + x + x 2 10) (5) What is the transition matrix from S1 to S 2 given 1 4 1 3 S1 = , and S 2 = , − 2 − 4 3 8 3 Math 410, Linear Algebra Dr. B Truong Name: _______________________ LESSONS, CHAPTER 4 I) Real Vector Space (VS) : V In this section , we pay attention the the 10 axioms for VS. Let V be a non-empty set of objects. Let u and v a pair of object in V and a scaler k Pay very attention to the axioms (1) closure under addition (u + v) V (6) closure under scalar multiplication kuV Examples of VS: a) Zero VS b) VS of R n c) VS{sequence of real numbers} d) VS of 2 2 matrices ( and m n matrices e) VS of {real valued functions} Examples of collectios which are VS: a) V = R 2 ; u = (u1 , u 2 ) and ku = (u1 ,0) b) V = {set of real vectors} ; u + v = uv and ku = u k 1I) Subspace (SS) W of a VS V: W is non-empty set W V and We need to show a SS by satisfying 2 conditions (u + v) W ( ku) W a) b) Examples: SS a) b) c) d) e) f) g) The Zero SS Lines through origine in R 2 and R 3 SS of matrices M nn Set { continuous functions in (-∞,∞)} {Continuous derivatives functions} SS of {all polynomials} SS of {all polynomials of degrees ≤ n} Examples: Not a SS a) b) c) d) {Positive real functions} {Set vectors (a,b) such that a,b > 0 } {Invertible matrices M nn } {Set of all polynomials of degrees = n} Subspace of a homogeneous linear system AX = 0 with m equations and n unknowns is a SS of R n Here A is a m n matrix and X is a m 1 vector III) Spanning sets: Let V be a VS (You will be familiarize the following notations: v j and w V and k j = constants, j = 1,2….r w = k j v j j = 1,2….r We say j=1 Is the linear combination with coefficients Further more, if S = (w1 , w2 ,…, wr ) V Then the set W of all possible linear combination vectors in V is a SP of V We say the SS W of V spanned by S Thus, W = span{w1 , w2 ,…, wr } or W = span{ S} Example 1: 1) Standard unit vectors in R n e1 = (1,0….), e2 = (0,1,….),…..en = (0,0,…n) span 2) 3) Rn The vector v = k ( a, b) , k = scalar, span the line passing through (0.0) The vectors u = (a, b), v = (c, d ) span the plane containing u and v I.e: The plane = span u, v 4) The polynomials 1, x, x 2 ….x n span the VS Pn Say : 5) 6) Example 2: Pn = span 1, x, x 2 ….x n Vectors u, v and w in R 2 we can find scalars C1 ,C 2 such that The linear combination satisfied: w = C1u + C2 v as long as u is not in span v ( or reverse) Vectors u, v and w in R 3 some time, we can’t find scalars C1 , C2 such that The linear combination satisfied: w = C1u + C2 v since w is not in spanu, v ( or reverse) Express w = (6,11,6) as the linear combination of u1 = (2,1,4) , u2 = (1,−1,3) , u 3 = (3,2,5) We set to solve the system: 2 1 4 c1 6 2 1 4 c1 6 1 − 1 3 c2 = 11 AC = b . Where A = 1 − 1 3 ; C = c2 and b = 11 3 2 5 c 6 3 2 5 c 6 3 3 1 We have C = A −1b = adj( A) . A 7 − 11 3 We find that the det(A) = 2 and adj( A) = 4 − 2 − 2 5 − 1 − 3 7 6 − 11 3 9 9 / 2 1 1 1 C = A −1b = adj( A) b C = 4 − 2 − 2 11 = − 10 = − 5 A 2 2 − 1 − 3 6 5 1 1/ 2 That means : 6 2 1 3 9 1 11 = 1 − 5 − 1 + 2 6 2 4 3 2 5 1 12 as a linear combination of Express C = 8 − 2 1 4 2 0 ; B = A = 3 0 1 2 Example3: We have to find x, y such that 1 4 2 0 1 12 + y = x 3 0 1 2 8 − 2 Or solve ( by equating all elements of both sides) x + 2 y = 1 (1) 4 x + 0 y = 12 (2) 3x + y = 8 (3) 0 x + 2 y = −2 (4) x = 3; y = −1 The equation (2) and (4) give These values also satisfy the eq (1) and (3). Thus we have 1 12 1 4 2 0 = 3 − 1 C = 3 A − B or 8 − 2 3 0 1 2 IV) Linear Independent: r Let v j 1 V a VS ; j = 1,2….r Form a vector equations for a linear combination as follows c v j j =0 j Conclusion: v j 1r is said to be Linearly Independent if all a) cj = 0 j stands for for all b) Otherwise, if ck 0 for some k the set v j 1 is Linear Dependent r Note: r a) The set v j 1 in which vk = 0 for some k , is Linear Dependeny (L/D) b) In R n , any 2 vectors u and v such that u kv are Linearly Independent (L/I) To show the v j 1 in R n or M ij or set of polynomials Pn always solve the equation r c v j j =0 j c M j j = 0 or j c j fj =0 j Usefull facts: The unit vector in Pn is (1,0,0….0), (0,1,0…..0),….(0,0,0,0,0..1) So the function f ( x) = 2 + 2 x + x 3 can be written as x0 1 x f ( x) = (2 2 0 3) 2 = 2 + 2 x + 0 x 2 + 3x 3 x x3 So the vector (2 2 0 3) represents the function f ( x) = 2 + 2 x + x 4 Definition: If the set of function f j 1 ( x) is continuous up to (n-1) th derivative Then the determinamt n f1 f1 ‘ • • • W ( f1 , f 2 ,…., f n ) = f1 ( n −1) f2 f2 ‘ • • • f2 ( n −1) • • • • • • • • • • • • • • • • • • fn • • • • fn ( n −1) is called the Wronskian of f1 , f 2 ,…., f n If W ( f1 , f 2 ,…., f n ) 0 on (−,) then f j 1 ( x) are L/I n Conclusion: a) Example 4: b) If W ( f1 , f 2 ,…., f n ) = 0 on (−,) then f j 1 ( x) are L/D a) Whether sin x and cos x are L/I? n W ( x, cos( x)) = sin x cos x =1 cos x − sin x L/I on on (−,) b) Whether functions f1 ( x) = −1 + 2 x − x 2 f 2 ( x) = x + x 2 f 3 ( x) = 2 + 2 x 2 are L/I − 1 + 2x − x 2 W ( f1 , f 2 , f 3 ) = 2 − 2x −2 x + x2 1 + 2x 2 2 + 2x 2 4x 4 (− 1 + 2x − x )(4(1 + 2 x) − 2(4x)) − ( x + x )(4(2 − 2x) + 8x) = 2 2 + (2 + 2 x 2 )(2(2 − 2 x) + 2(1 + 2 x) ) =8 So we conclude that the set f j 1 ( x) are linearly independent 3 Another way to do this problem is now, we consider these functions in P2 as vectors And find the Wronskian accordingly − 1 0 2 2 2 f1 ( x) = −1 + 2 x − x = 2 f 2 ( x) = x + x = 1 , f 3 ( x) = 2 + 2 x = 0 − 1 1 2 −1 0 2 W (v1 , v2 , v3 ) = 2 1 0 = (−2 + 0 + 4) − (−2 + 0 + 0) = 4 0 L / I −1 1 2 2 Note that the Wronskian are not the same. Math 410, Linear Algebra Dr. B Truong Name: _______________________ LESSONS, CHAPTER 4 CONTINUE I) Coordinates and Basis and Change of Basis: To make it easy, say the vector a a u = = is a vector in standard base b b S 1 0 0 1 a a 1 0 a Ie = = b b S 0 1 b Now, it is the same vector, but in different base, say B1 or in any base B j , then we can express u as a a 1 0 a = = = B j u j = b b B j 0 1 b C1 C 2 B (1) j Now, we have to determine C1 , C2 . 1 0 a = B j u j = 0 1 b C1 C = Bk u k = Bk 3 and so on….. C 2 B C 4 B j j And we are abe to determine C3 ,C 4 too. we want to transform u S uS to u B1 we say: to u B1 ( S ) ( B1 ) , then B1−1 To make it easy u S u S is to transfor m ( Su S ) = ( B1u B1 ) then the transition matrix is TS 1 = B1−1 NOTES Basically we have the transition matrix from base −1 B j to Bk is T jk = Bk B j (3) (2) 2 2 1 − 1 3 1 and B2 = Example 1: Given u = = . Write u in B1 = 1 1 S 1 3 4 2 We have to find C1 , C2 , C3 and C4 such that 2 1 − 1 C1 1 3 C 3 1 = 1 3 C = − 1 5 C S 2 B1 4 B2 (a) (b) (4) For (a) in (4) We have −1 u B1 = B1 Su S −1 C1 2 1 − 1 C1 1 − 1 2 −1 2 = = B = 1 C 1 1 S 1 3 C 2 B1 S 1 3 1 S 2 B1 C1 1 3 1 2 1 7 7/4 C = 4 − 1 1 1 = 4 − 1 = − 1 / 4 S 2 B1 So ve have C1 = 7 / 4 and C2 = −1/ 4 For part (b) in (4), we have two options, we either i) transform from S B2 ii) transform from B1 B2 If we choose option i), the transition matrix TS 2 is B2−1 S = B2−1 If we choose option ii), the transition matrix T12 = B2−1 B1 Now: i) If we choose option i), The the transition matrix T2 S is B2−1 S = B2−1 TS 2 = B −1 2 3 1 = 4 2 −1 = 1 2 − 1 2 − 4 3 therefore C 3 2 3 1 C 3 2 1 2 − 1 2 1 3 1 = 4 2 C C = TS 2 1 = 2 − 4 3 1 = 2 − 5 S 4 B2 S S B2 4 B2 ii) If we choose option ii), The the transition matrix T12 is B2−1 B1 T12 = B2−1 B1 = 1 2 − 11 − 1 1 1 − 5 = 2 − 4 3 1 3 2 − 1 13 Thus C 3 1 1 − 5 7 / 4 1 1 − 5 7 1 12 1 3 C = T12u B1 = 2 − 1 13 − 1 / 4 = 8 − 1 13 − 1 = 8 − 20 = 2 − 5 = B2 4 B2 II) Dimension: To make it easy, dimension of a set vectors in R n of a finite dimension spacet is the numbers of vectors which are linearly independent Example 2: a) 2 4 1 1 , 1 2 has Dim =3 ( they are L/I) cgeck it by find the det ≠0 1 2 3 b) 2 4 1 1 , 2 5 has Dim =2, since v2 = 2v1 L / D 1 2 3 c) 2 4 − 2 1 , 2 − 1 has Dim =1, since v2 = 2v1 L / D 1 2 − 1 c) d) e) This Spannig set spnans a line. So the basis for example a) is Span{v1 , v2 , v3 } the basis for example a) is Span{v1 , v3 } the basis for example a) is Span{v1} Base on examples above Let W be subspace of a finite dimensional vector space V, then i) W is finite dimensional ii) DimW ≤ Dim V iii) Dim W = Dim V W = V Example 2: a) 2 2 1 3 1 − 1 B1 = 1 , − 1 2 , B2 = 1 , 1 0 1 1 1 − 5 − 3 2 Find the transition matrix from B1 to B2 . As discussed above 5 2 3 2 1 − 1 2 2 1 3 1 = B2−1 B1 = 1 1 0 1 −1 2 = − 2 − 3 − 2 − 5 − 3 2 1 1 1 5 1 6 −1 T1→2 −1 b) 2 2 1 − 5 9 − 5 −1 If W = 8 , then W B1 = B1 W = 1 − 1 2 8 = − 9 1 1 1 − 5 − 5 − 5 c) Thn we can compute WB1 as d) 5 7 2 3 − 2 9 2 1 23 WB1 = T1→2W1 = − 2 − 3 − − 9 = 2 2 5 1 6 − 5 6 By direct computing WB2 IV 1 − 1 3 =B W = 1 1 0 − 5 − 3 2 −1 2 −1 1 − 5 8 = −1 − 5 1 1 2 1 2 2 1 7 − 2 − 5 2 1 23 − 8 = 2 2 1 − 5 6 Row space, Column space and Null space To be short: For finte dimentional space. Let Amn be a matrix containing m vectors in R n 1) The Row Space ( Row ( A) ) is the space spanned by the remaining horizontal vectors v j after elementary ro elementary operation. We say Row ( A) = Span{v j } 2) The Column Space ( Col ( A) ) is the space spanned by the remaining Vertical vector vk after elementary ro elementary operation. We say Col ( A) = Span{vk } 3) The Null Space ( Null ( A) ) is the solution of AX = 0 For further easy thought 1) The Row Space ( Row ( A) ) is space of independent horizontal vectors 2) The Column Space ( Col ( A) ) is the space independent vertical vectors 3) The Null Space ( Null ( A) ) is the solution of AX = 0 , X R n Null space of A is spanned by independent vectors in X Theorem: The vector b is said to be in the col(A) if there is asolution x such that Ax = b Or we can say that Ax = b is consistent if b Col ( A) Example 3: Assume A be a 4 by 6 matrix. After row elimination, we have 1 0 A is row equivalent to 0 0 0 2 −2 −3 0 1 3 0 0 1 0 0 0 0 0 0 0 1 2 0 2 − 2 0 0 0 0 1 0 0 2 0 0 − 2 1 0 Row ( A) = span , or any 3 original L/I row vectors − 3 3 1 0 2 2 1 0 − 2 1 − 2 − 3 0 1 3 Col ( A) = span 0 , 0 , 1 any 3 original L/I column vectors 0 0 0 0 0 0 For the Null (A), pay to the attention that this is X R 6 . If we solve the equation AX = b The number survival vectors are 3. The solution space is involved 3 parameters Let Then x 2 = s, x5 = t , x 6 = w x4 = −2 x5 + 2 x6 = −2t + 2w x3 = −3×4 − 2 x5 = −3(−2t + 2w) − 2t = 4t − 6w x1 = −2 x2 + 2 x3 + 3×4 − x6 = −2s + 2(4t − 6w) + 3(−2t + 2w) − w = −2s + 2t − 7w − 2s + 2t − 7 w 0 2 − 7 s − 2 0 0 4t − 6w 0 4 − 7 = s + t + w or Null ( A) = − 2t + 2w 0 − 2 2 0 t 0 1 0 0 1 w 0 2 − 7 − 2 0 0 0 4 − 6 Null ( A) = Span , , 0 − 2 2 0 1 0 0 0 1 CHECK: All vectors in Null(A) are orthogonal to all rows of equivalence of A or orthogonal to all the vectors of the original matrix A V) The Rank The Nullity Definition: 1) The Rank of a matrix A Rank ( A) is the number of vectors in Row(A) or in Col(A) 2) The Nullity of a matrix A is the number of vectors in Null(A) For the Example 3 above, the Rank ( A) = 3. This is also the “Dimension of A” (Some textbook denotes Rank ( A) = RA , The Null Space A = ( A) and Nullity = ) Note that, for matrix Amn DimA = Rank ( A) 1) 2) 3) 4) The Row(A) is a subspace in R n The Col (A) is a subspace in R m The Null (A) is a subspace in R n The Rank(A)+ Nullity = n If W is a subspace in R n then 1) W ⊥ is a subspace in R n 2) The only common vector of W and W ⊥ is 0 3) The orthogonal complement of W ⊥ is W We also see that 1) The Null (A) is the orthogonal complement of Row (A)in R n 2) The Null ( A T ) is the orthogonal complement of Col (A)in R m Math 410, Linear Algebra Dr. B Truong Name: _______________________ LESSONS, CHAPTER 4 CONTINUE I) Coordinates and Basis and Change of Basis: To make it easy, say the vector a a u = …

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