Intereting Posts

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Graph-theory exercise

I was thinking about the proof of the rank-nullity theorem and I thought about proving it as follows. I just wondered whether this proof worked?

**Lemma.** If $V$ is a finite-dimensional $F$-vector space and $U\leq V$, then $V/U$ is finite dimensional.

$\hspace{16.5mm}$ Moreover, we have that $\dim{V/U}=\dim{V}-\dim{U}$.

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- Gram matrix invertible iff set of vectors linearly independent

**Theorem (Rank-Nullity).** If $\alpha:V\to W$ is linear with $V$ finite-dimensional, then

$$\dim{V}=\dim(\text{im }\alpha)+\dim(\ker \alpha)$$

**Proof.** By the first isomorphism theorem we have

$$V/\ker{\alpha} \cong \text{im }{\alpha}.$$

Taking dimensions and applying the lemma we get

$$\dim V – \dim(\ker\alpha)=\dim(\text{im }\alpha)$$

which on rearrangement yields the result. //

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