These notes closely follow the presentation of the material given in David C. Lay’s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein. The Dot Product of Two Vectors in n Suppose that u1 v1 u and v u2 v2 are two vectors in 2. We define the dot product (also called the inner product or the scalar product) of u and v to be u v uTv. Thus, T v1 u v u v u1 u2 u1v1 u2v2 . v2 Note that u v is actually a 1 1 matrix. However, we will adopt the convention that u v is a scalar (equal to the single entry in this 1 1 matrix). For example, if 1 7 u and v , 3 3 then u v 17 33 2. Clearly, this definition can be extended to n for any positive integer n:If u1 v1 u2 v2 u and v un vn are any two vectors in n, then we define u v u1v1 u2v2 unvn. When convenient, we will simply write u v uTv (even though, technically, u v is a scalar and uTv is a 1 1 matrix). 1 Example Let 1 3 5 u 6 , v 4 , and w 5 . 8 1 0 Compute u v, v u, 3u v, u 3v, 3u v, u u, u v w, and u v u w. 2 Basic Properties of the Dot Product 1. If u and v are any two vectors in n, then u v v u. 2. If u and v are any two vectors in n and c is any scalar, then cu v u cv cu v. 3. If u, v, and w are any three vectors in n, then u v w u v u w. n 4. If u is any vector in , then u u 0. Also, u u 0 if and only if u 0n. 3 Length and Distance For any vector u n, we define the length (also called the norm or the magnitude)ofu to be u u u . Thus, if u1 u2 u , un then 2 2 2 u u u u1 u2 un . Example Find the length of the vector 1 u . 5 4 If u and v are any two vectors in n, then we define the distance between u and v to be u v. Example Find the distance between the vectors 1 3 u and v . 8 9 5 Basic Properties of Length and Distance n 1. If u is any vector in , then u 0. Also, u 0 if and only if u 0n. 2. If u is any vector in n and c is any scalar, then cu |c|u. 3. If u and v are any two vectors in n, then u v v u. 4. If u and v are any two vectors in n, then |u v| uv. (This is called the Cauchy–Schwarz Inequality.) 5. If u and v are any two vectors in n, then u v u v. (This is called the Triangle Inequality.) 6 n A unit vector is a vector of length 1.Ifv is any vector in with v 0n, then it is easy to find a unit vector that points in the same direction as v: In fact, the vector u 1 v v has length 1 and is a positive scalar multiple of v. Example Find a unit vector that points in the same direction as the vector 1 u 6 . 8 7 Orthogonality of Vectors Two vectors, u and v,inn are said to be orthogonal (or perpendicular) to each other if u v 0. This definition of orthogonality is inspired by what we can visualize in 2 or 3: Two non–zero vectors, u and v,in2 (or 3) are perpendicular to each other if and only if u v u v, and this is true if and only if u v 0. Example Show that the vectors 9 2 u and v 9 4 2 are orthogonal to each other, and show that the vectors 2 1 u and v 6 4 are not orthogonal to each other. 8 Example Describe the set of all vectors in 2 that are orthogonal to the vector 9 u . 4 9 Example Describe the set of all vectors in 3 that are orthogonal to the vector 4 u 4 . 7 10 Example Let 1 3 W Span 1 , 0 . 6 5 Note that W is a two–dimensional subspace of 3 (a plane in 3 that passes through the origin in 3). Describe the set of all vectors in 3 that are orthogonal to every vector in W. 11 Definition If W is a subspace of n, we define the orthogonal complement of W, denoted by W, to be the set of all vectors in n that are orthogonal to all of the vectors in W. Theorem Suppose that W is a subspace of n. Then: 1. W is a subspace of n. 2. W W 0n . 3. dimW dimW n and if BW is a basis for W and BW is a basis for W , n then BW BW a basis for . n Remark If W and U are any two subspaces of such that W U 0n and such that the union of a basis for W and a basis for U is a basis for n, then we say that n is the direct sum of W and U and we write n W U. Thus, the above theorem tells us that if W is any subspace of n, then n W W. We remark that n W U U W but that U W n W U. Example If 9 W Span , 4 then W is a subspace of 2. In fact, W is a line passing through the origin in 2. The subspace W consists of all vectors x1 x 2 x2 such that 9x1 4x2 0, or, to write this in a different way, x1 9 4 0. x2 Since 4 9 4 ~ 1 9 , 12 we see that 4 4 x1 t 4 x 9 t 9 t . x2 t 1 9 Therefore 4 W Span . 9 Note that 9 4 W W Span , 2. 4 9 Example If 4 W Span 4 , 7 then W is a subspace of 3. In fact, W is a line passing through the origin in 3. The subspace W consists of all vectors x1 2 x x2 x3 13 such that 4x1 4x2 7x3 0, or, to write this in a different way, x1 447 x2 0. x3 Since 7 447 ~ 1 1 4 , we see that 7 7 x1 t 4 s 1 4 1 7 x x2 t t 1 s 0 t 1 s 0 . x3 s 0 1 0 4 Therefore 1 7 W Span 1 , 0 . 0 4 Note that 4 1 7 W W Span 4 , 1 , 0 3. 7 0 4 Example If 1 3 W Span 1 , 0 , 6 5 then W is a subspace of 3. In fact, W is a plane passing through the origin in 3. The subspace W consists of all vectors x1 2 x x2 x3 such that 14 x1 x2 6x3 0 3x1 5x3 0 or, to write this in a different way, x1 1 16 0 x2 . 305 0 x3 Since 5 1 16 10 ~ 3 , 23 305 01 3 we see that 5 5 x1 3 t 3 5 x 23 t 23 t . x2 3 t 3 23 x3 t 1 3 Therefore 5 W Span 23 . 3 Note that 1 3 5 W W Span 1 , 0 , 23 3. 6 5 3 Theorem If A is any m n matrix, then rowA nulA and n rowA nulA. Also, colA nulAT and m colA nulAT . Proof Suppose that r rowA and n nulA. Then r ATx for some vector x m and An 0m. Now note that T T T T T T r n r n A x n x An x An x 0m 0. This shows that every vector in rowA is orthogonal to every vector in nulA. and hence that rowA nulA. Now, suppose that x rowA. Then r x 0 for all vectors r rowA. Thus, 15 r1 r1 x r2 r2 x Ax x 0m, rm rm x which shows that x nulA, and hence that rowA nulA. We conclude that rowA nulA and hence that n rowA nulA. To verify the second assertion of the theorem, note that colA rowAT nulAT . Example In order to illustrate the above theorem, suppose that A is the matrix 26 9 A 051 . 4719 We will show that 3 rowA nulA and that 3 colA nulAT . First, note that 51 26 9 10 10 10051 A ~ 1 ~ 051 01 5 0 51 4719 00 0 000 which shows that 10 0 rowA Span 0 , 5 . 51 1 We also see that nulA consists of all vectors x 3 such that 51 51 x1 10 t 10 51 x 1 t 1 t . x2 5 t 5 2 x3 t 1 10 Therefore, 51 nulA Span 2 . 10 Now, observe that 3 rowA nulA. 16 Next, note that 204 10 2 AT 657 ~ 011 9 119 00 0 which shows that 1 0 colA rowAT Span 0 , 1 2 1 and 2 nulAT Span 1 . 1 We observe that 3 colA nulAT .