2/26/2002, BASIS Math 21b, O. KnillHOMEWORK: 3.2 6,18,28,36*,38*,48(Preview: For next Tuesday: 3.3: 22,24,32,36,40,56*, 3.4: 4,14,16,22,32*,48)DEFINITION BASIS. A set of vectors ~v1, . . . , ~vmis a basis of a subspace Xof Rnif they are linear independent and if they span the space X. Linearindependent means that there are no nontrivial linear relations ai~v1+ . . . +am~vm= 0. Spanning the space means that very vector ~v can be written as alinear combination ~v = a1~v1+ . . . +am~vmof basis vectors. A linear subspaceis a set containing {0} which is closed under addition and scaling.EXAMPLE. Three vectors in space which are not contained in a plane form a basisFACT. If ~v1, ..., ~vnis a basis, then everyvector ~v can be represented uniquelyas a linear combination of the ~vi.~v = a1~v1+ . . . + an~vn.REASON. There is a representation because the vectors ~vispanthe space. If there were two different representations ~v = a1~v1+. . . + an~vnand ~v = b1~v1+ . . . + bn~vn, then subtraction would leadto 0 = (a1−b1)~v1+...+(an−bn)~vn. This nontrivial linear relationof the viis forbidden by assumption.EXAMPLE. The vectors ~v1=110, ~v2=011, ~v3=101form a basis in the three dimensional space. If~v =435, then ~v = ~v1+ 2~v2+ 3~v3and this representation is unique. We can find the coefficients by solvingA~x = ~v, where A has the vias column vectors. In our case, A =1 0 11 1 00 1 1xyz=435had the uniquesolution x = 1, y = 2, z = 3 leading to ~v = ~v1+ 2~v2+ 3~v3.FACT. If n vectors ~v1, ..., ~vnspan a space and ~w1, ..., ~wmare linear independent, then m ≤ n.REASON. Assume m > n. Because ~vispan, each vector ~wican be written asPjaij~vj= ~wi. After doingGauss-Jordan elimination of the m × n matrixa11. . . a1n| ~w1. . . . . . . . . | . . .am1. . . amn| ~wmwe end up with a matrix which hasin the last line0 ... 0 | b1~w1+ ... + bm~wm. The equation b1~w1+ ... + bm~wmis a nontrivial relationbetween the ~wi. Contradiction.DIMENSION. The number of elements in a basis is independent of the basis and called the dimension.EXAMPLES. The dimension of {0} is zero. The dimension of a line 1. The dimension of a plane is 2, thedimension of three dimensional space is 3. The dimension is independent on where the space is embedded in.For example: a line in the plane and a line in space have dimension 1.A BASIS DEFINES AN INVERTIBLE MATRIX. The n × n matrix A =| | |~v1~v2. . . ~vn| | |is invertible ifand only if ~v1, . . . , ~vndefine a basis in Rn.EXAMPLE. In the example above, the 3 × 3 matrix A is invertible.DIMENSION OF THE KERNEL. The number ofcolumn in rref(A) without leading 1’s is the dimen-sion of the kernel dim(ker(A)): we can introduce aparameter for each such column when solving Ax = 0using Gauss-Jordan elimination.DIMENSION OF THE IMAGE. The number ofleading 1 in rref(A), the rank of A is the dimen-sion of the image dim(im(A)) because every suchleading 1 produces a different column vector (calledpivot column vectors) and these column vectorsare linearly independent.DIMENSION FORMULA: (A : Rn→ Rm)dim(ker(A)) + dim(im(A)) = nEXAMPLE: A invertible ⇔ the dimension of theimage is n ⇔ the dimension of the kernel 0.PROOF. There are n columns. dim(ker(A)) is the number of columns without leading 1, dim(im(A)) is thenumber of columns with leading 1.FINDING A BASIS FOR THE KERNEL. To solve Ax = 0, we bring the matrix A into the reduced row echelonform rref(A). For every non-leading entry in rref(A), we will get a free variable ti. Writing the system Ax = 0with these free variables gives us an equation ~x =Piti~vi. The vectors ~viform a basis of the kernel of A.REMARK. The problem to find a basis for all vectors ~wiwhich are orthogonal to a given set of vectors, isequivalent to the problem to find a basis for the kernel of the matrix which has the vectors ~wiin its rows.FINDING A BASIS FOR THE IMAGE. Bring the m × n matrix A into the form rref(A). Call a column apivot column, if it contains a leading 1. The corresponding set of column vectors of the original matrix Aform a basis for the image because they are linearly independent and are in the image. Assume there are k ofthem. They span the image because there are (k − n) non-leading entries in the matrix.REMARK. The problem to find a basis of the subspace generated by ~v1, . . . , ~vn, is the problem to find a basisfor the image of the matrix A with column vectors ~v1, ..., ~vn.EXAMPLE. Two vectors on a line are linear dependent. The linear relation says that one is a multiple of theother. Three vectors in the plane are linear dependent. One can find a relation a~v1+ b~v2= ~v3by changing thesize of the lengths of the vectors ~v1, ~v2until ~v3becomes the diagonal of the parallelogram spanned by ~v1, ~v2.Four vectors in three dimensional space are linearly dependent. As in the plane one can change the length ofthe vectors to make ~v4a diagonal of the parallelepiped spanned by ~v1, ~v2, ~v3.EXAMPLE. Let A be the matrix A =0 0 11 1 01 1 1. In reduced row echelon form is rref(A) =1 1 00 0 10 0 0.There is one non-leading entry in the second row. The dimension of the kernel is 1.EXAMPLE. There are two column vectors with leading 1. The dimension of the image is 2. The dimensionformula 2 + 1 = 3 is satisfied.EXAMPLE. B = rref(A) =1 1 00 0 10 0 0. To determine a basis of the kernel we write Bx = 0 as a system oflinear equations: x + y = 0, z = 0. The variable y is the free variable. With y = t, x = −t is fixed. The linearsystem rref(A)x = 0 is solved by ~x =xyz= t−110. So, ~v =−110is a basis of the kernel.EXAMPLE. Because the first and third vectors in rref(A) are columns with leading 1’s, the first and thirdcolumns ~v1=011, ~v2=101of A form a basis of the image of A.WHY BASIS VECTORS? Would it not be easier just to look at the standard basis vectors ~e1, . . . , ~enonly?The reason for more general basis vectors is that they allow a more flexible adaptation at the situation. Aperson in Paris prefers a different set of basis vectors than a person in Boston. We will also see that in manyapplications, problems can be solved easier with the right basis.For example, to describe the reflection of aray at a plane, it is preferable to use
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