Prove that: $\dim(\ker(T+S)) \leq \dim(\ker(T) \cap \ker(S)) + \dim(\operatorname{Im}(T)\cap \operatorname{Im}(S))$ I have the following inequality I would like to prove the following linear algebra inequality.
Let $T,S : V \rightarrow W$ be two linear transformations. Prove that:
$\dim(\ker(T+S)) \leq \dim(\ker(T) \cap \ker(S)) + \dim(\operatorname{Im}(T)\cap \operatorname{Im}(S))$
I have started by taking arbitrary elements in each set as shown below, but not sure how to continue.
Let $x \in \ker(T+S)$. Then $(T+S)(x) = 0 \Leftrightarrow T(x)+S(x) = 0 \Leftrightarrow T(x) = -S(x)$
Let $y \in \ker(T \cap S)$. Then $T(y) = 0, S(y) = 0 \Leftrightarrow T(y) = S(y) = 0$
I want to now connect these two, to work towards the inequality but not entirely sure how. Any advice, how to proceed from here?
 A: Since $\ker T$ $\cap$ $\ker S \subseteq \ker S + T$, we can take a basis $(v_1,...,v_k)$ for $\ker T$ $\cap$ $\ker S$ and extend this to a basis $(v_1,...,v_k,w_1,...,w_m)$ for $\ker S + T$.
Now the $T(w_i)$ are independent. For we have
$$\sum_{i=1}^m c_iT(w_i) = 0 \iff T\left(\sum_{i=1}^m c_iw_i\right) = 0$$
And since $T(w_i) = -S(w_i)$, we also have
$$S\left(\sum_{i=1}^m c_iw_i\right) = 0$$
implying that $\sum_{i=1}^m c_iw_i \in \ker T$ $\cap$ $\ker S$. Thus $\sum_{i=1}^m c_iw_i = 0$ and so $c_i = 0 \ \forall i$.
This means that there are (at least) $m$ independent vectors in $\text{im }(T) \cap \text{im }(S)$. So $m \leq \dim\left(\text{im }(T) \cap \text{im }(S)\right)$ and we are done.
A: Let $X,Y,Z$ be subspaces of $V$ such that :
\begin{align}
\ker S &= \ker S\cap \ker T \oplus X \\
\ker T& = \ker S\cap \ker T \oplus Y \\
V &= \ker S \cap \ker T \oplus X \oplus Y \oplus Z
\end{align}
Let $v \in V$, which we write : $v = k + x + y +z$ with $k \in \ker S \cap \ker T$, $x\in X$, etc.
Then $v\in \ker (S+T)$ if, and only if :
\begin{align}
(S+T)(x) = 0 &\Longleftrightarrow S(y+z) + T(x+z) = 0 \qquad (*)
\end{align}
To finish the proof cleanly, we notice that we can define a linear map :
$$f \left\{\begin{array}{ccc}\ker (S+T) &\longrightarrow& \ker S \cap \ker T \times \operatorname{Im} S \cap\operatorname{Im}  T \\
v& \longmapsto& (k,T(x+z))
\end{array}\right.$$
$f$ is well defined by $(*)$ ($T(x+z)$ is also in $\operatorname{Im}  S$) and since $S|_{Y\oplus Z}$ and $T|_{X\oplus Z}$ are injective, we see that $f$ is injective. Therefore :
$$\dim (S+T) \leq \dim (\ker S \cap \ker T) + \dim (\operatorname{Im} S \cap \operatorname{Im} T)$$
