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- TL;DR Summary
- Let ##F## and ##G## be two non-zero linear functionals over a vector space ##V## of dimension ##n##. Assuming ##ker (F ) \neq \ker (G)##, determine the dimensions of the following subspaces: ##\ker (F )##, ##\ker (G)##, ##\ker (F ) + \ker (G)##, and ##\ker (F ) \cap \ker (G)##.

This is actually a solved exercise from a Brazilian book on Linear Algebra. The author presented the following solution:

The kernel and image theorem tells us that dimension ##\dim V=n=\dim\ker\left(F\right)+\dim \text{im}\left(F\right)=\dim\ker\left(G\right)+\dim\text{im}\left(G\right)##. As ##\text{im}\left(F\right)\subset R##, ##\dim\mathbb{R}=1## and ##F\neq0##, then ##\dim\text{im}\left(F\right)=1##. Similarly ##\dim\text{im}\left(G\right)=1##. Therefore, ##\dim\ker\left(F\right)=\dim\ker\left(G\right)=n-1##. On the other hand, the dimension theorem of the sum assures us that

$$

\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)+\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\dim\ker\left(F\right)+\dim\ker\left(G\right)=2n-2.

$$

In general, ##\ker\left(G\right)\subset\ker\left(F\right)+\ker\left(G\right)## and due to the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)##, we will have ##\ker\left(F\right)\begin{array}{c}

\subset \\ \neq \end{array}\ker\left(F\right)+\ker\left(G\right)##; then necessarily ##\ker\left(F\right)+\ker\left(G\right)=V##. So ##\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)=n## and hence

$$

\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\left(2n-2\right)=n-2.

$$

I am ok with almost everything he presented, but couldn't understand why

the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)## implies that ##\ker\left(F\right)+\ker\left(G\right)=V.##

Any ideas?

Thanks in advance.

The kernel and image theorem tells us that dimension ##\dim V=n=\dim\ker\left(F\right)+\dim \text{im}\left(F\right)=\dim\ker\left(G\right)+\dim\text{im}\left(G\right)##. As ##\text{im}\left(F\right)\subset R##, ##\dim\mathbb{R}=1## and ##F\neq0##, then ##\dim\text{im}\left(F\right)=1##. Similarly ##\dim\text{im}\left(G\right)=1##. Therefore, ##\dim\ker\left(F\right)=\dim\ker\left(G\right)=n-1##. On the other hand, the dimension theorem of the sum assures us that

$$

\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)+\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\dim\ker\left(F\right)+\dim\ker\left(G\right)=2n-2.

$$

In general, ##\ker\left(G\right)\subset\ker\left(F\right)+\ker\left(G\right)## and due to the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)##, we will have ##\ker\left(F\right)\begin{array}{c}

\subset \\ \neq \end{array}\ker\left(F\right)+\ker\left(G\right)##; then necessarily ##\ker\left(F\right)+\ker\left(G\right)=V##. So ##\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)=n## and hence

$$

\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\left(2n-2\right)=n-2.

$$

I am ok with almost everything he presented, but couldn't understand why

the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)## implies that ##\ker\left(F\right)+\ker\left(G\right)=V.##

Any ideas?

Thanks in advance.

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