The basis in a nutshell 2/28/2001 Math 21b, O. KnillDEFINITION BASIS. A set of vectors v1, . . . , vnis called a basis of a linear space if they arelinear independent and if they span the linear space. Linear independent means that there areno linear relations aiv1+ . . . + anvn= 0. Spanning the space means that very vector v can bewritten as a linear combination v = a1v1+ . . . + anvnof the basis.FACT. If v1, ..., vnis a basis, then everyvector v can be represented uniquelyas a linear combination of the vi.v = a1v1+ . . . + anvn.REASON. The representation is pos-sible because the vispan the space.If there were two different representa-tions, then subtracting these equationwould lead to a linear relation betweenthe viwhich is forbidden by the linearindependence.FACT. If n vectors v1, ..., vnspan aspace and w1, ..., wmare linear indepen-dent vectors, then m ≤ n. (see p. 162top in the book)DIMENSION. The number of elementsin a basis is independent of the basis.It is called the dimension of the linearspace.BASIS DEFINES INVERTIBLE MATRIX. The n ×n matrix A =| | |v1v2. . . vn| | |is invert-ible if and only if v1, . . . , vndefine a basis in Rn.DIMENSION OF THE KERNEL.The number of non-leading entriesrref(A) is the dimension of the ker-nel dim(ker(A)) because we can intro-duce a parameter for each of those whensolving Ax = 0 using Gauss elimina-tion.DIMENSION OF THE IMAGE. Thenumber of leading 1 in rref(A), therank of A is the dimension of theimage dim(im(A)) because every suchleading 1 produces a different columnvector (called pivot column vectors)and these column vectors are linearlyindependent.DIMENSION FORMULA:dim(ker(A)) + dim(im(A)) = nif A is a m × n matrix. The number nis the dimension of the domain RnofA : Rn→ Rm.EXAMPLE: A matrix A is invertible ifand only if the dimension of the imageis n. Therefore, the dimension of thekernel is 0.(We know that also becausewe have a unique solution to Ax = 0).FINDING A BASIS FOR THE KERNEL. To solve Ax = 0, we bring the matrix A into the reducedrow echelon form rref(A). For every non-leading entry in rref(A), we will get a free variable ti.Writing the system Ax = 0 with these free variables gives us an equation ~x =Piti~vi. The vectorsviform a basis of the kernel of A.REMARK. If we have the problem to find a basis for all vectors wiwhich are orthogonal to a givenset of vectors, this is the problem to find a basis for the kernel of the matrix which has the vectorswiin its rows.FINDING A BASIS FOR THE IMAGE. Bring the m × n matrix A into the form rref(A). Call acolumn a pivot column, if it contains a leading 1. The corresponding set of column vectors of theoriginal matrix A form a basis for the image because they are linearly independent and are in theimage. Assume there are k of them. They span the image because there are (k − n) non-leadingentries in the matrix.REMARK. If we have the problem to find a basis of the subspace generated by a few vectorsv1, . . . , vn, then this is the problem to find a basis for the matrix with column vectors v1, ...vn.Homework for Friday, March 2: Section: 3.3, Numbers 22,24,32,36,40,56*ExamplesEXAMPLE. The vectors v1=110, v2=011, v3=101form a basis in the three dimen-sional space.EXAMPLE. If v =435, then v = v1+ 2v2+ 3v3(with the basis above) and this representationis unique. We can find the coefficients by solving Ax = v, where A has the vias column vectors.In our case,1 0 11 1 00 1 1xyz=435had the unique solution x = 1, y = 2, z = 3 leading tov = v1+ 2v2+ 3v3. The matrix A is invertible because v1, v2, v3form a basis.EXAMPLE. Two vectors on a line are linear dependent. The linear relation says that one is amultiple of the other. Three vectors in the plane are linear dependent. One can find a relationav1+ bv2= v3by changing the size of the lengths of the vectos v1, v2until v3becomes thediagonal of the parallelogram spanned by v1, v2. Four vectors in three dimensional space arelinearly dependent. As in the plane one can change the length of the vectors to make v4a diagonalof the parallelepiped spanned by v1, v2, v3.EXAMPLES. The dimension of the point {0} is zero. The dimension of a line is 1. The dimensionof a plane is 2, the dimension of three dimensional space is 3. The dimension is independent onwhere the space is embedded in. A line in the plane and a line in space for example have the samedimension 1.EXAMPLE. Let A be the matrix A =0 0 11 1 01 1 1. In reduced row echelon form it is rref(A) =1 1 00 0 10 0 0. There is one non-leading entry A12= 1 in the first row. The dimension of thekernel is 1.EXAMPLE. In the example, there are two pivotal column vectors, the first and the third. There-fore, the dimension of the image is 2. The dimension formula 2 + 1 = 3 is satisfied.EXAMPLE. B = rref(A) =1 1 00 0 10 0 0. We determine a basis of the kernel by writing the Bx = 0as a system of linear equations x + y = 0, z = 0. The variable y is a free variable. With y = t,we have x = −t. The linear system rref(A)x = 0 is solved by ~x =xyz= t−110. So,v =−110is a basis of the kernel.EXAMPLE. In order to get a basis of the image, we chose the first and third column vectorsof the original matrix A because the first and third vectors in rref(A) are pivot columns. So,v1=011, v2=101form a basis of the image of
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