Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
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198 <strong>Linear</strong> <strong>Algebra</strong>, by Hefferon<br />
Again, there are three possibilities for the minimal polynomial m 1 (x) = (x − 3), m 2 (x) = (x − 3) 2 ,<br />
and m 3 (x) = (x − 3) 3 . We compute m 1 (T )<br />
⎛<br />
T − 3I = ⎝ 0 0 0 ⎞<br />
1 0 0⎠<br />
0 1 0<br />
and m 2 (T )<br />
⎛<br />
(T − 3I) 2 = ⎝ 0 0 0<br />
⎞ ⎛<br />
1 0 0⎠<br />
⎝ 0 0 0<br />
⎞ ⎛<br />
1 0 0⎠ = ⎝ 0 0 0<br />
⎞<br />
0 0 0⎠<br />
0 1 0 0 1 0 1 0 0<br />
and m 3 (T ).<br />
⎛<br />
(T − 3I) 3 = (T − 3I) 2 (T − 3I) = ⎝ 0 0 0 ⎞ ⎛<br />
0 0 0⎠<br />
⎝ 0 0 0 ⎞ ⎛<br />
1 0 0⎠ = ⎝ 0 0 0 ⎞<br />
0 0 0⎠<br />
1 0 0 0 1 0 0 0 0<br />
Therefore, the minimal polynomial is m(x) = m 3 (x) = (x − 3) 3 .<br />
(d) This case is also triangular, here upper triangular.<br />
2 − x 0 1<br />
c(x) = |T − xI| =<br />
0 6 − x 2<br />
∣ 0 0 2 − x∣ = (2 − x)2 (6 − x) = −(x − 2) 2 (x − 6)<br />
There are two possibilities for the minimal polynomial, m 1 (x) = (x − 2)(x − 6) and m 2 (x) =<br />
(x − 2) 2 (x − 6). Computation shows<br />
⎛<br />
that the minimal polynomial isn’t m 1 (x).<br />
(T − 2I)(T − 6I) = ⎝ 0 0 1<br />
⎞ ⎛<br />
0 4 2⎠<br />
⎝ −4 0 1<br />
⎞ ⎛<br />
0 0 2 ⎠ = ⎝ 0 0 −4<br />
⎞<br />
0 0 0 ⎠<br />
0 0 0 0 0 −4 0 0 0<br />
It therefore must be that m(x) = m 2 (x) = (x − 2) 2 (x −<br />
⎛<br />
6). Here is a verification.<br />
(T − 2I) 2 (T − 6I) = (T − 2I) · ((T<br />
− 2I)(T − 6I) ) = ⎝ 0 0 1 ⎞ ⎛<br />
0 4 2⎠<br />
⎝ 0 0 −4 ⎞ ⎛<br />
0 0 0 ⎠ = ⎝ 0 0 0 ⎞<br />
0 0 0⎠<br />
0 0 0 0 0 0 0 0 0<br />
(e) The characteristic polynomial is<br />
2 − x 2 1<br />
c(x) = |T − xI| =<br />
0 6 − x 2<br />
∣ 0 0 2 − x∣ = (2 − x)2 (6 − x) = −(x − 2) 2 (x − 6)<br />
and there are two possibilities for the minimal polynomial, m 1 (x) = (x − 2)(x − 6) and m 2 (x) =<br />
(x − 2) 2 (x − 6). Checking the first one<br />
⎛<br />
(T − 2I)(T − 6I) = ⎝ 0 2 1<br />
⎞ ⎛<br />
0 4 2⎠<br />
⎝ −4 2 1<br />
⎞ ⎛<br />
0 0 2 ⎠ = ⎝ 0 0 0<br />
⎞<br />
0 0 0⎠<br />
0 0 0 0 0 −4 0 0 0<br />
shows that the minimal polynomial is m(x) = m 1 (x) = (x − 2)(x − 6).<br />
(f) The characteristic polynomial is this.<br />
−1 − x 4 0 0 0<br />
0 3 − x 0 0 0<br />
c(x) = |T − xI| =<br />
0 −4 −1 − x 0 0<br />
= (x − 3) 3 (x + 1) 2<br />
3 −9 −4 2 − x −1<br />
∣ 1 5 4 1 4 − x∣<br />
There are a number of possibilities for the minimal polynomial, listed here by ascending degree:<br />
m 1 (x) = (x − 3)(x + 1), m 1 (x) = (x − 3) 2 (x + 1), m 1 (x) = (x − 3)(x + 1) 2 , m 1 (x) = (x − 3) 3 (x + 1),<br />
m 1 (x) = (x − 3) 2 (x + 1) 2 , and m 1 ⎛<br />
(x) = (x − 3) 3 (x + 1) 2 . The<br />
⎞ ⎛<br />
first one doesn’t pan<br />
⎞<br />
out<br />
−4 4 0 0 0 0 4 0 0 0<br />
0 0 0 0 0<br />
0 4 0 0 0<br />
(T − 3I)(T + 1I) =<br />
⎜ 0 −4 −4 0 0<br />
⎟ ⎜0 −4 0 0 0<br />
⎟<br />
⎝ 3 −9 −4 −1 −1⎠<br />
⎝3 −9 −4 3 −1⎠<br />
1 5 4 1 1 1 5 4 1 5<br />
⎛<br />
⎞<br />
0 0 0 0 0<br />
0 0 0 0 0<br />
=<br />
⎜ 0 0 0 0 0<br />
⎟<br />
⎝−4 −4 0 −4 −4⎠<br />
4 4 0 4 4
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