- #1

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- TL;DR Summary
- Let ##F## and ##G## be non-zero linear forms in vector space ##V##, linearly dependent. Prove that ##\ker\left(F\right)=\ker\left(G\right)## and that its dimension is ##n-1## if ##\dim V=n##.

Attempt of a solution.

By the Rank–nullity theorem,

$$

\dim V=\dim Im_{F}+\dim\ker\left(F\right)

\Rightarrow n=1+\dim\ker\left(F\right)

\Rightarrow \dim\ker\left(F\right)=n-1.

$$

Similarly, it follows that $$\dim\ker\left(G\right)=n-1.$$

This first part, for obvious reasons, is very clear.

The next one offered me a terrible challenge.

I saw a solution, but it didn't convince me for sure.

Here it is.

As ##F## and ##G## are linearly dependent,

$$

\alpha F\left(u\right)+\beta G\left(u\right)=0

\Rightarrow G\left(u\right)=\frac{-\alpha}{\beta}F\left(u\right),

$$

since in this expression, ##\alpha## and ##\beta## have to be different from 0, because the two linear forms are non-zero. It follows that every vector in the image of ##F## is also a vector in the image of ##G##, which implies that ##Im_{F}=Im_{G}##. As for any subspace ##V##, ##V=\ker\left(F\right) \oplus Im_{F}##, where ##Im_{F}=Im_{G}##, then

$$

\ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{G}

\Rightarrow \ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{F}

\Rightarrow \ker\left(F\right)=\ker\left(G\right),

$$

as there is no intersection between the two sets.

I would like to know if this solution is correct and if is the best one, because it seems not elegant for me.

By the Rank–nullity theorem,

$$

\dim V=\dim Im_{F}+\dim\ker\left(F\right)

\Rightarrow n=1+\dim\ker\left(F\right)

\Rightarrow \dim\ker\left(F\right)=n-1.

$$

Similarly, it follows that $$\dim\ker\left(G\right)=n-1.$$

This first part, for obvious reasons, is very clear.

The next one offered me a terrible challenge.

I saw a solution, but it didn't convince me for sure.

Here it is.

As ##F## and ##G## are linearly dependent,

$$

\alpha F\left(u\right)+\beta G\left(u\right)=0

\Rightarrow G\left(u\right)=\frac{-\alpha}{\beta}F\left(u\right),

$$

since in this expression, ##\alpha## and ##\beta## have to be different from 0, because the two linear forms are non-zero. It follows that every vector in the image of ##F## is also a vector in the image of ##G##, which implies that ##Im_{F}=Im_{G}##. As for any subspace ##V##, ##V=\ker\left(F\right) \oplus Im_{F}##, where ##Im_{F}=Im_{G}##, then

$$

\ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{G}

\Rightarrow \ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{F}

\Rightarrow \ker\left(F\right)=\ker\left(G\right),

$$

as there is no intersection between the two sets.

I would like to know if this solution is correct and if is the best one, because it seems not elegant for me.

Last edited: