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Exercise 5.4.41

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Let $v_{i}$ be the $i$ -th row vector of $A$ . We have ${v_{1}, v_{2}}$ is linearly independent and $v_{i} = (i - 1) (v_{2} - v_{1}) + v_{1}$ . So the rank of $A$ is $2$ . This means that $t^{n - 2}$ is a factor of the characteristic polynomial of $A$ . Finally, set

β = {(1, 1, \dots, 1), (1, 2, \dots, n)}

and check that $W = span (β)$ is a $L_{A}$ -invariant subspace by computing

A (\begin{matrix} 1 \\ 1 \\ ⋮ \\ 1 \end{matrix}) = (\frac{n (n + 1)}{2} - n^{2}) (\begin{matrix} 1 \\ 1 \\ ⋮ \\ 1 \end{matrix}) + n^{2} (\begin{matrix} 1 \\ 2 \\ ⋮ \\ n \end{matrix})

and

A (\begin{matrix} 1 \\ 2 \\ ⋮ \\ n \end{matrix}) = (\frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} (n + 1)}{2}) (\begin{matrix} 1 \\ 1 \\ ⋮ \\ 1 \end{matrix}) + \frac{n^{2} (n + 1)}{2} (\begin{matrix} 1 \\ 2 \\ ⋮ \\ n \end{matrix}) .

So we know the characteristic polynomial is

{(- 1)}^{n} t^{n - 2} \det (\begin{matrix} \frac{n (n + 1)}{2} - n^{2} - t & \frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} (n + 1)}{2} \\ n^{2} & \frac{n^{2} (n + 1)}{2} - t \end{matrix})

= {(- 1)}^{n} t^{n - 2} \frac{12 t^{2} + (- 6 n^{3} - 6 n) t - n^{5} + n^{3}}{12} .

But this is the formula for $n \geq 2$ . It’s natural that when $n = 1$ the characteristic polynomial is $1 - t$ . Ur...I admit that I computed this strange answer by wxMaxima.

jephianlin

2011-06-27 00:00

Exercise 5.4.41

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