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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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    The other relation you need is that the dimensions of kernel and image add up to the dimension of the domain.2012-09-30

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

\begin{align*} \newcommand{ker}{\operatorname{ker}} \newcommand{dim}{\operatorname{dim}} \dim T &= \dim \ker(f_T) + \dim f_T(T) \\ \dim S &= \dim \ker(f_S) + \dim f_S(S) \end{align*}

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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    @celtschk You're right. Fixed.2012-09-30