Answer to Question #205404 in Linear Algebra for sabelo Zwelakhe Xu

Question

show that any g€l(v, c) and u€v with g(u) not equal to 0 v=null g i{£u:£€c

Answer 1

To show that V = null g ⊕ {$\lambda$u | $\lambda$ ∈ C}, we need to show that V = null g + {$\lambda$u | $\lambda$ ∈ C} and that null g ∩ {$\lambda$u | $\lambda$ ∈ C} = {0}.

First we show that V = null g + {$\lambda$u | $\lambda$ ∈ C}. Let $v\isin V$. We need to find $n\isin null\ g$ and

$w\isin$ {$\lambda$u | $\lambda$ ∈ C} for which v=n+w.  Suppose $g(v)=\lambda\isin C$, and that $g(u)=\mu\isin C$. We know that $\mu \neq 0$ because u is not in null g. Thus, $\mu$ has an inverse in C. Let $\alpha=\lambda \mu^{-1}\isin C$. Note that this means $\alpha$ times $\mu$ gives back $\lambda$ . So,

$g(\alpha u)=\alpha g(u)=\alpha\mu=\lambda$

Thus, $g(\alpha u)=g(v)$. Let $n=v-\alpha u$. Then $g(n)=g(v)-g(\alpha u)=0$. Therefore, n ∈ null g. Set w = $\alpha$u. Then $w\isin$ {$\lambda$u | $\lambda$ ∈ C}, and v = n + w. So we can write v as the sum of elements of null g and {$\lambda$u | $\lambda$ ∈ C}. Therefore, V = null g + {$\lambda$u | $\lambda$ ∈ C}.

Next we show that null g ∩ {$\lambda$u | $\lambda$ ∈ C} = {0}.  Suppose $v\isin$ null g ∩ {$\lambda$u | $\lambda$ ∈ C}. Then

$v\isin$ {$\lambda$u | $\lambda$ ∈ C}, so $v=\lambda u$ for some $\lambda\isin C$.

Since v ∈ null g, g(v) = 0. So g($\lambda$u) = 0. But g($\lambda$u) = $\lambda$g(u). Since $\lambda$g(u) = 0, either $\lambda$ = 0 or g(u) = 0. But u is not in null g, so g(u) $\neq$ 0. This means $\lambda$ must equal 0. So v = $\lambda$u implies that v = 0. Therefore, null g ∩ {$\lambda$u | $\lambda$ ∈ C} = {0}. Since V = null g + {$\lambda$u | $\lambda$ ∈ C} and that

null g ∩ {$\lambda$u | $\lambda$ ∈ C} = {0}, we have that V = null g ⊕ {$\lambda$u | $\lambda$ ∈ C}.