The Matrix of a Linear Transformation
Finding the Matrix We have seen how to find the matrix that changes from one basis to another. We have also seen how to find the matrix for a linear transformation from R^{m} to R^{n}. Now we will show how to find the matrix of a general linear transformation when the bases are given.
Let L be a linear transformation from V to W and let S = {v_{1}, ... ,v_{m}} and T = {w_{1}, ... ,w_{n}} be bases for V and W respectively. Then the matrix A representing L with respect to the bases S and T has ij^{th} component that is the j^{th} coordinate of the vector L(v_{i}). We start with the example when both bases are "standard". For this section, we will use the following conventions. We will use the letter "E" to denote the standard basis. For P_{n}, we will call the basis (P_{n})_{E} = {1, t, t^{2}, ..., t^{n}} the "standard" basis. For M^{mxn}, we will call the basis M_{E}^{mxn} = {A_{11}, A_{12}, ..., A_{1n}, A_{21}, ..., A_{2n},... ..., A_{m1}, ..., A_{mn}} the standard basis where A_{ij} is the matrix with the ij^{th} entry equal to 1 and all the rest are zero. With this convention, the standard basis for M^{3x2} is given by
Example Let L be the linear transformation from R^{2} to P_{2} defined by L(x,y) = x + yt + (x + y)t^{2} Find the matrix representing L with respect to the standard bases.
Solution Notice that L(1,0) = 1 + t^{2} = (1,0,1) L(0,1) = t + t^{2} = (0,1,1) hence the matrix is given by
Now we will proceed with a more complicated example.
Example Let L be the linear transformation from R^{2} to R^{2} such that L(x,y) = (x  2y, y  2x) and let S = {(2, 3), (1, 2)} be a basis for R^{2}. Find the matrix for L that sends a vector from the S basis to the standard basis.
Solution This involves two parts. The first is to find the matrix for L from the standard basis to the standard basis. This matrix is found by finding L(1, 0) = (1, 2) and L(0,1) = (2, 1) The matrix is
Next we find the matrix from the S basis to the standard basis E. This matrix is
Now consider the diagram below
The matrix that we want is the composition of these two mappings. Remembering that composition of functions is written from right to left we get
Example Let L be the linear transformation from P_{2} to P_{2} with such that L(a + bt + ct^{2}) = (a + c) + (a + 2b)t + (a + b + 3c)t^{2} and let S = {1  t, 1  t^{2}, t  t^{2}) and T = {2 + t + t^{2}, 1 + t, 1 + t + t^{2}} Find the matrix of L with respect to the bases S and T.
Solution We first find the matrix for L from the standard basis to the standard basis. We have L(1,0,0) = L(1 + 0t + 0t^{2}) = 1 + t + t^{2} = (1,1,1) L(0,1,0) = L(0 + t + 0t^{2}) = 2t + t^{2} = (0,2,1) L(0,0,1) = L(0 + 0t + t^{2}) = 1 + 3t^{2} = (1,0,3) Hence the matrix for L with the standard bases is
Next we find the transition matrices
Now we consider the diagram below
Since matrices are functions, we compose the functions by multiplying the matrices from right to left. We get
Example Let L be the linear transformation from M^{2x2} to M^{2x2} and let
and
Find the matrix for L from S to S.
Solution First we find the matrix for L in the standard basis. We have
so that
Next we have
We use the diagram
and our matrix is
Diagonalizing The last example showed us that the matrix for L was of the form P^{1}AP This was the definition of a matrix that is similar to A. If A is an n x n matrix and L is the linear transformation L(v) = Av and if the eigenvectors {v_{1}, ... ,v_{n}}are linearly independent then they form a basis for R^{n}. With S = {v_{1}, ... ,v_{n}} then we have seen that if P is the matrix whose i^{th} column is v_{i}, then D = P^{1}AP is a diagonal matrix. But this P is the transition matrix from the standard basis to the basis S. We have proven the following theorem.
Theorem Let L be the linear transformation L(v) = Av then A is diagonalizable with n linearly independent eigenvectors S = {v_{1}, ... ,v_{n}} if and only if the matrix of L with respect to S is diagonal. Orthogonal Matrices We finish this discussion with a little geometry. A linear transformation L called an isometry if for any u and v in V, L(u) ^{.} L(v) = u ^{.} v Notice that isometries preserve angles. Theorem L is an isometry if and only if the matrix of L with respect to the natural basis is orthogonal.
Proof We use the fact that for any vectors x and y and matrix A, (Ax) . y = x . (A^{T}y) If L is an isometry, then u ^{.} v = L(u) ^{.} L(v) = Au ^{.} Av = u ^{.} (A^{T}Av) for all u and v. But then v = A^{T}Av so that A^{T}A = I that is, A is orthogonal Conversely, if A is orthogonal, then L(u) ^{.} L(v) = Au ^{.} Av = u ^{.} (A^{T}Av) = u ^{.} v so that L is an isometry. Back to the Linear Algebra Home Page Back to the Math Department Home Page email Questions and Suggestions

