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15.
a) Let’s ﬁnd the eigenvectors for the eigenvalue 1. We need to solve (A − I )X = 0.
The three equations in the system are the same: x + y + z = 0, so the solution set
is two dimensional and has basis {[1, 0, − 1], [0, 1, − 1]}.
Now let’s ﬁnd an eigenvector for 4. Let’s reduce the matrix A − 4I : −2 1 1
1 −2 1
1 − 2 1 R3 + R2 1 − 2 1
R2 + 2R1 1 − 2 1 R1 ↔ R2 − 2 1 1 0 −3 3 1 0 1 −1 R3 − R1
1 1 −2
1 1 −2
0 3 − 3 − 3 R2 0 0 0 which means that the solution space is given by z = t, y = t, x = t, so [1, 1, 1] is a
basis for it. 1 01
Therefore,
writing eigenvectors as columns, P = 0 1 1 , and we’ll have
our
−1 −1 1
100
P − 1A P = 0 1 0 .
004
301
b) Let’s look for eigenvector for the eigenvalue 1: the matrix B − I = 0 0 1 is al
000
ready rowreduced, so we can easily see that [0, 1, 0] is a basis for the null space.
But then we only have one eigenvector corresponding to the eigenvalue 1 which
has algebraic multiplicity two. Therefore B is not diagonalizable.
c) He’s right: if they were similar, since A is diagonalizable, B would be too. 2...
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This note was uploaded on 01/19/2010 for the course MAT MAT223 taught by Professor Uppal during the Spring '09 term at University of Toronto.
 Spring '09
 UPPAL
 Linear Algebra, Algebra, Vectors

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