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**Instructor's Solution Manual for "Applied Linear Algebra" (with Errata)**

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**Sample text**

We have already seen the examples U (1), U (2) and SU (2). For general values of n, the study of U (n) can be split into that of its determinant, which lies in U (1) so is easy to deal with, followed by the subgroup SU (n), which is a much more complicated story. Digression. Note that it is not quite true that the group U (n) is the product group SU (n) × U (1). If one tries to identify the U (1) as the subgroup of U (n) of elements of the form eiθ 1, then matrices of the form m ei n 2π 1 for m an integer will lie in both SU (n) and U (1), so U (n) is not a product of those two groups (it is an example of a semi-direct product, these will be discussed in section 18).

The map L ∈ U (n) → det(L) ∈ U (1) is a group homomorphism. We have already seen the examples U (1), U (2) and SU (2). For general values of n, the study of U (n) can be split into that of its determinant, which lies in U (1) so is easy to deal with, followed by the subgroup SU (n), which is a much more complicated story. Digression. Note that it is not quite true that the group U (n) is the product group SU (n) × U (1). If one tries to identify the U (1) as the subgroup of U (n) of elements of the form eiθ 1, then matrices of the form m ei n 2π 1 for m an integer will lie in both SU (n) and U (1), so U (n) is not a product of those two groups (it is an example of a semi-direct product, these will be discussed in section 18).

To get some more insight into the structure of the group SU (2), consider an arbitrary 2 by 2 complex matrix α γ β δ Unitarity implies that the rows are orthonormal. This results from the condition that the matrix times its conjugate-transpose is the identity α γ β δ α β γ δ = 1 0 0 1 Orthogonality of the two rows gives the relation γα + δβ = 0 =⇒ δ = − 30 γα β The condition that the first row has length one gives αα + ββ = |α|2 + |β|2 = 1 Using these two relations and computing the determinant (which has to be 1) gives γ ααγ γ − βγ = − (αα + ββ) = − = 1 αδ − βγ = − β β β so one must have γ = −β, δ = α and an SU (2) matrix will have the form α −β β α where (α, β) ∈ C2 and |α|2 + |β|2 = 1 The elements of SU (2) are thus parametrized by two complex numbers, with the sum of their length-squareds equal to one.