5 Change of Basis In many applications, we may need to switch between two or more different bases for a vector space. So it would be helpful to have formulas for converting the components of a vector with respect to one basis into the corresponding components of the vector (or matrix of the operator) with respect to the other basis. The theory and tools for quickly determining these “change of basis formulas” will be developed in these notes. 5.1 Unitary and Orthogonal Matrices Definitions Unitary and orthogonal matrices will naturallyariseinchangeofbasisformulas. Theyaredefined as follows: 1 † A is a unitary matrix A is an invertible matrix with A− A , ⇐⇒ = and 1 T A is an orthogonal matrix A is an invertible real matrix with A− A ⇐⇒ = A is a real unitary matrix . ⇐⇒ Because an orthogonal matrix is simply a unitary matrix with real-valued entries, we will mainly consider unitary matrices (keeping in mind that anything derived for unitary matrices will also hold for orthogonal matrices after replacing A† with AT ). The basic test for determining if a square matrix is unitary is to simply compute A† and see if it is the inverse of A ; that is, see if AA† I . = !◮Example 5.1: Let 3 4 i 5 5 A = 4 3 i 5 5 Then 3 4 i 5 − 5 A† = 4 3 i − 5 5 9/22/2013 Chapter & Page: 5–2 Change of Basis and 3 4 3 4 i i 5 5 5 − 5 AA† I . = 4 3 4 3 = ··· = i i 5 5 − 5 5 So A is unitary. Obviously, for a matrix to be unitary, it must be square. It should also be fairly clear that, if T † 1 A is a unitary (or orthogonal) matrix, then so are A∗ , A , A and A− . ◮ T † ? Exercise 5.1: Prove that, if A is a unitary (or orthogonal) matrix, then so are A∗ , A , A 1 and A− . The term “unitary” comes from the value of the determinant. To see this, first observe that, if A is unitary then 1 † I AA− AA . = = Using this and already discussed properties of determinants, we have † † 2 1 det(I) det(AA ) det(A) det(A ) det(A) det(A)∗ det(A) . = = = = = | | Thus, A is unitary det A 1 . H⇒ | | = And since there are only two real numbers which have magnitude 1, it immediately follows that A is orthogonal det A 1 . H⇒ = ± An immediate consequence of this is that if the absolute value of the determinant of a matrix is not 1, then that matrix cannot be unitary. Why the term “orthogonal” is appropriate will become obvious later. Rows and Columns of Unitary Matrices Let u11 u12 u13 u1N ··· u21 u22 u23 u2N ··· U . = . .. . u N1 u N2 u N3 u N N ··· be a square matrix. By definition, then, u11∗ u21∗ u N1∗ ··· u12∗ u22∗ u N2∗ ··· U† u13∗ u23∗ u N3∗ . = ··· . .. . u1N ∗ u2N ∗ u N N ∗ ··· Unitary and Orthogonal Matrices Chapter & Page: 5–3 More concisely, † U jk u jk and U U kj ∗ ukj ∗ . [ ] = jk = [ ] = Now observe: U is unitary † 1 U U− ⇐⇒ = U†U I ⇐⇒ = † U U I jk ⇐⇒ jk = [ ] † U jm U mk δ jk “for all ( j, k)” ⇐⇒ m [ ] = X umj ∗ umk δ jk “for all ( j, k)” ⇐⇒ m = X The righthand side of the last equation is simply the formula for computing the standard matrix inner product of the column matrices u1 j u1k u2 j u2k and , . . . u N j u Nk and the last line tells us that 1 if j k this inner product = . = ( 0 if j k 6= In other words, that line states that u11 u12 u13 u1N u21 u22 u23 u2N , , ,..., . . . . . u N1 u N2 u N3 u N N is an orthonormal set of column matrices. But these column matrices are simply the columns of our original matrix U . Consequently, the above set of observations(startingwith“ U is unitary”) reduces to A square matrix U is unitary if and only if its columns form an orthonormal set of column matrices (using the standard column matrix inner product). You can verify that a similar statement holds using the rows of U instead of its columns. ?◮Exercise 5.2: Show that a square matrix U is unitary if and only if its rows form an orthonormal set of row matrices (using the row matrix inner product). (Hints: Either consider how the above derivation would have changed if we had used UU† instead of U†U , or use the fact just derived along with the fact that U is unitary if and only if UT is unitary.) version: 9/22/2013 Chapter & Page: 5–4 Change of Basis In summary, we have just proven: Theorem 5.1 (The Big Theorem on Unitary Matrices) Let u11 u12 u13 u1N ··· u21 u22 u23 u2N ··· U . = . .. . u N1 u N2 u N3 u N N ··· be a square matrix. Then the following statements are equivalent (that is, if any one statement is true, then all the statements are true). 1. U is unitary. 2. The columns of U form an orthonormal set of column matrices (with respect to the usual matrix inner product). That is, 1 if j k umj ∗ umk = m = ( 0 if j k X 6= . 3. The rows of U form an orthonormal set of row matrices (with respect to the usual matrix inner product). That is, 1 if j k u jm∗ ukm = m = ( 0 if j k X 6= . ?◮Exercise 5.3: What is the corresponding “Big Theorem on Orthogonal Matrices”? An Important Consequence and Exercise You can now verify a result that will be important in our change of basis formulas involving orthonormal bases. ?◮Exercise 5.4: Let u11 u12 u13 u1N ··· u21 u22 u23 u2N ··· U . = . .. . u N1 u N2 u N3 u N N ··· be a square matrix, and let S e1, e2, ..., eN and B b1, b2, ..., bN = { } = { } be two sets of vectors (in some vector space) related by b1 b2 ... bN e1 e2 ... eN U . = Change of Basis for Vector Components: The General Case Chapter & Page: 5–5 (I.e., b j ekukj for j 1, 2,..., N .) = = k X a: Show that S is orthonormal and U is a unitary matrix B is also orthonormal . H⇒ b: Show that S and B are both orthonormal sets U is a unitarymatrix . H⇒ 5.2 Change of Basis for Vector Components: The General Case Given the tools and theory we’ve developed, finding and describing the “most general formulas for changing the basis of a vector space” is disgustingly easy (assuming the space is finite dimensional). So let’s assume we have a vector space V of finite dimension N and with an inner product . Let h·|·i A a1, a2, ..., aN and B b1, b2, ..., bN . = { } = { } be two bases for V , and define the corresponding four N N matrices × MAA , MAB , MBB and MBA by [MAA] a j ak , jk = [MAB] a j bk , jk = [MBB] b j bk , jk = and [MBA] b j ak . jk = (See the pattern?) These matrices describe how the vectors in A and B are related to each other. Two quick observations should be made about these matrices: 1. The first concerns the relation between MAB and MBA . Observe that † [MAB] a j bk bk a j ∗ [MBA] ∗ MBA . jk = = = kj = jk So † MAB MBA . (5.1a) = Likewise, of course, † MBA MAB . (5.1b) = version: 9/22/2013 Chapter & Page: 5–6 Change of Basis 2. The second observation is that MAA and MBB greatly simplify if the bases are orthonor- mal. If A is orthonormal, then [MAA] a j ak δ jk I jk . jk = = = [ ] So MAA I if A isorthonormal . (5.2a) = By exactly the same reasoning, it should be clear that MBB I if B isorthonormal . (5.2b) = Now let v be any vector in V and, for convenience, let us denote the components of v with respect to A and B using α j ’s and β j ’s respectively, α1 β1 α β 2 2 v A . and v B . | i = . | i = . . . αN βN Remember, this means v αkak βkbk . (5.3) = = k k X X Our goal is to find the relations between the α j ’s and the β j ’s so that we can find one set given the other. One set of relations can be found by taking the inner product of v with each a j and using equations (5.3). Doing so: a j v a j αkak a j βkbk = * + = * + k k X X a j v αk a j ak β k a j bk H⇒ = = k k X X a j v a j ak αk a j bk βk H⇒ = = k k X X a j v [MAA] αk [MAB] βk H⇒ = jk = jk k k X X The formulas in the second equation in the last line are simply formulas for the j th entry in the products of MAA and MAB with the column matrices of αk’s and β’s. So that equation tells us that a1 v α1 β1 h | i a v α β 2 2 2 h .| i MAA . MAB . = . = . . . . aN v αN βN h | i Change of Basis for Vector Components: The General Case Chapter & Page: 5–7 Recalling what the column matrices of αk’s and β’s are, we see that this reduces to a1 v h | i a2 v h | i . MAA v A MAB v B . (5.4) . = | i = | i . aN v h | i ?◮Exercise 5.5 (semi-optional): Using the same assumptions as were used to derive (5.4), derive that b1 v h | i b2 v h | i .