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A First Course in Linear Algebra, 2017a

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572 Selected Exercise Answers<br />

7.2.2 The eigenvectors and eigenvalues are:<br />

⎧⎡<br />

⎤⎫<br />

⎧⎡<br />

⎨ 2 ⎬ ⎨<br />

⎣ 0 ⎦<br />

⎩ ⎭ ↔ 1, ⎣<br />

⎩<br />

1<br />

−2<br />

1<br />

0<br />

⎤⎫<br />

⎧⎡<br />

⎬ ⎨<br />

⎦<br />

⎭ ↔ 1, ⎣<br />

⎩<br />

7<br />

−2<br />

2<br />

⎤⎫<br />

⎬<br />

⎦<br />

⎭ ↔ 3<br />

The matrix P needed to diagonalize the above matrix is<br />

⎡<br />

2 −2<br />

⎤<br />

7<br />

⎣ 0 1 −2 ⎦<br />

1 0 2<br />

and the diagonal matrix D is<br />

⎡<br />

⎣<br />

1 0 0<br />

0 1 0<br />

0 0 3<br />

⎤<br />

⎦<br />

7.2.3 The eigenvectors and eigenvalues are:<br />

⎧⎡<br />

⎤⎫<br />

⎧⎡<br />

⎨ −6 ⎬ ⎨<br />

⎣ −1 ⎦<br />

⎩ ⎭ ↔ 6, ⎣<br />

⎩<br />

−2<br />

−5<br />

−2<br />

2<br />

⎤⎫<br />

⎧⎡<br />

⎬ ⎨<br />

⎦<br />

⎭ ↔−3, ⎣<br />

⎩<br />

The matrix P needed to diagonalize the above matrix is<br />

⎡<br />

−6 −5<br />

⎤<br />

−8<br />

⎣ −1 −2 −2 ⎦<br />

2 2 3<br />

−8<br />

−2<br />

3<br />

⎤⎫<br />

⎬<br />

⎦<br />

⎭ ↔−2<br />

and the diagonal matrix D is<br />

⎡<br />

⎣<br />

6 0 0<br />

0 −3 0<br />

0 0 −2<br />

⎤<br />

⎦<br />

7.2.8 The eigenvalues are dist<strong>in</strong>ct because they are the n th roots of 1. Hence if X is a given vector with<br />

X =<br />

n<br />

∑ a j V j<br />

j=1<br />

then<br />

so A nm = I.<br />

A nm X = A nm<br />

∑<br />

n a j V j =<br />

j=1<br />

n<br />

∑ a j A nm V j =<br />

j=1<br />

n<br />

∑ a j V j = X<br />

j=1<br />

7.2.13 AX =(a + ib)X. Now take conjugates of both sides. S<strong>in</strong>ce A is real,<br />

AX =(a − ib)X

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