Exercise 4.3.28

Answers

1.

For brevity, we write

v (y (t)) = {(y (t), y^{'} (t), \dots, y^{(n)} (t))}^{t}

and

v_{i} (t) = {(y (t), y^{'} (t), \dots, y^{(n)} (t))}^{t} .

Since the defferential operator is linear, we have

v ((x + cy) (t)) = v (x (t)) + cv (y (t)) .

Now we have that

[T (x + cy)] (t) = \det (\begin{matrix} v ((x + cy) (t)) & v_{1} (t) & v_{2} (t) & \dots & v_{n} (t) \end{matrix})

= \det (\begin{matrix} v (x (t)) + cv (y (t)) & v_{1} (t) & v_{2} (t) & \dots & v_{n} (t) \end{matrix})

= [T (x)] (t) + c [T (y)] (t)

since determinant is a linear function of the first column when all other columns are held fixed.

2.

Since

N (T)

is a space, it enough to say that

y_{i} \in N (T)

for all

i

. But this is easy since

[T (y)] (t) = \det (v_{i} (t), v_{1} (t), \dots, v_{n} (t)) = 0 .

The determinant is zero since the matrix has two identity columns.

jephianlin

2011-06-27 00:00