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Exercise 3.4.10

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1.

It’s easy to check that

0 - 2 + 3 - 1 - 0 + 0 = 0 .

So the vector $(0, 1, 1, 1, 0)$ is an element in $V$ . Since the set contains only one nonzero vector, it’s linearly independent.

2.

As usual we can find a basis

β = {(2, 1, 0, 0, 0), (- 3, 0, 1, 0, 0), (1, 0, 0, 1, 0), (- 2, 0, 0, 0, 1)} .

So we know that ${(0, 1, 1, 1, 0)} \cup β$ can generate the space $V$ . Do the same thing to this new set and remember to put $(0, 1, 1, 1, 0)$ on the first column in order to keep it as an element when we do Gaussian elimination.

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

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

Now we know that

β^{'} = {(0, 1, 1, 1, 0), (2, 1, 0, 0, 0), (- 3, 0, 1, 0, 0), (- 2, 0, 0, 0, 1)}

forms a basis for $V$ .

jephianlin

2011-06-27 00:00

Exercise 3.4.10

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