Solution to Exercise 4.3.22 from „Linear Algebra“

We have that

{\begin{matrix} T (1) & = & 1 e_{1} + 1 e_{2} + \dots + 1 e_{n + 1} \\ T (x) & = & c_{0} e_{1} + c_{1} e_{2} + \dots + c_{n} e_{n + 1} \\ ⋮ & = & ⋮ \\ T (x^{n}) & = & c_{0}^{n} e_{1} + c_{1}^{n} e_{2} + \dots + c_{n}^{n} e_{n + 1}, \end{matrix}

where ${e_{1}, e_{2}, \dots, e_{n + 1}}$ is the standard basis for $𝔽^{n + 1}$ . So we get the desired conclusion.

By Exercise 2.4.22

T

is isomorphism and hence invertible. So the matrix

M

is also invertible and hence

\det (M) \neq 0

.

We induction on

n

. For

n = 1

, we have

\det (\begin{matrix} 1 & c_{0} \\ 1 & c_{1} \end{matrix}) = (c_{1} - c_{0})

. Suppose the statement of this question holds for

n = k - 1

, consider the case for

n = k

. To continue the proof, we remark a fomula first below.

x^{k} - y^{k} = (x - y) (x^{k - 1} + x^{k - 2} y + \dots + y^{k - 1})

For brevity we write

p (x, y, k) = x^{k - 1} + x^{k - 2} y + \dots + y^{k - 1} .

Now to use the induction hypothesis we can add $- 1$ time the first row to all other rows without changing the determinant.

\det (\begin{matrix} 1 & c_{0} & c_{0}^{2} & \dots & c_{0}^{n} \\ 1 & c_{1} & c_{1}^{2} & \dots & c_{1}^{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & c_{n} & c_{n}^{2} & \dots & c_{n}^{n} \end{matrix})

= \det (\begin{matrix} 1 & c_{0} & c_{0}^{2} & \dots & c_{0}^{n} \\ 0 & c_{1} - c_{0} & (c_{1} - c_{0}) p (c_{1}, c_{0}, 2) & \dots & (c_{1} - c_{0}) p (c_{1}, c_{0}, n) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & c_{n} - c_{0} & (c_{n} - c_{0}) p (c_{n}, c_{0}, 2) & \dots & (c_{n} - c_{0}) p (c_{n}, c_{0}, n) \end{matrix})

= \prod_{j = 1}^{n} (c_{j} - c_{0}) \det (\begin{matrix} 1 & p (c_{1}, c_{0}, 2) & \dots & p (c_{1}, c_{0}, n) \\ ⋮ & ⋮ & ⋮ \\ 1 & p (c_{n}, c_{0}, 2) & \dots & p (c_{n}, c_{0}, n) \end{matrix})

Now we write $e_{i} = {(c_{1}^{i}, c_{2}^{i}, \dots, c_{n}^{i})}^{t}$ for $i = 0, 1, \dots n - 1$ . So the determinant of the last matrix in the equality above can be written as

\det (\begin{matrix} e_{0} & e_{1} + c_{0} e_{0} & e_{2} + c_{0} e_{1} + c_{0}^{2} e_{1} & \dots & e_{n - 1} + c_{0} e_{n - 2} + \dots + c_{0}^{n - 1} e_{0} \end{matrix})

= \det (\begin{matrix} e_{0} & e_{1} & e_{2} + c_{0} e_{1} & \dots & e_{n - 1} + c_{0} e_{n - 2} + \dots \end{matrix})

= \det (\begin{matrix} e_{0} & e_{1} & e_{2} & \dots & e_{n - 1} + c_{0} e_{n - 2} + \dots \end{matrix})

= \dots = \det (\begin{matrix} e_{0} & e_{1} & e_{2} & \dots & e_{n - 1} \end{matrix}) .

And by induction hypothesis, the value of it would be

\prod_{1 \leq i \leq j \leq n} (c_{j} - c_{i}) .

Combining the two equalities above, we get the desired conclusion.

Exercise 4.3.22

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Exercise 4.3.22

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