BASIS Math 21b, O. KnillLINEAR SUBSPACE. A subset X of Rnwhich is closed under addition and scalar multiplication is called alinear subspace of Rn.WHICH OF THE FOLLOWING SETS ARE LINEAR SPACES?a) The kernel of a linear map.b) The image of a linear map.c) The upper half plane.d) the line x + y = 0.e) The plane x + y + z = 1.f) The unit circle.BASIS. A set of vectors ~v1, . . . , ~vmis a basis of a linear subspace X of Rnif they arelinear independent and if they span the space X. Linear independent means thatthere are no nontrivial linear relations ai~v1+ . . . + am~vm= 0. Spanning the spacemeans that very vector ~v can be written as a linear combination ~v = a1~v1+. . .+am~vmof basis vectors. A linear subspace is a set containing {~0} which is closed underaddition and scaling.EXAMPLE 1) 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.EXAMPLE 2) Two nonzero vectors in the plane which are not parallel form a basis.EXAMPLE 3) Three vectors in R3which are in a plane form not a basis.EXAMPLE 4) Two vectors in R3do not form a basis.EXAMPLE 5) Three nonzero vectors in R3which are not contained in a single plane form a basis in R3.EXAMPLE 6) The columns of an invertible n × n matrix form a basis in Rnas we will see.FACT. 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 at least one representation because the vectors~vispan the space. If there were two different representations ~v =a1~v1+. . .+an~vnand ~v = b1~v1+. . .+bn~vn, then subtraction wouldlead to 0 = (a1− b1)~v1+ ... + (an− bn)~vn. This nontrivial linearrelation of the viis forbidden by assumption.FACT. If n vectors ~v1, ..., ~vnspan a space and ~w1, ..., ~wmare linear independent, then m ≤ n.REASON. This is intuitively clear in dimensions up to 3. You can not have more then 4 vectors in space whichare linearly independent. We will give a precise reason later.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 1), the 3 × 3 matrix A is invertible.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.EXAMPLES.1) Two vectors on a line are linear dependent. One is a multiple of the other.2) Three vectors in the plane are linear dependent. One can find a relation a~v1+ b~v2= ~v3by changing the sizeof the lengths of the vectors ~v1, ~v2until ~v3becomes the diagonal of the parallelogram spanned by ~v1, ~v2.3) Four vectors in three dimensional space are linearly dependent. As in the plane one can change the lengthof the 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 B = rref(A) =1 1 00 0 10 0 0.To determine a basis of the kernel we write Bx = 0 as a system of linear equations: x + y = 0, z = 0. Thevariable y is the free variable. With y = t, x = −t is fixed. The linear system 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 DO WE INTRODUCE BASIS VECTORS? Wouldn’t it be justeasier to look at the standard basis vectors ~e1, . . . , ~enonly? The rea-son for more general basis vectors is that they allow a more flexibleadaptation to the situation. A person in Paris prefers a different setof 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 a ray at aplane or at a curve, it is preferable to use basis vec-tors which are tangent or orthogonal. When lookingat a rotation, it is good to have one basis vector inthe axis of rotation, the other two orthogonal to theaxis. Choosing the right basis will be especially im-portant when studying differential equations.A PROBLEM. Let A =1 2 31 1 10 1 2. Find a basis for ker(A) and im(A).SOLUTION. From rref(A) =1 0 −10 1 20 0 0we see that = ~v =1−21is in the kernel. The two column vectors110,2, 1, 1 of A form a basis of the image because the first and third column are pivot
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