We recall Theorem 4, Sec 1.3.LetA be an m × n matrix. Then the following is equivalent: a. For each b ∈ Rm,theequationAx = b has a solution. b. Each b ∈ Rm is a linear combination of the columns of A. c. The columns of A span Rm. Subspaces, Basis, Dimension and Rank d. A has a pivot position in every row. n Theorem 8, Sec 1.7 Any {v1,...vp} in R is linearly dependent if p > n. 1 A consequence of Theorem 4 above is that if A has less than m column vectors, the set of columns of A cannot span Rm 2 A consequence of Theorem 8 above is that The maximum number of independent vectors in Rn is n. Any other vector is linear combination of the vectors in the maximum independent set of vectors. Subspaces Definition. A subspace of Rn is any set H in Rn that has three properties: a. 0 ∈ H. b. u + v ∈ H for all u, v ∈ H. c. cu ∈ H for all c ∈ Rn and u ∈ H. A subspace is closed under addition and scalar multiplication. Example 1. - Planes and lines through the origin in R3 are R3 subspaces of . Plane: H = Span{u, v} is a subspace of R3. Rn is a subspace of itself. Basis for a Subspace Basis for a Subspace n Definition. A basis for a subspace H of R is a linearly It is consequence that the vectors independent set in H that spans H. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ n 1 0 0 Example 2. Let V = {a1,...,an} be a set of vectors in R . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎢ . ⎥ Suppose that they are the columns of an invertible n × n matrix A. e1 = ⎢ ⎥ , e2 = ⎢ ⎥ , ··· , en = ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ ⎣ ⎦ Then V is a basis for Rn. 0 Solution. By the big Invertible Matrix Theorem - IMT we have 0 0 1 that columns of A form a basis for Rn, because the identity matrix is invertible. form a linearly independent set Thm 8 (e) and We also have the converse. If set V is a basis for Rn, i.e., V is span Rn Thm8 (h). linearly independent set that spans Rn, then matrix A is invertible. V Rn So is a basis for . That also follows from IMT. Basis for a Subspace Vector Coordinates relative to a Basis B {b , ··· , b } H Given a basis = 1 p for a subspace suppose Definition. Let B be a basis for H. For each x in H,the x that vector can be generated in two ways: coordinates of x relative to the basis B are the weights p c1,...,cp such that x = c1b1 + ···cpbp, and the vector in R x = c1b1 + ···cpbp and x = d1b1 + ···+ dpbp ⎡ ⎤ c1 . ⎢ . ⎥ [x]B = ⎣ . ⎦ Then subtracting gives . cp 0=x − x =(c1 − d1)b1 + ···(cp − dp)bp (1) is called the coordinate vector of x (relative to B)orthe Since B is linearly independent all weights in (1) must be B-coordinate vector of x. zero. So cj = dj for j =1, ··· , p. Coordinate systems Coordinate systems T T T Example 3. Let v1 =[3, 6, 2] , v2 =[−1, 0, 1] , x =[3, 12, 7] , Rowreductionshowsthat and B = {v1, v2}.ThenB is a basis for H = Span{v1, v2} because ⎡ ⎤ ⎡ ⎤ 3 −13 102 v1 and v2 are linearly independent. ⎣ 6012⎦ ⎣ 013⎦ (a) Is x ∈ H? (b) If so, find the B-coordinate vector of x. 21 7 000 (a) x ∈ H if and only if the system Thus c1 =2, c2 = 3 is the solution and x is in H. T T T T c1[3, 6, 2] + c2[−1, 0, 1 ]=[3, 12, 7] (b) The B-coordinates of x is [x]B =[2, 3] . The basis B determines a coordinate system on H. See figure is consistent. below. Coordinate systems Coordinate systems The correspondence x → [x]B is a one-to-one correspondence between H and R2 that preserves linear combinations. Such a correspondence is called an isomorphism.WesaythatH is isomorphic to R2. In general, if B = {b1, ··· , bk } is a basis of H, then the mapping k x → [x]B is a one-to-one correspondence between H and R ,even though the vectors in H mayhavemorethankentries.SoH is isomorphic to Rk . Basis and the number of basic vectors Dimension and Rank Proposition 1. IfasubspaceH has a basis with k elements so every other basis has exactly k vectors. Sketch of the proof. Remark. Any set of k linearly independent vectors form a basis for Rk . H k Rk A subspace with a basis with vectors is isomorphic to . Definition. The dimension of a nonzero subspace H, denoted by k So we need to show that all basis for R has exactly k vectors. dim H, is the number of vectors in any basis for H. The dimension By Theorem 8, Section 1.7, the maximum number of vectors of the zero space is zero. Rk p Rk abasisfor can have is .Soabasisfor must have at Definition. Given an m × n matrix A,therank of A is the k most (independent) vectors. maximum number of linearly independent column vectors in A. k Suppose a basis for R has l vectors with l < k.ByTheorem This number is denoted by rankA = r. 4, placing the vectors of the basis as the column vectors of a It can be shown that the number of independent columns in a matrix A, the matrix must have one pivot at each row to span matrix A is also equal to the number os independent row of A. Rk , i.e., k pivots. But A can have at most l < k pivots, because it has l columns. A contradiction. Therefore, a basis for Rk must have exactly k vectors. Dimension and Rank Theorem 1. - The Basis Theorem Let H be a k dimensional subspace of Rn. Any linearly independent set of exactly k vectors in H is automatically a basis for H.Alsoasetofk elements that spans H form a basis for H. Theorem 2. - Invertible Matrix Theorem (continued)Let A be an n × n square matrix. Then the following are equivalent. a. A is invertible m. Columns of A form a basis for Rn. n. rank A = n .