Let $V$ be an $n$-dimensional vector space. If $f,g\in V^*$ are linearly independent, then find $\dim(\ker f\cap \ker g)$.
$\dim(\ker f)=\dim(\ker g)=n-1$ because $f,g\neq 0$, correct?
Hint for answer, please.
Let $V$ be an $n$-dimensional vector space. If $f,g\in V^*$ are linearly independent, then find $\dim(\ker f\cap \ker g)$.
-1
$\begingroup$
linear-algebra
2 Answers
0
By Grassmann’s formula, $$ \dim(\ker f\cap\ker g)=\dim\ker f+\dim\ker g-\dim(\ker f+\ker g) $$ Now, since $f$ and $g$ are nonzero (because the zero vector does not belong to a linearly independent set), \begin{align} \dim\ker f&=n-1\\ \dim\ker g&=n-1 \end{align}
So you're correct, until now.
Since $f$ and $g$ are linearly independent, $$ \ker f+\ker g=… $$ because $\ker f\ne\ker g$.
-
0I don't know $ker f + ker g =...$ ${v+u / f(v)=g(u)=0 }?$ – 2017-02-11
-
0@Magering Since $f$ and $g$ are linearly independent, $\ker f\ne\ker g$: can you show this? Therefore $\ker f+\ker g$ properly contains both. But both have dimension $n-1$, so… – 2017-02-11
0
Yes to 1.
As to 2, consider the restriction of $f$ to $U = \ker(g)$. It is clearly an element of $U^{\star}$. Can it be zero?