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Coordinates, Criteria for a Set of Vectors to Be a Basis and Transition Matrices

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Coordinates, Criteria for a Set of Vectors to Be a Basis and Transition Matrices Coordinates, Criteria for a set of vectors to be a basis and Transition matrices Coordinate vectors. Suppose that V is a vector space, with a basis β = fv1; : : : ; vng. Suppose that v 2 V . Then there is a unique expansion v = c1v1 + ··· + cnvn with ci 2 R. The coordinate vector of v with respect to the basis β is 0 1 c1 B . C n (1) (v)β = @ . A 2 R : cn Some standard bases. n 1. The vector space R of n × 1 column vectors has the standard basis 8 0 1 1 0 0 1 0 0 19 > > <> B 0 C B 1 C B . C=> ∗ 1 B C 2 B C n B . C β = e = B . C ; e = B . C ; : : : ; e = B C : > @ . A @ . A @ 0 A> :> 0 0 1 ;> 2. The vector space Rn of 1 × n row vectors has the standard basis ∗ β = fe1 = (1; 0;:::; 0); e2 = (0; 1; 0;:::; 0); : : : ; en = (0;:::; 0; 1)g: m×n 3. The vector space R of m × n matrices has the standard basis ∗ β = fe1;1; e1;2; : : : ; em;ng where eij is the m × n matrix whose entry in the i-th row and j-th column is a 1, and all other entries are zero. As an example, we see that the standard basis of 2×3 R is 8 9 > 1 0 0 0 1 0 0 0 1 > > e1;1 = ; e1;2 = ; e1;3 = ; > <> 0 0 0 0 0 0 0 0 0 => β∗ = > 0 0 0 0 0 0 0 0 0 > > e = ; e = ; e = > :> 2;1 1 0 0 2;2 0 1 0 2;3 0 0 1 ;> 4. The vector space Pn of polynomials of degree < n has the standard basis β∗ = f1; x; x2; : : : ; xn−1g: Counting the number of elements in these bases, we see that n m×n dim (R ) = n; dim (Rn) = n; dim (R ) = mn; dim (Pn) = n: A problem on change of perspective. In this section we consider the following problem. Suppose that we rotate the coordinate system 1 0 e1 = ; e2 = 0 1 2 in R counterclockwise by 45 degrees, to obtain a new coordinate system fv1; v2g. The fact that this gives us a coordinate system is the statement that β = fv1; v2g is a basis 1 2 2 of R . The coordinates of a vector ~v 2 R , from the perspective of the new coordinate system fv1; v2g are given by the coordinate vector (~v)β, as c1 (~v)β = c2 if ~v = c1v1 + c2v2. The coordinate vector (~v)β represents the way the point ~v will be perceived by a person who turns counterclockwise an angle of 45 degrees. We have (the new coordinate vectors should have length 1) p1 ! p−1 ! v = 2 ; v = 2 : 1 p1 2 p1 2 2 The mathematical problem that we must solve is then: 2 a) Show that β = fv1; v2g is a basis of R . b) Compute the coordinate vector (~v)β of an arbitrary element x ~v = 2 2: y R We will now give a solution to this problem, working directly from the definition of the basis. Later in this note, we will develop some more techniques, which will allow us to give simpler (but less conceptual) solutions to a) and b). We establish a), by verifying that β satisfies the definition of a basis. We first show that v1; v2 are linearly independent. We must show that c1 = c2 = 0 is the only solution to 1 ! −1 ! p p 0 c v + c v = c 2 + c 2 = : 1 1 2 2 1 p1 2 p1 0 2 2 We first rewrite this equation as 1 1 ! p − p c 0 2 2 1 = ; p1 p1 c 0 2 2 2 and then solve this system for c1; c2. We now calculate the RRE form of the matrix ! p1 − p1 2 2 p1 p1 2 2 and find that it is 1 0 : 0 1 Thus c1 = 0; c2 = 0 is the only solution, and so v1; v2 are LI. 2 We will next show that span(v1; v2) = R . To do this, we must establish that any vector x ~u = 2 2 y R can be expressed as a linear combination ~u = c1v1 + c2v2 2 for some c1; c2 2 R. That is, we must solve the equation 1 1 ! p − p c x (2) 2 2 1 = ; p1 p1 c y 2 2 2 for c1; c2 (with fixed x; y). The augmented matrix of this system is ! p1 − p1 j x 2 2 ; p1 p1 j y 2 2 which has the RRE form p p ! 1 0 j 2 x + 2 y 2p 2p : 2 2 0 1 j − 2 x + 2 y The unique solution to (2) is thus p p p p 2 2 2 2 (3) c = x + y; c = − x + y: 1 2 2 2 2 2 2 this gives us a (unique) solution to ~u = c1v1 + c2v2, so span(v1; v2) = R , and thus β is a 2 basis of R , establishing a). Our last calculation has already given us a solution to part b). We have shown that for x ~v = 2 2; y R the coordinate vector is p p ! 2 x + 2 y (~v) = 2p 2p ; β 2 2 − 2 x + 2 y as follows from (3). We can rewrite the coordinate vector as a matrix multiplication: p p 2 2 ! (~v) = 2p p2 ~v: β 2 2 − 2 2 We will learn later in this note that the 2 × 2 matrix appearing in this formula is the ∗ 1 2 2 transition matrix from the standard basis β = fe ; e g of R to the basis β = fv1; v2g of 2 R , written as p p ! ∗ 2 2 (4) M β = 2p p2 : β 2 2 − 2 2 Theorem 0.1. Suppose that β = fv1; : : : ; vng is a set of vectors in a vector space V . Suppose that one of the following holds: 1) V = Span(v1; : : : ; vn). 2) dim V = n, the number of vectors in β. Then β is a basis of V if and only if v1; : : : ; vn are Linearly Independent 1) of the theorem follows from the definition of a basis, since we are given that v1; : : : ; vn span. Suppose that 2) holds. By 1) we have that β is a basis of Span(v1; : : : ; vn) if and only if v1; : : : ; vn are LI, which holds if and only if dim Span(v1; : : : ; vn) = n. The subspace 3 Span(v1; : : : ; vn) of V is then equal to V if and only if dim Span(v1; : : : ; vn) = dim V by 2) of Theorem 4.10. A criterion for a set of n vectors to be a basis in an n-dimensional vector space Suppose that A = (A1;:::;An) is an m × n matrix. From the identity (equation (1) of Lecture Note 2) 0 1 c1 B . C 1 n A @ . A = c1A + ··· + cnA ; cn we see that the relations 1 n c1A + ··· + cnA = 0m correspond to the nonzero solutions to 0 1 c1 B . C A @ . A = 0m: cn 1 n m Thus A ;:::;A 2 R are linearly independent if and only if A~x = 0m has only the trivial solution, where A is the m × n matrix A = (A1;:::;An). In the case when m = n, we have that A~x = 0n has only the trivial solution if and only if Det(A) 6= 0 (Theorem 6 of Lecture Note 3). Now recall Theorem 4.10 from Lecture Note 4, which tell us that n vectors v1; : : : ; vn in an n-dimensional vector space V span V if they are LI, and thus are a basis of V . This n gives us the following criterion for a set of vectors in R to be a basis. n Theorem 0.2. Suppose that v1; : : : ; vn are elements of R . Then fv1; : : : ; vng is a basis n of R if and only if Det(v1; : : : ; vn) 6= 0. We can rephrase this criterion in a form which is applicable to any finite dimensional vector space that we know the dimension of. Theorem 0.3. Suppose that V is an n-dimensional vector space and v1; : : : ; vn are ele- ments of V . Suppose that β is a basis of V . Then fv1; : : : ; vng is a basis of V if and only if Det((v1)β;:::; (vn)β) 6= 0: Warning: The method of Theorem 0.3 cannot be used if we do not know the dimension of V . A necessary part of a solution using this theorem is to verify and state that the number n of given vectors fv1; : : : ; vng is equal to the dimension of the vector space V . We now return to problem a) from the previous section. Theorem 0.2 gives us a simple 2 method to show that fv1; v2g is a basis of R . 2 2 fv1; v2g is a basis of R since dim R = 2 and ! p1 − p1 1 1 Det(v ; v ) = Det 2 2 = + = 1 6= 0: 1 2 p1 p1 2 2 2 2 Some Examples 4 Example Determine if 8 0 1 0 1 0 19 < 2 1 3 = β = v1 = @ 1 A ; v2 = @ 1 A ; v3 = @ 2 A : 3 1 4 ; 3 is a basis of R . 3 3 Solution: Since we are given 3 = dim R vectors v1; v2; v3, β is a basis of R if and only if Det(v1; v2; v3) 6= 0. We compute (showing work) 0 2 1 3 1 Det @ 1 1 2 A = 0: 3 1 4 3 Thus β is not a basis of R .
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