1E (Linear Algebra)

Solution

T : V \to W

dim V = dim ker T + dim im T

. Apply it to

β ∣_{im α}

. Its image is

im (β α)

and its kernel is

im α \cap ker β

, so

dim im α = dim im (β α) + dim (im α \cap ker β) .

Since

im (α γ) \subseteq im α

dim (im (α γ) \cap ker β) \leq dim (im α \cap ker β) .

Using the preceding identity for both restrictions and rearranging gives

dim im (β α) + dim im (α γ) \leq dim im α + dim im (β α γ) .

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