Exercise 3.4.11

Answers

1.

Similarly check

1 - 4 + 3 + 0 - 0 = 0 .

So the set containing only the vector $(1, 2, 1, 0, 0)$ is linearly independent by the same reason.

2.

Do the same thing to the set

{(1, 2, 1, 0, 0)} \cup β

(\begin{matrix} 1 & 2 & - 3 & 1 & - 2 \\ 2 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix})

⇝ (\begin{matrix} 1 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 1 & - 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix})

Now we know that the set

{(1, 2, 1, 0, 0), (2, 1, 0, 0, 0), (1, 0, 0, 1, 0), (- 2, 0, 0, 0, 1)}

forms a basis for $V$ .

jephianlin

2011-06-27 00:00