# Proving statement about dimensions of vector spaces

While studying for my linear algebra test I came across the following problem:

Let $f: \mathbb{V} \to \mathbb{W}$ be a linear transformation and let $S$ and $T$ subspaces of $\mathbb{V}$ such that $S \cap T = \text{Ker}(f)$. Show that

$$\dim(S+T) = \dim(\text{Ker}(f)) + \dim(f(S)) + \dim(f(T)).$$

I don't really know how to approach this. There's a good change I have to use that $\dim(S+T) = \dim(S) + \dim(T) - \dim(S \cap T)$, but all I can do with that is transform this into $\dim(f(S)) + \dim(f(T)) = \dim(S) + \dim(T)$. I think the problem here is that I don't know anything about $f(S)$ or $f(T)$. How would one go about proving this?

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Do you mean $S\cap T\subset ker(f)$? –  Shahab Sep 30 '12 at 14:23
@Shahab: fixed it. –  Javier Badia Sep 30 '12 at 14:26
The other relation you need is that the dimensions of kernel and image add up to the dimension of the domain. –  celtschk Sep 30 '12 at 14:41

Let $f_T$ be the restriction of $f$ on $T$. Likewise for $f_S$. Using the rank-nullity theorem:


It's clear that: $$f_T(T) = f(T) \\ f_S(S) = f(S)$$

Since $\ker(f) = T \cap S$, $\ker(f) \subset T$ and $\ker(f) \subset S$. Therefore: \begin{align*} \ker(f_T) = \ker(f) \\ \ker(f_S) = \ker(f) \end{align*}

Hence: \begin{align*} \dim T &= \dim \ker(f) + \dim f(T) \\ \dim S &= \dim \ker(f) + \dim f(S) \end{align*}

Adding side by side, we get: $$\dim T + \dim S = 2 \dim \ker(f) + \dim f(T) + \dim f(S)$$

By rearranging, we get: $$\dim T + \dim S - \dim \ker(f) = \dim \ker(f) + \dim f(T) + \dim f(S)$$

Since $\ker(f) = T \cap S$, we conclude: $$\dim (S + T) = \dim \ker(f) + \dim f(T) + \dim f(S)$$

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You mean $\dim S+T$. $S\cup T$ will in general not be a subspace. –  celtschk Sep 30 '12 at 14:43
@celtschk You're right. Fixed. –  Ayman Hourieh Sep 30 '12 at 14:44