Components and change of basis

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1 Math 20F Linear Algebra Lecture 16 1 Components and change of basis Slide 1 Review: Isomorphism Review: Components in a basis Unique representation in a basis Change of basis Review: Isomorphism Definition 1 (Isomorphism) The linear transformation T : V W is an isomorphism if T is one-to-one and onto Slide 2 Example: T : P 1 IR 3 given by T (a + bt + ct 2 ) = a b c is an isomorphism

2 Math 20F Linear Algebra Lecture 16 2 Basis and components Definition 2 (Dimension) A vector space V has dimension n if the maximum number of li vectors is n Slide 3 Definition 3 (Basis) A basis of an n-dimensional vector space V is any set {u 1,, u n } of n li vectors Theorem 1 (Basis) Let V be an n-dimensional vector space The set {u 1,, u n } is a basis of V {u 1,, u n } are li and they span V Theorem 2 Let {u 1,, u n } be a basis of V Then, each vector v V has a unique decomposition v = a 1 u a n v n Proof of Theorem 1: ( ) Suppose that {u 1,, u n } does not span V Then there exists v V that it is not a linear combination of {u 1,, u n } That is, bv + a 1 u a n u n = 0 implies that b = a 1 = = a n = 0 This in turn says that {v, u 1,, u n } is a li set But this is a contradiction with the assumption that n is the maximum number of li vectors in V ( ) {u 1,, u n } is li and spans V Then, for all v V there exists numbers a 1, a n such that v = a 1 u a n u n That is, the set {v, u 1,, u n } is ld for all v V That says that n is the maximum number of li vectors in V Proof of Theorem 2: The set {u 1,, u n } is a basis of V so they span V Then, there exists scalars a i, for 1 i n such that the following decomposition holds, v = n a i u i This decomposition is unique Because, if there is another decomposition v = i=1 n b i u i i=1 then the difference has the form n (a i b i )u i = 0 i=1 Because the vectors {u 1,, u n } are li this implies that a i = b i for all 0 i n

3 Math 20F Linear Algebra Lecture 16 3 Exercises Consider the basis {e 1, e 2 } given by Slide 4 e 1 = 1, e 2 = Consider a second basis {u 1, u 2 } given by u 1 = 1, u 2 = Find the components of x = e 1 + 2e 2 in the basis {u 1, u 2 } [ ] [ ] 1 1 x = e 1 + 2e 2 =, [x] 2 e = 2 e The vectors {u 1, u 2 } form a basis so there exists constants c 1, c 2 such that [ ] [ ] c1 c1 x = c 1 u 1 + c 2 u 2 =, [x] c u = 2 c 2 Therefore, [x] e = c 1 [u 1 ] e + c 2 [u 2 ] e u e e That is, [ 1 2 ] e = [ Then one has to solve the augmented matrix [ ] ] [ c1 c 2 ] u [ ], so c 1 = 3/2 and c 2 = 1/2, and then [ 1 [x] y = 2 ] [, [x] u = e ] u

4 Math 20F Linear Algebra Lecture 16 4 Change of basis Slide 5 Theorem 3 (Change of basis) Let {u 1,, u n } and {v 1,, v n } be basis of V Then, there exists a unique n n invertible matrix P v u such that [x] v = P v u [x] u, for all x V Furthermore, the matrix P v u has the form P v u = [[u 1 ] v,, [u n ] v ] Proof of Theorem 3: Both sets {u 1,, u n } and {v 1,, v n } are basis of V, then there exist a unique set of numbers {u 1,, u n } and {v 1,, v n } such that x = u 1 u u n u n = u 1, x = v 1 v v n v n = v 1 u n u v n v Therefore, v 1 = [[u 1 ] v,, [u n ] v ] u 1 v n v This system of equations for (u 1, u n ) has a unique solution solutions for all (v 1, v n ), because the u s and v s are basis That is, P v u = [[u 1 ] v,, [u n ] v ] is invertible u n u

5 Math 20F Linear Algebra Lecture 17 5 Exercises Slide 6 (2, Sec 47) Let {b 1, b 2 }, {c 1, c 2 } be basis of IR 2 Let b 1 = c 1 + 4c 2 and b 2 = 5c 1 3c 2 Find [x] c for [x] b = (5, 3) b Find [x] b for [x] c = (1, 1) c In P 2 find the change of coordinate matrix from the basis B = {1 2t + t 2, 3 5t + 4t 2, 2t + t 2 } to the standard basis {1, t, t 2 } Find the B-coordinates of x = 1 2t Determinants Slide 7 Determinants of 2 2 matrices Definition Properties

6 Math 20F Linear Algebra Lecture 17 6 Slide determinant Definition 4 The determinant of a 2 2 matrix A = a c given by a b = det(a) = = ad bc c d b d is The determinant appears in the computation of the inverse matrix Properties Theorem 4 (Main properties of 2 2 determinants) Let A = [a 1, a 2 ] be a 2 2 matrix Let c be a 2-vector det([a 1 + c, a 2 ]) = det([a 1, a 2 ]) + det([c, a 2 ]) Slide 9 det([ca 1, a 2 ]) = c det([a 1, a 2 ]) det([a 1, a 2 ]) = det([a 2, a 1 ]) det([a 1, a 1 ]) = 0 a 1, a 2 are ld det([a 1, a 2 ]) = 0 A is invertible det(a) 0 det(a) = det(a T )

7 Math 20F Linear Algebra Lecture 17 7 Properties Theorem 5 (Determinants and elementary row operations) Let A be a 2 2 matrix Slide 10 Let B be the result of adding to a row in A a multiple of another row in A Then, det(b) = det(a) Let B be the result of interchanging two rows in A Then, det(b) = det(a) Let B be the result of multiply a row in A by a number k Then, det(b) = k det(a) Determinants and areas Theorem 6 Let A = [a 1, a 2 ] be a 2 2 matrix, with a 1 and a 2 being nonzero and noncollinear Then, det([a 1, a 2 ]) is the area of the parallelogram formed by a 1 and a 2 Slide 11 Proof: Choose a basis e 1, e 2 such that a 1 = be 1, for some number b 0 Because a 1 is not collinear to a 2, there exists a c 0 such that the vector u = ca 1 + a 2 is collinear to e 2 For that vector u holds that u = he 2, where h is the height of the parallelogram Summarizing: [ ] [ ] b 0 a 1 =, u = 0 h Now, the determinant of A is the following: det(a) = det([a 1, a 2]) = det([a 1, a 2 + ca 1]) = det([a 1, u])

8 Math 20F Linear Algebra Lecture 18 8 Slide 12 Therefore, b 0 det(a) = 0 h = hb Then, det(a) = h b, where h is the height and b the length of the base of the parallelogram Determinants Slide 13 Determinant of 3 3 matrices Determinant of n n matrices Some properties

9 Math 20F Linear Algebra Lecture 18 9 Determinant of 3 3 matrices Slide 14 Definition 5 The determinant of a 3 3 matrix A is given by a 11 a 12 a 13 det(a) = a 21 a 22 a 23 a 31 a 32 a 33 a = 22 a 23 a 32 a 33 a a a 23 a 31 a 33 a a a 22 a 31 a 32 a 13, = (a 22 a 33 a 32 a 23 )a 11 (a 21 a 33 a 31 a 23 )a 12 +(a 21 a 32 a 31 a 22 )a 13 This formula is called expansion by the first row Determinant of 3 3 matrices Theorem 7 (Expansions by rows) The determinant of a 3 3 matrix A can also be computed with an expansion by the second row or by the third row Slide 15 The proof is just do the calculation For example, the expansion by the second row is the following: a 12 a 13 a 32 a 33 a21 + a 11 a 13 a 31 a 33 a22 a 11 a 12 a 31 a 32 a23 = (a 12a 33 a 32a 13)a 21 + (a 11a 33 a 31a 13)a 22 (a 11a 32 a 31a 12)a 23 = det(a)

10 Math 20F Linear Algebra Lecture Determinant of 3 3 matrices Theorem 8 (Expansions by columns) The determinant of a 3 3 matrix A can also be computed with an expansion by the any of its columns Slide 16 The proof is again to do the calculation For example, the expansion by the first column is the following: a 22 a 23 a 32 a 33 a11 a 12 a 13 a 32 a 33 a21 + a 12 a 13 a 22 a 23 a31 = (a 22a 33 a 32a 23)a 11 (a 12a 33 a 32a 13)a 21 + (a 12a 23 a 22a 13)a 31 = det(a) Slide 17 Notation: A = Determinant of n n matrices a 11 a 1j a 1n A ij = a i1 a ij a in a n1 a nj a nn , Sign of coefficient a ij is ( 1) i+j

11 Math 20F Linear Algebra Lecture Determinant of n n matrices Slide 18 Definition 6 The determinant of an n n matrix A = [a ij is given by det(a) = det(a 11 )a 11 det(a 12 )a ( 1) 1+n det(a 1n )a 1n, n = ( 1) 1+j det(a 1j )a 1j j=1 This formula is called expansion by the first row Determinant of n n matrices Slide 19 Theorem 9 The determinant of an n n matrix A = [a ij ] can be computed by an expansion along any row or along any column That is, det(a) = = n ( 1) i+j det(a ij )a ij, for any i = 1,, n, j=1 n ( 1) i+j det(a ij )a ij, for any j = 1,, n i=1 Notation: The cofactor C ij of a matrix A is the number given by C ij = ( 1) i+j det(a ij ) Theorem 10 det(a) = det(a T )

12 Math 20F Linear Algebra Lecture Slide 20 Determinant of n n matrices Suggestion: If a matrix has a row or a column with several zeros, then it is simpler to compute its determinant by an expansion along that row or column Theorem 11 The determinant of a triangular matrix is the product of its diagonal elements Examples: = (1) = = = (1) = =

Formula for the inverse matrix. Cramer s rule. Review: 3 3 determinants can be computed expanding by any row or column

Formula for the inverse matrix. Cramer s rule. Review: 3 3 determinants can be computed expanding by any row or column Math 20F Linear Algebra Lecture 18 1 Determinants, n n Review: The 3 3 case Slide 1 Determinants n n (Expansions by rows and columns Relation with Gauss elimination matrices: Properties) Formula for the