Exercise 3.2.20

Answers

1.

Just like the skill we learned in the Exercise 1.4.2 we can solve the system of linear equation

Ax = 0

and get the solution space

{(x_{3} + 3 x_{5}, - 2 x_{3} + x_{5}, x_{3}, - 2 x_{5}, x_{5}) : x_{i} \in 𝔽} .

And we know that ${(1, - 2, 1, 0, 0), (3, 1, 0, - 2, 1)}$ is a basis for the solution space. Now we can construct the desired matrix

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

2.

AB = O

, this means that every column vector of

B

is a solution of

Ax = 0

. If rank of

B

is greater than

2

, we can find at least three independent vectors from columns of

B

. But this is is impossible since by Dimension Theorem we know that

\dim (𝔽^{5}) = \dim (R (L_{A})) + \dim (N (L_{A}))

and so dim $(N (L_{A})) = 5 - 3 = 2$ .

jephianlin

2011-06-27 00:00