Exercise 1.6.14

Answers

1.
Since $Au = 0$ , so $u$ is a basis for the nullspace. Since $Av = 3 v$ and $Aw = 5 w$ , also $v$ and $w$ are independent, so both $v$ and $w$ are bases for the column space.
1.
We have $x = \frac{1}{3} v + \frac{1}{5} w$ , so $Ax = A (\frac{1}{3} v + \frac{1}{5} w) = \frac{1}{3} Av + \frac{1}{5} Aw = \frac{1}{3} 3 v + \frac{1}{5} 5 w = v + w$
1.
$Ax = u$ has no solution. If it did then $u$ would be in the column space, which contradicts with $Au = 0$ , meaning $u$ is in the nullspace.

niuers

2020-03-20 00:00