Review of linear algebra


 Myra Robinson
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1 Review of linear algebra 1 Vectors and matrices We will just touch very briefly on certain aspects of linear algebra, most of which should be familiar. Recall that we deal with vectors, i.e. elements of R n, which here we will denote with bold face letters such as v, and scalars, in other words elements of R. We could also use C n, or Q n as needed, with scalars respectively C or Q. The main point is that, for the scalars, we need to be able to add, subtract, multiply and divide (except by 0). We can then add two vectors: if v = (v 1,..., v n ) and w = (w 1,..., w n ), then v + w = (v 1 + w 1,..., v n + w n ). Scalar multiplication is similarly defined: given t R and v = (v 1,..., v n ) R n, tv = (tv 1,..., tv n ). Vector addition is commutative and associative, there is a zero vector 0 = (0,..., 0), and every vector v has an additive inverse v = ( 1)v = ( v 1,..., v n ). Scalar multiplication satisfies: for all s, t R and v R n, s(tv) = (st)v and 1v = v. Finally there are two analogues of the distributive law: s, t R and v R n, (s + t)v = sv + tv, and for all t R and v, w R n, t(v + w) = tv + tw. These are easily check by using the usual properties of addition and multiplication for real numbers. We will not discuss the standard definition of linear independence, span, dimension or basis here. However, we will frequently use the standard basis e 1,..., e n, where the components of e i are 0 except for the i th component, which is 1. Thus, any vector v = (v 1,..., v n ) can be uniquely written in terms of the standard basis: v = n i=1 v ie i. There is also the dot product or scalar product or inner product of two vectors v, w R n, which we shall call the inner product and write as v, w, although if called the dot product it is usually written as v w. Note that the product of two vectors is a scalar, whence the name scalar product. It 1
2 is bilinear and symmetric: for all v, w, u R n and t R, v + w, u = v, u + w, u ; tv, w = t v, w ; u, v + w = u, v + u, w ; v, tw = t v, w ; v, w = w, v Of course, the third and fourth identities are a consequence of the first two and the symmetry condition. We can also define the length or norm of v: v = ( v, v ) 1/2. The standard basis e 1,..., e n is orthonormal: for all i, j, { 0, if i j; e i, e j = 1, if i = j. Any basis u 1,..., u n with this property, that u i, u j = 1 if i j and u i, u i = u i 2 = 1, will be called an orthonormal basis. Our main interest will be interesting sets of matrices. Recall that an m n matrix is a rectangular array A = a 11 a a 1n a m1 a m2... a mn We will often abbreviate this as A = (a ij ). The above matrix consists of m rows and n columns. We refer to the number a ij as the ij th entry. This means that a ij is the number in the i th row and j th column. In particular a vector (x 1,..., x n ) is also a matrix, in this case a 1 n matrix. We will call such a matrix a row vector. We can also think of a vector as an n 1 matrix, which we shall refer to as a column vector. (We will often have to think of vectors as column vectors because of our conventions on the way we write functions.) The set of all m n matrices is written M m,n (R). M m,n (C), M m,n (Q), and even M m,n (Z) are defined similarly. We can add two matrices in M m,n (R) and multiply a matrix by a scalar. The zero matrix O = O m,n M m,n (R) is the matrix all of whose entries are 0. In case m = n, we abbreviate M n,n (R) by M n (R) and call such a matrix a square (n n) matrix. An important element of M n (R) is the identity matrix I n = I, 2.
3 whose diagonal entries a ii are equal to 1 and whose other entries a ij, i j, are equal to 0. It is easy to see that, for all A M m,n (R), I m A = AI n = A. Given an m n matrix A and an n k matrix B, we can form the matrix product AB, an m k matrix whose ij th entry is given by n t=1 a itb tj. Thus the ij th entry is the inner product of the i th row of A with the j th column of B. Matrix multiplication is associative and distributes over matrix addition, where defined, but for A, B M n (R) (the only case where AB and BA are both defined and of the same shape), it is rarely the case that AB = BA: matrix multiplication is not commutative. Recall that a linear function F : R n R m is a function F such that, for all v, w R n and t R, F (v + w) = F (v) + F (w) and F (tv) = tf (v). A linear function is completely specified by its values on the standard basis vectors e 1,..., e n. Conversely, given any set of vectors v 1,..., v n R m, there is a unique linear function F : R n R m such that F (e i ) = v i for all i, namely F (x 1,..., x n ) = i x iv i. In this case, recall that we can associate an m n matrix to F as follows: write the vectors v i = (a 1i,..., a mi ). Then to F we associate the matrix A = a 11 a a 1n a m1 a m2... a mn. Here the columns of A are the vectors v i, written vertically, and the linear map F (x 1,..., x n ) corresponds to the matrix product A x, where A x is the n 1 matrix (column vector) whose j th entry is n i=1 a jix i. In particular A e i = v i, written as a column vector; its j th entry is a ji and it is equal to k j=1 a jie j, where in the equality A e i = k a ji e j the e i on the left is a basis vector in R n and the e j on the right is a basis vector in R k. Note the reversal of the indices! The case F : R n R n corresponds to square (n n) matrices. For example the linear function Id R n corresponds to the identity matrix I n. Then we have: Proposition 1.1. If G: R k R n and F : R n R m are linear maps, and A and B are the matrices corresponding to F and G respectively, then F G is again linear and the matrix corresponding to F G is the matrix product A B. j=1 3
4 This gives another, more conceptual proof of the associativity of matrix multiplication. Let A be an m n matrix A = (a ij ). Recall that the transpose matrix t A is the n m matrix whose (i, j) th entry is a ji. For example, if A is a square (n n) matrix, then t A is the reflection of A along the diagonal running from upper left to lower right. Clearly, t ( t A) = A. A calculation shows that, for all standard basis vectors e i R m and e j R n, e i, Ae j = t Ae i, e j. (Here of course the first inner product is of vectors in R m and the second is of vectors in R n.) Using bilinearity, it follows that, for all v R m and w R n, v, Aw = t Av, w. From this (or directly from the definitions) one can prove: If A is an m n matrix and B is an n k matrix, then t (AB) = t B t A. 2 Invertible matrices We will write linear maps F : R n R m as matrices A: F (v) = Av, with the understanding that, for the right hand side, v must be viewed as a column vector. Define the nullspace or kernel of A to be the set {v R n : Av = 0}. We then have the basic result: Proposition 2.1. The linear function A: R n R m is injective the nullspace of A is {0} the columns of A are linearly independent. The linear function A: R n R m is surjective the columns of A span R m. Corollary 2.2. Let F : R n R m be a linear function, corresponding to the matrix A. (i) If F is injective, then n m. (ii) If F is surjective, then n m. (iii) If n = m, then F is injective F is surjective F is a bijection, and in this case the inverse function F 1 corresponds to a matrix, denoted A 1, with the property that AA 1 = A 1 A = I n. 4
5 We call a matrix A M n (R) invertible if an inverse A 1 exists. The problem of deciding when a given n n matrix A is invertible can be answered by determinants. Recall that, for every n, we have a function det: M n (R) R with the following properties: 1. For all A, B M n (R), det(ab) = (det A)(det B). 2. det I n = A is invertible det A 0. In this case, 4. det t A = det A. det(a 1 ) = (det A) 1. Define the general linear group GL n (R) to be the subset of M n (R) consisting of invertible matrices. Equivalently, by (3) above, GL n (R) = {A M n (R) : det A 0}. The subset GL n (R) of M n (R) is closed under products, I n GL n (R), and if A GL n (R), then by definition A 1 exists and A 1 GL n (R); note that A 1 is invertible and that (A 1 ) 1 = A. Define the special linear group SL n (R) via: SL n (R) = {A M n (R) : det A = 1}. Clearly SL n (R) GL n (R). By (1) above, SL n (R) is closed under multiplication and by (2) above, I n SL n (R). Finally, if A SL n (R), then A is invertible and A 1 SL n (R) by (3). 3 Orthogonal matrices Orthogonal matrices are invertible matrices with very special geometric properties. Definition 3.1. A linear function A: R n R n is an isometry if, for all v R n, Av = v. In other words, A preserves length. Proposition 3.2. Given A M n (R), the following conditions on A are equivalent. (i) A is an isometry, i.e. for all v R n, Av = v. 5
6 (ii) For all v, w R n, Av, Aw = v, w. In other words, A preserves inner product. (iii) The columns of A are an orthonormal basis of R n. (iv) A is invertible and t A = A 1. (v) The rows of A are an orthonormal basis of R n. Proof. (i) = (ii): This follows from the polarization identity: For all v, w R n, v + w 2 v w 2 = 4 v, w. This in turn follows from the bilinearity and symmetry of inner product and expansion: For example, v + w 2 = v + w, v + w = v, v + 2 v, w + w, w, and similarly for v w 2. Then, if A is an isometry, 4 Av, Aw = Av + Aw 2 Av Aw 2 Hence Av, Aw = v, w. = A(v + w) 2 A(v w) 2 = v + w 2 v w 2 = 4 v, w. (ii) = (i): If Av, Aw = v, w for all v, w R n, then take v = w, so that Av 2 = Av, Av = v, v = v 2. (ii) = (iii): The columns of A are equal to u i = Ae i. By (ii), u i, u j = Ae i, Ae j = e i, e j. Thus u 1,..., u n is an orthonormal basis of R n. (iii) (iv): The ij th entry of t AA is the inner product u i, u j. Hence t AA = I n u i, u j is 0 if i j and 1 if i = j the columns of A are an orthonormal basis of R n. (iv) (v): Similar to the above, using A t A instead of t AA. (iv) = (ii): If t A = A 1, then for all v, w R n, Av, Aw = v, t AAw = v, A 1 Aw = v, w. We see that any of the five statements in the proposition implies any other, so they are all equivalent. 6
7 Definition 3.3. A matrix A M n (R) satisfying any (and hence all) of the equivalent properties above is called an orthogonal matrix. The set of all orthogonal n n matrices is denoted O n, the orthogonal group. The set of all orthogonal matrices with determinant 1 is denoted SO n, the special orthogonal group. Proposition 3.4. If A, B O n, then AB O n. Moreover I n SO n and hence I n O n. Finally, if A O n, then A 1 O n. Similar statements hold with O n replaced by SO n. Proof. If A, B O n, then t (AB) = t B t A = B 1 A 1 = (AB) 1. Thus AB O n. Clearly I n SO n. Finally, note that, in general, if A is an n n matrix with an inverse A 1, then t (A 1 ) = ( t A) 1, by applying the identity t (AB) = t B t A to the product AA 1 = I. Thus, if A is orthogonal, It follows that A 1 O n. t (A 1 ) = ( t A) 1 = (A 1 ) 1. The following says that there is not a big difference between O n and SO n : Proposition 3.5. If A O n, then det A = ±1. Proof. Using t A = A 1, we see that det A = det t A = det A 1 = (det A) 1. Thus (det A) 2 = 1, so that det A = ±1. We sometimes think of SO n as the set of rigid motions of R n (fixing the origin). More details about SO 2 and O 2 are in the homework. 7
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