Question

if the finite dimensional vector space V(F) be the direct sum of it's two subspaces W1 and W2 then show that dim. V= dim.W1 + dim. W2
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Answer 1

Answer:

Let $U_1$ and $U_2$ be bases of $W_1$ and $W_2$. Then $B = U_1 \cup U_2$ is a basis of $V$.

In fact, it is easy to see that $B$ is spanning $V$. Let's show it's linear independent. Take

$\alpha_1a_1+...+ \alpha_na_n+\beta_1b_1+\dots \beta_mb_m = 0$

where $a_i \in U_1$ and $b_i \in U_2$.

Since, by hipothesis, $W_1 \cap W_2 = \{0\}$, we must have

$\alpha_1a_1+...+ \alpha_na_n = \beta_1b_1+\dots \beta_mb_m = 0$

But once both the sets $(a_i)$ and $(b_j)$ are linear independent (they form bases), we must have $\alpha_i = \beta_j = 0$ for every $i,j.$

Then $B$ is linear independent and is a basis of $V$.

Therefore

$dim_V = |B| = |U_1\cup U_2| = |U_1|+|U_2| = dim_{W_1}+dim_{W_2}$