Proving Dimensions Inequality: Linear Algebra

Question

Given a vector space $W$ and subspaces $V$, $V'$ and $V''$, such that $W = V\oplus V' = V\oplus V''$.

Prove: $$\dim(V'\cap V'')\geq \dim(W) - 2\cdot \dim(V)$$

Answer 1

From the direct sums, we have \begin{align*} \text{dim}(W)&=\text{dim}(V)+\text{dim}(V')\\ \text{dim}(W)&=\text{dim}(V)+\text{dim}(V'') \end{align*} Adding these, we get $$2\,\text{dim}(W)=2\,\text{dim}(V)+\color{blue}{\text{dim}(V')+\text{dim}(V'')}.$$ Are you familiar with $\text{dim}(V'+V'')+\text{dim}(V' \cap V'')=\text{dim}(V')+\text{dim}(V'')$? (you may want to check out the result here)

If yes, then we get, \begin{align*} 2\,\text{dim}(W)& =2\,\text{dim}(V)+\color{blue}{\text{dim}(V'+V'')+\text{dim}(V' \cap V'')} \end{align*} Since $V'+V''$ is a subspace of $W$, therefore $\text{dim}(V'+V'') \leq \text{dim}(W)$. We get \begin{align*} 2\,\text{dim}(W)& \leq 2\,\text{dim}(V)+\text{dim}(W)+\text{dim}(V' \cap V'')\\ \text{dim}(W)- 2\,\text{dim}(V)& \leq \text{dim}(V' \cap V'')\\ \end{align*}