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Linear Transformations and Matrices

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Linear Transformations and Matrices Vector Spaces–Review Recall that a vector space is a set equipped with ◮ addition ◮ scalar multiplication. Some Examples of Vector Spaces ◮ Rn, whose elements are n-tuples of numbers, such as (1, π, 3). ◮ Pn, whose elements are polynomials of degree at most n. ◮ The set of solutions of some (fixed) linear homogeneous differential equation (that is, solutions defined on a certain interval) ◮ The set of matrices of a given dimension with real entries Linear combinations and bases ◮ z is a linear combination of x1,..., xn if it is a sum of multiples of the xi . ◮ That is, z = a1x1 + . + anxn for some constants ai . ◮ The xi are linearly dependent if some linear combination of them (other than the trivial one with all ai = 0) is zero ◮ They are linearly independent if no linear combination of them is zero, except for the trivial case when all ai = 0. ◮ If the xi are linearly dependent, that means one of them can be written as a linear combination of the others. (Pick one with a non-zero ai and solve the equation a1x1 . anxn = 0 for xi .) Spanning ◮ Vectors x1,..., xn are said to span a space (or subspace) V if every element of V is a linear combination of the xi . ◮ Example: (0, 0, 1), (0, 1, 0), and (1, 0, 0) span R3. ◮ Example: (1, 1, 0), (0, 1, 1) do not span R3. Why not? ◮ Question: When do three vectors in R3 not span R3? Dimension ◮ A space (or subspace) has dimension n if it can be spanned by n vectors but not by fewer than n vectors. ◮ What are the subspaces of R3 of dimension 1? ◮ What are the subspaces of R3 of dimension 2? ◮ What are the subspaces of R3 of dimension 3? ◮ What are the subspaces of R3 of dimension 4? Spaces whose elements are functions ◮ What is the dimension of Pn, the space of polynomials of degree at most n? ◮ What is the dimension of the space of solutions of y ′′ = y? ◮ What is the dimension of the space of solutions of y ′′′ = y? ◮ The space of all smooth functions on [0, 1] does not have any finite dimension, because it contains every Pn. It is “infinite-dimensional”. What is a basis? A basis is a minimal spanning set. Specifically x1,..., xn is a basis for V if ◮ it spans V ◮ No proper subset of it spans V Question: Give a basis for R3. Linear independence of basis elements The elements of a basis x1,..., xn for a space V are linearly independent. Otherwise, one of them could be written as a linear combination of the others, and we would get a smaller spanning set. Understanding the concept of basis Questions: ◮ How many vectors will there be in a basis of R3? ◮ Do the vectors in a basis have to be linearly independent? ◮ Give an example of a basis of R3 in which the vectors are not all perpendicular. ◮ If I have 3 linearly independent vectors in R3 do they have to be a basis? ◮ What is a basis for the space of solutions of y ′′ = −y? Linear Transformations In this lecture we use T for a function, also called a “map” or “transformation”, from one vector space to another. An important example of such a map is when T is defined by multiplication by a matrix. For example, the map that takes (x, y) onto 1 2 x 3 4 y maps R2 into R2. If the matrix is not square then it might map, for example, R3 into R2. That would require a matrix with three rows and 2 columns. Every linear transformation of Rn is given by a matrix ◮ a linear transformation is determined by what it does to a basis. ◮ Example. Suppose T takes (1, 0, 0) to (3, 5, 2), and (0, 1, 0) to (4, 4, 3), and (0, 0, 1) to (6, 7, 9). Then it is represented by a matrix: x 3 4 6 x T y = 5 4 7 y z 2 3 9 z ◮ The columns of the matrix are the images of the standard basis vectors Kernel of a linear transformation A linear transformation can collapse part of the space. For example, consider the transformation that takes (x, y, z) to (x, y, 0). It is called “projection on the xy plane.” The dimension of the range is 2. When this happens there will always be some nonzero vectors x such that Tx = 0. For example, in this case, x = (0, 0, 1). ◮ The kernel of a linear transformation T is defined to be the set of vectors x such that Tx = 0. ◮ The kernel is zero (technically 0) if and only if T is one-to-one, because if Tx = Ty then T (x − y)=0 and vice-versa. ◮ The kernel is a subspace of the domain space of T , since if Tx = 0 and Ty = 0 then T (ax + by) =0 too. ◮ If the subspace has big dimension that means T is very far from one-one. ◮ Example: T (x, y, z)= x. The kernel has dimension 2. Dim (Ker T) + Dim(range T) = Dim (Domain T) ◮ Dim Ker T measures how much T collapses things. ◮ Dim range T measures how much T does not collapse things. ◮ Take a basis for KerT and extend it to a basis for V (the domain of T ). Then the image of these last vectors (the ones you used to extend the basis) are a basis for Range T . Look: ◮ Let x1,... xk be a basis for KerT and extend it with xk+1,..., xn to a basis for V . Then an arbitrary element of TV is T applied to a linear combination of the xi , so it’s a linear combination of the Txi ; but the first k of these are zero, so it’s a linear combination of Txk+1,..., Txn. Therefore these vectors span the range space TV . But if they were linearly dependent, some combination of the Txk+1,..., Txn would be zero, so T of some linear combination of the xk+1,..., xn would be zero, so the x1,..., xn would not be linearly independent, contradiction. ◮ That implies the theorem in the slide title Transformations and Matrices ◮ Transformation + basis implies matrix representation ◮ To represent a transformation by a matrix, you need a basis. ◮ Different bases, different representation Example. let T take (x, y, z) to (x + y, z, x + z). In the standard basis (1, 0, 0), (0, 1, 0), (0, 0, 1), T has the matrix 1 1 0 0 0 1 1 0 1 But in the basis (1, 1, 0), (−1, 1, 0), (0, 0, 1), it has a different matrix; let’s find that matrix. Sample problem Let T take (x, y, z) to (x + y, z, x + z). Find the matrix of T in the basis (1, 1, 0), (−1, 1, 0), (0, 0, 1). Solution. Let a = (1, 1, 0), b = (−1, 1, 0), and c = (0, 0, 1). Then we have to express Ta as a linear combination of a, b, c to get the first row of the matrix. Well, Ta = T (1, 1, 0) = (2, 0, 1). Now, how can we express that as a combination of a, b, and c? So our solution will be Ta = Aa + Bb + Cc = (A, A, 0) + (−B, B, 0) + (0, 0, C). If that is going to be (2, 0, 1), we need C = 1 and A + B = 0 and A − B = 2, so A = 1 and B = −1. Check: a − b + c = (2, 0, 1)? Yes. That just gives us one column of the answer. In the textbook, it is shown how to solve three such equations at once to find the matrix of T , but that is not the best way. We will see a better way in the next lecture..
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