Linear Algebra Test 2 Review by JC McNamara

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1 Linear Algebra Test 2 Review by JC McNamara 2.3 Properties of determinants: det(a T ) = det(a) det(ka) = k n det(a) det(a + B) det(a) + det(b) (In some cases this is true but not always) A is invertible iff det(a) 0 (Thm 2.3.3) det(ab) = det(a) det(b) (Thm 2.3.4) AB is invertible A and B are invertible det(a 1 ) = 1 det(a) (Thm 2.3.5) A 1 = 1 det(a) adj(a) The matrix of cofactors is the following matrix C 11 C C 1n C 21 C C 2n C n1 C n2... C nn The adjoint of A is denoted by adj(a) and is equal the transpose of the matrix of cofactors i.e. C 11 C C n1 C 12 C C n2 adj(a) = C 1n C 2n... C nn Cramer s Rule Let A be an n n invertible matrix. Then the solutions of Ax = b are x 1 = det(a 1) det(a), x 2 = det(a 2) det(a),..., x n = det(a n) det(a) where A j is a matrix obtained from A by replacing the j th columb by b. Theorem A square matrix A is invertible det(a) 0. Proof. Let R be the reduced row-echelon form of A (i.e. R = E k...e 2 E 1 A, where E 1, E 2,..., E k are the elementary matrices used to obtain R from A). By Lemma 2.3.2, det(r) = det(e k ) det(e k 1 )... det(e 2 ) det(e 1 ) det(a). Since det(e n ) 0, then let a = det(e k ) det(e k 1 )... det(e 2 ) det(e 1 ) 0. So det(r) = a det(a). If A is invertible, then R = I n, hence det(r) = 1. Thus det(a) 0. On the other hand, assume det(a) 0, then det(r) 0 since it does not have a row of zeros. Hence R = I n which means that A is invertible. Theorem If A and B are square matrices of the same size, then det(ab) = det(a) det(b). Proof. There are two cases to look at, if A is invertible or A is not invertible. Case 1. If A is not invertible, then neither is AB. Thus from the previous theorem, det(ab) = 0 and det(a) = 0, so det(ab) = det(a) det(b) Case 2. If A is invertible, then A is expressible as a product of elementary matrices such that A = E 1 E 2...E r. Thus AB = E 1 E 2...E r B. Therefore, By Lemma 2.3.2, det(ab) = det(e 1 ) det(e 2 )... det(e r ) det(b) = det(e 1 E 2...E r ) det(b) = det(a) det(b). Theorem If A is invertible, then det(a 1 ) = 1 det(a). Proof. Since A 1 A = I, then det(a 1 A) = det(i). Hence det(a 1 ) det(a) = 1. Therefore, det(a 1 ) = 1 det(a).

2 2 3.1 A lot of the basics on vectors are pretty trivial at this point so I won t bother putting them on. Theorem If u, v, and w are vectors in R n, and if k and m are scalars, then (a) u + v = v + u (b) (u + v) + w = u + (v + w) (c) u + 0 = 0 + u = u (d) u + ( u) = 0 (e) k(u + v) = ku + kv (f) (k + m)u = ku + mu (g) k(mu) = (km)u (h) 1u = u The proofs for these are pretty straightforward, just turn each vector into ordered n-tuples and it will work out. 3.2 The norm of v (also called the length or magnitude of v) is denoted by v and is v = v1 2 + v2 2 + v v2 n If v is a vector in R n, and if k is any scalar, then: (a) v 0 (b) v = 0 v = 0 (c) kv = k v The dot product of vectors u and v is denoted by u v and is defined as u v = u v cos θ Also, if u = 0 or v = 0, then u v = 0. If u = (u 1, u 2,..., u n ) and v = (v 1, v 2,..., v n ), then u v = u 1 v 1 + u 2 v u n v n. The dot product is a scalar. Note: v = v v u v Since u v = u v cos θ, then θ = arccos( u v ). Theorem If u, v, and w are vectors in R n, and if k is a scalar, then: (a) u v = v u (b) u (v + w) = u v + u w (c) k(u v) = (ku) v (d) v v 0 and v v = 0 iff v = 0 The proofs for (a) and (b) are straightforward. Proof (c). Let u = (u 1, u 2,..., u n ) and v = (v 1, v 2,..., v n ). Then k(u v) = k(u 1 v 1 + u 2 v u n v n ) = (ku 1 )v 1 + (ku 2 )v (ku n )v n = (ku) v. Proof (d). v v = v v v 2 n = v 2. Since v 0 and v = 0 v = 0, the same applies for v v. Theorem (Cauchy-Schwarz Inequality) u v u v Verification in 2-space. Let u = (u 1, u 2 ) and v = (v 1, v 2 ). Squaring both sides gives us u v 2 ( u v ) 2 (u 1 v 1 + u 2 v 2 ) 2 (u u 2 2)(v v 2 2) u 2 1v u 1 v 1 u 2 v 2 + u 2 2v 2 2 u 2 1v u 2 1v u 2 2v u 2 2v 2 2 2u 1 v 1 u 2 v 2 u 2 1v2 2 + u 2 2v1 2 0 u 2 1v2 2 + u 2 2v1 2 2u 1 v 1 u 2 v 2 0 (u 1 v 2 u 2 v 1 ) 2.

3 Theorem (a) If u and v are vectors in R n, then u + v u + v Proof. u + v 2 = (u + v) (u + v) = (u u) + 2(u v) + (v v) = u 2 + 2(u v) + v 2 u u v + v 2 (Property of absolute value) u u v + v 2 (Cauchy-Schwartz) = ( u + v ) 2 u + v u + v. Theorem (Parallelogram Equation for Vectors) If u and v are vectors in R n, then Proof. u + v 2 + u v 2 = 2( u 2 + v 2 ) u + v 2 + u v 2 = (u + v) (u + v) + (u v) (u v) = 2(u u) + 2(v v) = 2( u 2 + v 2 ) 3.3 -Orthogonal means perpendicular. -Two vectors are orthogonal iff their dot product = 0. -The vector component of u along a is and the vector component of u orthogonal to a is proj a u = u a a 2 a u proj a u = u u a a 2 a. -If u and v are equal, then Thm is an equality. -In R 2, the distance between a point p 0 (x 0, y 0 ) and line ax + by + c = 0 is D = ax 0 + by 0 + c a2 + b 2 -In R 3, the distance between a point p 0 (x 0, y 0, z 0 ) and plane ax + by + cz + d = 0 is D = ax 0 + by 0 + cz 0 + d a2 + b 2 + c 2

4 If u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) are vectors in 3-space, then the cross product u v is the vector defined by i j k u 1 u 2 u 3 v 1 v 2 v 3. -u v u v - u v = u v sin θ i j = k, j k = i, etc. -The area of a parallelogram is the norm of the cross product determined by the two vectors from the same vertex of a parallelogram. -The area of a triangle is half of the norm of the cross product determined by the two vectors from the same vertex. -If u, v, w R 3, then the scalar triple product of u, v, and w is u (v w). -The norm of the scalar product is the volume of the parallelepiped formed by the three vectors. Theorem 3.5.1(a) If u, v, and w are vectors in 3-space, then u (u v) = 0 Proof. (b) v (u v) = 0 (c) u v 2 = u 2 v 2 (u v) 2 (d) u (v w) = (u w)v (u v)w (e) (u v) w = (u w)v (v w)u u (u v) = (u 1, u 2, u 3 ) (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ) = u 1 (u 2 v 3 u 3 v 2 ) + u 2 (u 3 v 1 u 1 v 3 ) + u 3 (u 1 v 2 u 2 v 1 ) = 0 Theorem (a) and (f) If u, v, and w are vectors in 3-space and k is any scalar, then (a) u v = (v u) (f) u u = 0 Proof. (a) Since the cross product is a determinant, interchanging two rows changes the sign of the determinant. Thus u v = (v u). Proof. (f) Since the cross product is a determinant, the determinant of a matrix with two of the same rows is always zero. Thus u u = 0. (b) u (v + w) = (u v) + (u w) (c) (u + v) w = (u w) + (v w) (d) k(u v) = (ku) v = u (kv) (e) u 0 = 0 u = 0 Theorem If u and v are vectors in 3-space, then u v is equal to the area of the parallelogram determined by u and v. Proof. The area of a parellelogram is A = bh = u v sin θ = u v.

5 5 4.1 A vector space V is a non-empty set defined under addition and scalar multiplication satisfying the following: Note: u, v, w V and k, m R 1. If u and v V, then u + v V (Closure under addition) 2. u + v = v + u (Commutativity) 3. u + (v + w) = (u + v) + w (Associativity) 4. 0 V called the zero vector 0 + u = u + 0 = u u V (Additive Identity) 5. u V, u V u + ( u) = ( u) + u = 0 (Additive Inverse) 6. If k is any scalar and u is any object in V, then ku is in V. (Closure under scalar multiplication) 7. k(u + v) = ku + mu (Distributivity) 8. (k + m)u = ku + mu (Distributivity) 9. k(mu) = (km)(u) 10. 1u = u Theorem Let V be a vector space, u V and k R, then: (a) 0u = 0 (b) k0 = 0 (c) ( 1)u = u (d) ku = 0 k = 0 or u = 0 Proof. I m omitting the reasons since I m sure you guys can figure it out yourselves. (a) (b) (c) (d) Let ku = 0 and k 0. 0u + 0u = (0 + 0)u = 0u (0u + 0u) + ( 0u) = 0u + ( 0u) 0u + (0u 0u) = 0 0u + 0 = 0 0u = 0 k0 = k(0 + 0) = k0 + k0 k0 + ( k0) = (k0 + k0) + ( k0) 0 = k0 + (k0 + ( k0)) 0 = k = k0 u + ( 1)u = 1u + ( 1)u = (1 + ( 1))u = 0u = 0 ku = 0 1 k (ku) = 1 k 0 ( 1 k k)u = 0 1u = 0 u = 0

6 4.2 -A subset W of a vector space V is called a subspace of V if W is itself a vector space. -A set W is a subspace of V if the following conditions hold: (a) If u and v are vectors in W, then u + v W. (b) If k is any scalar and u is any vector in W, then ku W. -Every space has at least one subspace, the zero space W = {0} and the whole space W = V. These are called trivial subspaces. A subspace that is neither one of those two is called a proper subspace. -If W 1, W 2,..., W n are subspaces of a vector space V, then W 1 W 2... W n is also a subspace of V. -If w is a vector in a vector space V, then w is said to be a linear combination of the vectors v 1, v 2,..., v n in V if w can be expressed in the form w = k 1 v 1 + k 2 v k n v n where k 1, k 2,..., k n are scalars. These scalars are called the coefficients of the linear combination. -Span{w 1, w 2,..., w n } is the set of all linear combinations of the vectors w 1, w 2,..., w r (which itself is a subspace of V ). Theorem The solution set of Ax = 0 (with n-unknowns) forms a subspace R n. Proof. Let W be the solution set for the system. The set W is non-empty because it contains the trivial solution x = 0. To show that W is a subspace of R n, we must show that it is closed under addition and scalar multiplication. Let x 2 and x 2 be solutions such that Ax 1 = 0 and Ax 2 = 0 It follows that 4.3 -Let v 1, v 2,..., v n be vectors in vector space V. These vectors are called linearly independent if has only the trivial solution A(x 1 + x 2 ) = Ax 1 + Ax 2 = = 0 A(kx 1 ) = ka(x 1 ) = k0 = 0. k 1 v 1 + k 2 v k n v n = 0 k 1 = k 2 =... = k n = 0. Otherwise, they are linearly dependent. -A finite set that contains 0 is linearly dependent. -A set with exactly one vector is linearly independent iff that vector is not 0. -A set with exactly two vectors is linearly independent iff neither vector is a scalar multiple of the other. -If f 1 (x), f 2 (x),..., f n (x) are functions that are n 1times differentiable on the interval (, ), then the determinant f 1 (x) f 2 (x)... f n (x) f 1(x) f 2(x)... f n(x) W (x) = f (n 1) 1 (x) f (n 1) 2 (x)... f n (n 1) (x) is called the Wronskian of f 1, f 2,..., f n. -If W (x) 0, then f 1, f 2,..., f n are linearly independent. Theorem 4.3.1(a) A set S with two or more vectors is linearly dependent iff at least one of the vectors in S is expressible as a linear combination of the other vectors in S. Proof. Let S be a linearly dependent set. Then k 1 v 1 + k 2 v k n v n = 0 has only the trivial solution. Suppose k 1 0 : k 1 v 1 = (k 2 v k n v n ) v 1 = k 2 k 1 v 2 k 3 k 1 v 3... k n k 1 v n. Thus at least one of the vectors in S is expressible as a linear combination of the other vectors in S. Notice that as long as k j 0 for some 1 j n, then v j can be expressed a linear combination of the other vectors in S. On the other hand, assume that at least one of the vectors in S can be expressed as a linear combination of the other vectors such that v 1 c 2 v 2 c 3 v 3... c r v r = 0 v 1 = c 2 v 2 + c 3 v c r v r which is clearly linearly dependent since v 1 has a non-zero coefficient. (b) A set S with two or more vectors is linearly independent iff no vector in S is expressible as a linear combination of the other vectors in S.

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