Question A.3: Let A = |-2 2 -3 2 1 -6 -1 -2 0|...
Let \overline{A} = \left|\begin{matrix}-2 &2 &-3 \\ 2& 1& -6 \\ -1& -2& 0 \end{matrix} \right|
Its characteristics equation is
\left|\begin{matrix}-2-λ &2 &-3 \\ 2& 1-λ & -6 \\ -1& -2& 0-λ \end{matrix} \right|(-2 – λ)(-λ + λ² – 12) – 2(-2λ – 6) – 3(-4 + 1 – λ) = 0
λ³ + λ² – 21λ – 45 = 0
(λ – 5)(λ – 3)(λ – 3) = 0
λ = 5, – 3, -3
The eigenvector is found by substituting the eigenvalues. For λ = 5
By manipulation of the rows, this can be reduced to
\left|\begin{matrix}-1 &-2 &-5 \\ 0& 16 & 32 \\ 0& 0& 0 \end{matrix} \right|\left|\begin{matrix} x \\ y \\ z \end{matrix} \right|=\left|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right|Therefore
-x – 2y – 5z = 0
16y + 32z = 0
As eigenvectors are not unique, by assuming that z = 1, and solving, one eigenvector is
(-1, -2, 1)^t
Similarly, eigenvectors for λ = -3 can be found. This gives the following values:
Choose arbitrarily x_{1} = 2, x_{2} = -1 and then
(2, -1, 0)^t
is another eigenvector. Similarly,
(3, 0, 1)^t
can be the third eigenvector. A matrix formed of these vectors is
and the diagonalization is obtained:
\overline{P}^{-1} \overline{A}\overline{P}= \left|\begin{matrix}5 &0 &0 \\ 0& -3& 0 \\ 0& 0& -3 \end{matrix} \right|This contains the eigenvalues as the diagonal elements.
Now choose some other eigenvectors and form a new matrix, say,
Again with these values, \overline{P}^{-1} A\overline{P} is
\left|\begin{matrix}1 &1 &3 \\ 2& 1& 0 \\ -1& 1& 1 \end{matrix} \right|^{-1} \left|\begin{matrix}-2 &2 &-3 \\ 2& 1&-6 \\ -1& -2& 0 \end{matrix} \right| \left|\begin{matrix}1 &1 &3 \\ 2& 1& 0 \\ -1& 1& 1 \end{matrix} \right|=\left|\begin{matrix}5&0 &0 \\ 0& -3& 0 \\ 0& 0& -3 \end{matrix} \right|This is the same result as before.