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Notes on Change of Basis and Linear Transformations [PDF] CHANGE OF BASIS AND ALL OF THAT LANCE D. DRAGER 1. Introduction The goal of these notes is to provide an apparatus for dealing with change of basis in vector spaces, matrices of linear transformations, and how the matrix changes when the basis changes. We hope this apparatus will make these computations easier to remember and work with. To introduce the basic idea, suppose that V is vector space and v1, v2,..., vn is an ordered list of vectors from V . If a vector x is a linear combination of v1, v2,..., vn, we have (1.1) x = c1v1 + c2v2 + ··· + cnvn for some scalars c1, c2, . , cn. The formula (1.1) can be written in matrix notation as c1 c2 (1.2) x = v v ... v . 1 2 n . . cn The object v1 v2 ... vn might seem a bit strange, since it is a row vector whose entries are vectors from V , rather than scalars. Nonetheless the matrix product in (1.2) makes sense. If you think about it, it makes sense to multiply a matrix of vectors with a (compatibly sized) matrix of scalars, since the entries in the product are linear combinations of vectors. It would not make sense to multiply two matrices of vectors together, unless you have some way to multiply two vectors together and get some object. You don’t have that in a general vector space. More generally than (1.1), suppose that we have vectors x1, x2,..., xm that are all linear combinations of v1, v2,..., vn. The we can write x1 = a11v1 + a21v2 + ··· + an1vn x2 = a12v1 + a22v2 + ··· + an2vn (1.3) . xm = a1mv1 + a2mv2 + ··· + anmvn for scalars aij. The indices tell you that aij occurs as the coefficient of vi in the expression of xj. This choice of the order of the indices makes things work out nicely later. These equations can be summarized as n X (1.4) xj = viaij, j = 1, 2, . , m, i=1 Version Time-stamp: ”2007-04-25 21:55:05 drager”. 1 2 LANCE D. DRAGER where I’m writing the scalars aij to the right of the vector to make the equation easier to remember; it means the same thing as writing the scalars on the left. Using our previous idea, we can write (1.3) in matrix form as a11 a12 . a1m a21 a22 . a2m (1.5) x x ... x = v v ... v 1 2 m 1 2 n . .. . an1 an2 . anm To write (1.5) in briefer form, let X = x1 x2 ... xm , V = v1 v2 ... vn and let A be the n × m matrix a11 a12 . a1m a21 a22 . a2m A = . . .. . an1 an2 . anm Then we can write (1.5) as (1.6) X = VA. The object V is a row vector whose entries are vectors. For brevity, we’ll refer to this as “a row of vectors.” We’ll usually use upper case script letters for a row of vectors, with the corresponding lower case bold letters for the entries. Thus, if we say that U is a row of k vectors, you know that U = u1 u2 ... uk . You want to remember that the short form (1.6) is equivalent to (1.3) and and (1.4). It is also good to recall, from the usual definition of matrix multiplication, that (1.6) means (1.7) xj = V colj(A), where colj(A) is the column vector formed by the j-th column of A. The multiplication between rows of vectors and matrices of scalars has the fol- lowing properties, similar to matrix algebra with matrices of scalars. Theorem 1.1. Let V be a vector space and let U be a row of vectors from V . Let A and B denotes scalar matrices, with sizes appropriate to the operation being considered. Then we have the following. (1) UI = U, where I is the identity matrix of the right size. (2) (UA)B = U(AB), the “associative” law. (3) (sU)A = U(sA) = s(UA) for any scalar s. (4) U(A + B) = UA + UB, the distributive law. The proofs of these properties are the same as the proof of the corresponding statements for the algebra of scalar matrices. They’re not illuminating enough to write out here. The associative law will be particularly important in what follows. 2. Ordered Bases Let V be a vector space. An ordered basis of V is an ordered list of vectors, say v1, v2,..., vn, that span V and are linearly independent. We can think of this CHANGE OF BASIS AND ALL OF THAT 3 ordered list as a row of vectors, say V = v1 v2 ... vn . Since V is an ordered basis, we know that any vector v ∈ V can be written uniquely as v = c1v1 + c2v2 + ··· + cnvn c1 c2 = v v ... v 1 2 n . . cn = Vc, where c is the column vector c1 c2 c = . . . cn Thus, another way to state the properties of an ordered basis V is that for every vector v ∈ V there is a unique column vector c ∈ Rn such that v = Vc. This unique column vector is called the coordinate vector of v with respect to the basis V. We will use the notation [v] for this vector. Thus, the coordinate vector V [v] is defined to be the unique column vector such that V (2.1) v = V[v] . V The following proposition records some facts (obvious, we hope) about coordinate vectors. Proving this proposition for yourself would be an instructive exercise. Proposition 2.1. Let V be a vector space of dimension n and let V be an ordered basis of V . Then, the following statements hold. (1) [sv] = s[v] , for all v ∈ V and scalars s. V V (2) [v + w] = [v] + [w] for all vectors v, w ∈ V . V V V (3) v ↔ [v] is a one-to-one correspondence between V and n. V R More generally, if U is a row of m vectors in V , our discussion of (1.7) shows (2.2) U = VA ⇐⇒ col (A) = [u ] , j = 1, 2, . , m j j V Since the coordinates of a vector with respect to the basis V are unique, we get the following fact, which is stated as a proposition for later reference. Proposition 2.2. Let V be a vector space of dimension n, and let V be an ordered basis of V . Then, if U is a row of m vectors from V , there is a unique n×m matrix A so that U = VA. We can apply the machine we’ve built up to prove the following theorem. 4 LANCE D. DRAGER Theorem 2.3. Let V be a vector space of dimension n and let V be an ordered basis of V . Let U be a row of n vectors from V and let A be the n × n matrix such that U = VA. Then U is an ordered basis of V if and only if A is invertible. Proof. Since U contains n vectors, U is a basis if and only if its entries are linearly independent. To check this, we want to consider the equation (2.3) c1u1 + c2u2 + ··· + cnun = 0 If we let c be the column vector with entries cj, we have 0 = c1u1 + c2u2 + ··· + cnun = Uc = (VA)c = V(Ac) By Proposition 2.2, V(Ac) can be zero if and only if Ac = 0. Suppose that A is invertible. Then Ac = 0 implies that c = 0. Thus, all the ci’s in (2.3) must be zero, so u1, u2,..., un are linearly independent. Conversely, suppose that u1, u2,..., un are linearly independent. Let c be a vector such that Ac = 0. Then (2.3) holds. By linear independence, c must be 0. Thus, the equation Ac = 0 has only the trivial solution, and so A is invertible. 3. Change of Basis Suppose that we have two ordered bases U and V of an n-dimensional vector space V . From the last section, there is a unique matrix A so that U = VA. We will denote this matrix by SVU . Thus, the defining equation of SVU is (3.1) U = VSVU . Proposition 3.1. Let V be an n-dimensional vector space. Let U, V, and W be ordered bases of V . Then, we have the following. (1) SUU = I. (2) SUW = SUV SVW . −1 (3) SUV = [SVU ] . Proof. For the first statement, note that U = UI so, by uniqueness, SUU = I. For the second statement, recall the defining equations W = USUW , V = USUV , W = VSVW . Compute as follows W = VSVW = (USUV )SVW = U(SUV SVW ) and so, by uniqueness, SUW = SUV SVW . For the last statement, set W = U in the second statement. This yields SUV SVU = SUU = I, so SUV and SVU are inverses of each other. CHANGE OF BASIS AND ALL OF THAT 5 The matrices SUV tell you how to change coordinates from one basis to another, as detailed in the following Proposition. Proposition 3.2. Let V be a vector space of dimension n and let U and V be ordered bases of V . Then we have [v] = S [v] , U UV V for all vectors v ∈ V . Proof. The defining equations are v = U[v] , v = V[v] . U V Thus, we have v = V[v] V = (US )[v] UV V = U(S [v] ), UV V and so, by uniqueness, [v] = S [v] . U UV V Thus, SUV tells you how to compute the U-coordinates of a vector from the V-coordinates. Another way to look at this is the following diagram.
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