Page 248 - 35Linear Algebra
P. 248
248 Diagonalization
Figure 13.1: This theorem answers the question: “What is diagonalization?”
is invertible because its determinant is −1. Therefore, the eigenvectors of M form a
basis of R, and so M is diagonalizable. Moreover, because the columns of P are the
components of eigenvectors,
−1 0 0
MP = Mv 1 Mv 2 Mv 3 = −1.v 1 0.v 2 2.v 3 = v 1 v 2 v 3 0 0 0 .
0 0 2
Hence, the matrix P of eigenvectors is a change of basis matrix that diagonalizes M;
−1 0 0
P −1 MP = 0 0 0 .
0 0 2
2 × 2 Example
13.4 Review Problems
Reading Problems 1 , 2
Webwork: No real eigenvalues 3
Diagonalization 4, 5, 6, 7
1. Let P n (t) be the vector space of polynomials of degree n or less, and
d : P n (t) → P n (t) be the derivative operator. Find the matrix of
dt
d in the ordered bases E = (1, t, . . . , t ) for the domain and F =
n
dt
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