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Linear Algebra, Theory And Applications, 2012a

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272 LINEAR TRANSFORMATIONS CANONICAL FORMS<br />

where the second factor is irreducible over Q. Consider φ 2 (λ) first. Messy computations<br />

yield<br />

⎛<br />

⎞<br />

−16 −16 −16 −16 −32<br />

0 0 0 0 0<br />

φ 2 (A) =<br />

⎜ 0 0 0 0 0<br />

⎟<br />

⎝ 0 0 0 0 0 ⎠<br />

16 16 16 16 32<br />

and so<br />

⎛<br />

ker (φ 2 (A)) = a<br />

⎜<br />

⎝<br />

−1<br />

1<br />

0<br />

0<br />

0<br />

⎞ ⎛<br />

⎟<br />

⎠ + b ⎜<br />

⎝<br />

−1<br />

0<br />

1<br />

0<br />

0<br />

⎞ ⎛<br />

⎟<br />

⎠ + c ⎜<br />

⎝<br />

−1<br />

0<br />

0<br />

1<br />

0<br />

⎞ ⎛<br />

⎟<br />

⎠ + d ⎜<br />

⎝<br />

Now start with one of these basis vectors and look for an A cycle. Picking the first one, you<br />

obtain the cycle<br />

⎛ ⎞ ⎛ ⎞<br />

−1 −15<br />

1<br />

⎜ 0<br />

⎟<br />

⎝ 0 ⎠ , 5<br />

⎜ 1<br />

⎟<br />

⎝ −5 ⎠<br />

0 7<br />

because the next vector involving A 2 yields a vector which is in the span of the above two.<br />

You check this by making the vectors the columns of a matrix and finding the row reduced<br />

echelon form. Clearly this cycle does not span ker (φ 2 (A)) , so look for another cycle. Begin<br />

with a vector which is not in the span of these two. The last one works well. Thus another<br />

A cycle is<br />

⎛<br />

⎜<br />

⎝<br />

It follows a basis for ker (φ 2 (A)) is<br />

⎧⎛<br />

⎞ ⎛<br />

−2<br />

⎪⎨<br />

0<br />

⎜ 0<br />

⎟<br />

⎝ 0 ⎠ , ⎜<br />

⎝<br />

⎪⎩<br />

1<br />

−2<br />

0<br />

0<br />

0<br />

1<br />

−16<br />

4<br />

−4<br />

0<br />

8<br />

⎞ ⎛<br />

⎟<br />

⎠ , ⎜<br />

⎝<br />

⎞ ⎛<br />

⎟<br />

⎠ , ⎜<br />

⎝<br />

−16<br />

4<br />

−4<br />

0<br />

8<br />

−1<br />

1<br />

0<br />

0<br />

0<br />

⎞<br />

⎟<br />

⎠<br />

⎞ ⎛<br />

⎟<br />

⎠ , ⎜<br />

⎝<br />

−15<br />

5<br />

1<br />

−5<br />

7<br />

⎞⎫<br />

⎪⎬<br />

⎟<br />

⎠<br />

⎪⎭<br />

From the above theory, these vectors are linearly independent. Finally consider a cycle<br />

coming from ker (φ 1 (A)). This amounts to nothing more than finding an eigenvector for A<br />

corresponding to the eigenvalue 4. An eigenvector is ( −1 0 0 0 1 ) T<br />

.Now the desired<br />

matrix for the similarity transformation is<br />

⎛<br />

⎞<br />

−2 −16 −1 −15 −1<br />

0 4 1 5 0<br />

S ≡<br />

⎜ 0 −4 0 1 0<br />

⎟<br />

⎝ 0 0 0 −5 0 ⎠<br />

1 8 0 7 1<br />

−2<br />

0<br />

0<br />

0<br />

1<br />

⎞<br />

⎟<br />

⎠ .

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