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A First Course in Linear Algebra, 2017a

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7.2. Diagonalization 365<br />

Exercise 7.2.3 F<strong>in</strong>d the eigenvalues and eigenvectors of the matrix<br />

⎡<br />

89 38<br />

⎤<br />

268<br />

⎣ 14 2 40 ⎦<br />

−30 −12 −90<br />

One eigenvalue is −3. Diagonalize if possible.<br />

Exercise 7.2.4 F<strong>in</strong>d the eigenvalues and eigenvectors of the matrix<br />

⎡<br />

1 90<br />

⎤<br />

0<br />

⎣ 0 −2 0 ⎦<br />

3 89 −2<br />

One eigenvalue is 1. Diagonalize if possible.<br />

Exercise 7.2.5 F<strong>in</strong>d the eigenvalues and eigenvectors of the matrix<br />

⎡<br />

11 45<br />

⎤<br />

30<br />

⎣ 10 26 20 ⎦<br />

−20 −60 −44<br />

One eigenvalue is 1. Diagonalize if possible.<br />

Exercise 7.2.6 F<strong>in</strong>d the eigenvalues and eigenvectors of the matrix<br />

⎡<br />

95 25<br />

⎤<br />

24<br />

⎣ −196 −53 −48 ⎦<br />

−164 −42 −43<br />

One eigenvalue is 5. Diagonalize if possible.<br />

Exercise 7.2.7 Suppose A is an n×n matrix and let V be an eigenvector such that AV = λV. Also suppose<br />

the characteristic polynomial of A is<br />

det(xI − A)=x n + a n−1 x n−1 + ···+ a 1 x + a 0<br />

Expla<strong>in</strong> why<br />

(<br />

A n + a n−1 A n−1 + ···+ a 1 A + a 0 I ) V = 0<br />

If A is diagonalizable, give a proof of the Cayley Hamilton theorem based on this. This theorem says A<br />

satisfies its characteristic equation,<br />

A n + a n−1 A n−1 + ···+ a 1 A + a 0 I = 0<br />

Exercise 7.2.8 Suppose the characteristic polynomial of an n × nmatrixAis1 − X n .F<strong>in</strong>dA mn where m<br />

is an <strong>in</strong>teger.

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