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Q. 3.12

Let $\overrightarrow{a_{1} }=\left(-1,1,1\right), \overrightarrow{a_{2} }=\left(1,-1,1\right)$ and $\overrightarrow{a_{3} }=\left(1,1,-1\right).$

(1) Try to find linear operators mapping the tetrahedron $\Delta \overrightarrow{0}\overrightarrow{a_{1} }\overrightarrow{a_{2} } \overrightarrow{a_{3} }$ onto the tetrahedron $\Delta \overrightarrow{0} \left(-\overrightarrow{a_{1} }\right) \left(-\overrightarrow{a_{2} }\right) \left(-\overrightarrow{a_{3} }\right).$ See Fig. 3.54(a).

(2) Try to find a linear operator mapping the tetrahedron $\Delta \overrightarrow{0}\overrightarrow{a_{1} }\overrightarrow{a_{2} } \overrightarrow{a_{3} }$ onto the parallelogram $\overrightarrow{a_{1} }\overrightarrow{a_{2} }.$ See Fig. 3.54(b).

Verified Solution

(1) There are six such possible linear operators. The simplest one, say $f_{1},$ among them is the one that satisfies

$f_{1} \left(\overrightarrow{a_{i} }\right)=-\overrightarrow{a_{i} }$ for 1 ≤ i ≤ 3.

In the natural basis $N=\left\{\overrightarrow{e_{1} },\overrightarrow{e_{2} },\overrightarrow{e_{3} }\right\},$

$\left[ f_{1}\right] _{N} =\left[\begin{matrix} \overrightarrow{a_{1} } \\ \overrightarrow{a_{2} } \\ \overrightarrow{a_{3} } \end{matrix} \right] ^{-1} \left[\begin{matrix} -1 & & 0 \\ & -1 & \\ 0 & &-1 \end{matrix} \right] \left[\begin{matrix} \overrightarrow{a_{1} } \\ \overrightarrow{a_{2} } \\ \overrightarrow{a_{3} } \end{matrix} \right]=-I_{3}$

$\Rightarrow f_{1}\left(\overrightarrow{x} \right) =-\overrightarrow{x}=-\overrightarrow{x}I_{3}.$

It is possible that $\overrightarrow{a_{1} }$ and $\overrightarrow{a_{2} }$ are mapped into $-\overrightarrow{a_{2} }$ and $-\overrightarrow{a_{1} }$ respectively while $\overrightarrow{a_{3} }$ is to $-\overrightarrow{a_{3} }.$ Denote by $f_{2}$ such a linear operator. Then

$f_{2}\left(\overrightarrow{a_{1} }\right) =-\overrightarrow{a_{2} },$

$f_{2}\left(\overrightarrow{a_{2} }\right) =-\overrightarrow{a_{1} },$

$f_{2}\left(\overrightarrow{a_{3} }\right) =-\overrightarrow{a_{3} }. \left(*_{1} \right)$

$\Rightarrow \left[ f_{2}\right] _{N} = P^{-1} \left[\begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 &-1 \end{matrix} \right] P, where P=\left[\begin{matrix} \overrightarrow{a_{1} } \\ \overrightarrow{a_{2} } \\ \overrightarrow{a_{3} } \end{matrix} \right]=\left[\begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 &-1 \end{matrix} \right]$

$= \frac{1}{2} \left[\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right]\left[\begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right]\left[\begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 &-1 \end{matrix} \right]=\left[\begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right]$

$\Rightarrow f_{2}\left(\overrightarrow{x} \right) = \overrightarrow{x}\left[ f_{2}\right] _{N} =\overrightarrow{x}\left[\begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right]=-\overrightarrow{x}\left[\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right]. \left(*_{2} \right)$

Notice that, $\left(*_{1} \right)$ is equivalent to

$-f_{2}\left(\overrightarrow{e}_{1} \right)+f_{2}\left(\overrightarrow{e}_{2} \right)+f_{2}\left(\overrightarrow{e}_{3} \right)=-\overrightarrow{a_{2} }, f_{2}\left(\overrightarrow{e}_{1} \right)-f_{2}\left(\overrightarrow{e}_{2} \right)+f_{2}\left(\overrightarrow{e}_{3} \right)=-\overrightarrow{a_{1} }, f_{2}\left(\overrightarrow{e}_{1} \right)+f_{2}\left(\overrightarrow{e}_{2} \right)-f_{2}\left(\overrightarrow{e}_{3} \right)=-\overrightarrow{a_{3} }$

$\Rightarrow f_{2}\left(\overrightarrow{e}_{1} \right)+f_{2}\left(\overrightarrow{e}_{2} \right)+f_{2}\left(\overrightarrow{e}_{3} \right)=-\left(\overrightarrow{a_{1} }+\overrightarrow{a_{2} }+\overrightarrow{a_{3} }\right)=-\left(1,1,1\right)$

$\Rightarrow f_{2}\left(\overrightarrow{e}_{1} \right)=-\overrightarrow{e}_{2}, f_{2}\left(\overrightarrow{e}_{2} \right)=-\overrightarrow{e}_{1} , f_{2}\left(\overrightarrow{e}_{3} \right)=-\overrightarrow{e}_{3} . \left(*_{3} \right)$

This is just $\left(*_{2} \right). f_{2}$ is diagonalizable. Similarly, both

$f_{3}\left(\overrightarrow{x} \right) = -\overrightarrow{x} \left[\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} \right]$  and $f_{4}\left(\overrightarrow{x} \right) = -\overrightarrow{x} \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right]$

are another two such linear operators.

The last two linear operators are

$f_{5}\left(\overrightarrow{x} \right) = -\overrightarrow{x} \left[\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} \right]$  and $f_{6}\left(\overrightarrow{x} \right) = -\overrightarrow{x} \left[\begin{matrix} 0& 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right].$

Both are not diagonalizable. For details, see Sec. 3.7.8.

(2) The parallelogram $\overrightarrow{a_{1} }\overrightarrow{a_{2} }$ has the vertices at $\overrightarrow{0},\overrightarrow{a_{1} },\overrightarrow{a_{1} }+\overrightarrow{a_{2} }=2 \overrightarrow{e_{3} }$ and $\overrightarrow{a_{2} }.$

Define a linear operator g: R³ → R³ as

$g\left(\overrightarrow{a_{1} }\right) =\overrightarrow{a_{1} },$

$g\left(\overrightarrow{a_{2} }\right) =\overrightarrow{a_{2} },$

$g\left(\overrightarrow{a_{3} }\right) =\overrightarrow{a_{1} }+\overrightarrow{a_{2} }=2\overrightarrow{e_{3} }.$

The process like $\left(*_{2} \right)$ or $\left(*_{3} \right)$ will lead to

$g\left(\overrightarrow{x }\right)=\overrightarrow{x }\left[ g\right] _{N} =\overrightarrow{x }\left[\begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ \\ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ \\ 0 & 0 & 1 \end{matrix} \right]$ or

$\left[ g\right] _{B} =P\left[ g\right] _{N}P^{-1}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} \right],$

where P is as above and $B=\left\{\overrightarrow{a_{1} },\overrightarrow{a_{2} },\overrightarrow{a_{3} }\right\} .$ g is diagonalizable and

$Q\left[ g\right] _{N}Q^{-1}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right], where Q=\left[\begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -3 \end{matrix} \right].$

g is a projection of R³ onto the subspace $\ll \overrightarrow{a_{1} },\overrightarrow{a_{2} }\gg$ along $\ll \left(1,1,-3\right)\gg$ as can be visualized in Fig. 3.54(b).

The readers are urged to find more such linear operators.

One of the main advantages of diagonalizable linear operators or matrices A is that it is easy to compute the power

$A^{n}$

for n ≥ 1 and n < 0 if A is invertible. More precisely, suppose

$A=P^{-1}\left[\begin{matrix} \lambda _{1} & & 0 \\ &\lambda _{2} & \\ 0 & &\lambda _{3} \end{matrix} \right]P$

⇒ 1. det(A) = $\lambda _{1} \lambda _{2} \lambda _{3} .$

2. A is invertible ⇔ $\lambda _{1} \lambda _{2} \lambda _{3} \neq 0 .$ In this case,

$A^{-1}=P^{-1}\left[\begin{matrix} \lambda^{-1} _{1} & & 0 \\ &\lambda^{-1} _{2} & \\ 0 & &\lambda^{-1} _{3} \end{matrix} \right]P.$

3. Hence

$A^{n}=P^{-1}\left[\begin{matrix} \lambda^{n} _{1} & & 0 \\ &\lambda^{n} _{2} & \\ 0 & &\lambda^{n} _{3} \end{matrix} \right]P.$

4. tr(A) = $\lambda _{1}+\lambda _{2}+\lambda _{3}.$

5. For any polynomial $g\left(t\right)\in P _{n}\left(R\right),$

$g\left(A\right)=P^{-1}\left[\begin{matrix} g\left(\lambda _{1} \right)& & 0 \\ &g\left(\lambda _{2} \right) & \\ 0 & &g\left(\lambda _{3} \right) \end{matrix} \right]P.$    (3.7.49)

These results still hold for any diagonalizable matrix of finite order.