MATH 311-504Topics in Applied MathematicsLecture 2-9:Basis and dimension (continued).Matrix of a linear transformation.Basis and dimensionDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Theorem Any vector space V has a basis. If Vhas a finite basis, then all bases for V are finite andhave the same number of elements.Definition. The dimension of a vector space V ,denoted dim V , is the number of elements in any ofits bases.Examples. • dim Rn= n• Mm,n(R): the space of m×n matricesdim Mm,n(R) = mn• Pn: polynomials of degree at most ndim Pn= n + 1• P: the space of all polynomialsdim P = ∞• {0}: the trivial vector spacedim {0} = 0Bases for RnLet v1, v2, . . . , vmbe vectors in Rn.Theorem 1 If m < n then the vectorsv1, v2, . . . , vmdo not span Rn.Theorem 2 If m > n then the vectorsv1, v2, . . . , vmare linearly dependent.Theorem 3 If m = n then the followingconditions are equivalent:(i) {v1, v2, . . . , vn} is a basis for Rn;(ii) {v1, v2, . . . , vn} is a spanning set for Rn;(iii) {v1, v2, . . . , vn} is a linearly independent set.Theorem Let S be a subset of a vector space V .Then the following conditions are equivalent:(i) S is a linearly independent spanning set for V ,i.e., a basis;(ii) S is a minimal spanning set for V ;(iii) S is a maximal linearly independent subset of V .“Minimal spanning set” means “remove any elementfrom this set, and it is no longer a spanning set”.“Maximal linearly independent subset” means“add any element of V to this set, and it willbecome linearly dependent”.How to find a basis?Theorem Let V be a vector space. T hen(i) any spanning set for V can be reduced to aminimal spanning set;(ii) any linearly independent subset of V can beextended to a maximal linearly independent set.That is, any spanning set contains a basis, while anylinearly independent set is contained in a basis.Approach 1. Get a spanning set for the vectorspace, then reduce this set to a basis.Approach 2. Build a maximal linearly independentset adding one vector at a time.Example. f : R3→ R2, f (x) =1 1 −12 1 0x.Find the dimension of the image of f .The image of f is spanned by columns of thematrix: v1= (1, 2), v2= (1, 1), and v3= (−1, 0).Observe that v3= v1− 2v2. It follows that Im f isspanned by vectors v1and v2alone. Clearly, v1andv2are linearly independent. Hence {v1, v2} is abasis for Im f and dim Im f = 2.Alternatively, since v1and v2are linearlyindependent, they constitute a basis for R2.It follows that Im f = R2and dim Im f = 2.Example. f : R3→ R2, f (x) =1 1 −12 1 0x.Find the dimension of the null-space of f .The null-space of f is the solution set of the system1 1 −12 1 0x = 0.To solve the system, we convert the matrix toreduced form:1 1 −12 1 0→1 1 −10 −1 2→1 0 101 −2Hence (x, y, z) ∈ Null f if x + z = y − 2z = 0.General solution: (x, y , z) = (−t, 2t, t), t ∈ R.Thus Null f is the line t(−1, 2, 1) and dim Null f = 1.Example. L : P4→ P4, (Lp)(x) = p(x) + p(−x).Find the dimensions of Im L and Null L.p(x) = a0+ a1x + a2x2+ a3x3+ a4x4=⇒ (Lp)(x) = 2a0+ 2a2x2+ 2a4x4.Since {1, x, x2, x3, x4} is a basis for P4, the imageof L is spanned by polynomials L1, Lx, Lx2, Lx3, Lx4.L1 = 2, Lx2= 2x2, Lx4= 2x4, Lx = Lx3= 0.Hence Im L is spanned by 1, x2, x4. Clearly, 1, x2, x4are linearly independent so that they form a basisfor Im L and dim Im L = 3.The null-space of L consists of polynomialsa1x + a3x3. That is, it is spanned by x and x3. Thus{x, x3} is a basis for Null L and dim Null L = 2.Basis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates of v with respec t to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)provides a one-to-one correspondence between Vand Rn. Besides, this mapping is linear.Matrix of a linear mappingLet V , W be vector spaces and f : V → W be a linear map.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let w1, w2, . . . , wmbe a basis for W and g2: W → Rmbethe coordinate mapping corresponding to this basis.Vf−→ Wg1yyg2Rn−→ RmThe composition g2◦f ◦g−11is a linear mapping of Rnto Rm.It is represented as v 7→ Av, where A is an m×n matrix.A is called the matrix of f with resp ect to bases v1, . . . , vnand w1, . . . , wm. Columns of A are coordinates of vectorsf (v1), . . . , f (vn) with respect to the basis w1, . . . , wm.Examples. • D : P2→ P1, (Dp)(x) = p′(x).Let ADbe the matrix of D with respect to the bases1, x, x2and 1, x. Columns of ADare coordinatesof polynomials D1, Dx, Dx2w.r.t. the basis 1, x.D1 = 0, Dx = 1, Dx2= 2x =⇒ AD=0 1 00 0 2• L : P2→ P2, (Lp)(x) = p(x + 1).Let ALbe the matrix of L w.r.t. the basis 1, x, x2.L1 = 1, Lx = 1 + x, Lx2= (x + 1)2= 1 + 2x + x2.=⇒ AL=1 1 10 1 20 0 1Problem. Consider a linear operator L : R2→ R2,Lxy=1 10 1xy.Find the matrix of L with respect to the basisv1= (3, 1), v2= (2, 1).Let N be the desired matrix. Columns of N are coordinates ofthe vectors L(v1) and L(v2) w.r.t. the basis v1, v2.L(v1) =1 10 131=41, L(v2) =1 10 121=31.Clearly, L(v2) = v1= 1v1+ 0v2.L(v1) = αv1+ βv2⇐⇒3α + 2β = 4α + β = 1⇐⇒α = 2β = −1Thus N =2 1−1