Prove that if $W_1$ and $W_2$ are finite-dimensional subspaces of a vector space V, then the subspace $W_1+W_2$ is finite-dimensional, and $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$.
Proof: let $\{a_1,a_2,…,a_i,v_1,v_2…v_n\}$ be basis for $W_1$, then $\dim(W_1)= i+n$ and $\{b_1,b_2,…,b_j,v_1,v_2…v_n\}$ be basis for $W_2$, then $\dim(W_2)= j+n$ . Since both of them are finite-dimensional, and the definition of sum of two subspaces tells us that the basis of the sum is a combination of those two subspaces, which is $\dim(W_1+W_2)= i +j+n$. Hence we can arrive that $W_1+W_2$ is finite-dimensional.
To prove $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$. Since the basis of the sum of two subspaces is a combination of both subspaces, $\dim(W_1+W_2) = i +j+n$. Since the both subspaces have n elements in common, so $\dim(W_1 \cap W_2)= n$.
Hence $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ is $ i +j+n = (i +n) + (j+n)- n$. Q.E.D
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Best Answer
Some problematic points in your proof:
You are missing some fundamental information. What are $i,j,n$? Why do both bases contain the same vectors $v_1,\dots,v_n$?
Presumably, you mean "a combination of those two bases". In any case, the term "a combination of" is too vague for this statement to be correct.
which is $\dim(W_1+W_2)= i +j+n$. Hence we can arrive that $W_1+W_2$ is finite-dimensional.
It is not true that the two subspace have $n$ elements in common. If we're talking about vector spaces over $\Bbb R$ or $\Bbb C$, then the subspaces should have either infinitely many elements or one element in common.
A correct proof, in which I have attempted to parallel yours as much as possible.
Let $v_1,\dots,v_n$ be a basis of $W_1 \cap W_2$. Since $W_1 \cap W_2 \subseteq W_1$, we can extend this to a basis $v_1,\dots,v_n,a_1,\dots,a_i$ of $W_1$. Similarly, let $v_1,\dots,v_n,b_1,\dots,b_j$ be a basis of $W_2$. It is clear that the union of these bases, $$ \mathcal B = \{v_1,\dots,v_n,a_1,\dots,a_i,b_1,\dots,b_j\} $$ is a spanning set of $W_1 + W_2$. In order to show that this is a basis, we must also show that $\mathcal B$ is linearly independent.
One we have proven the claim that $\mathcal B$ is indeed a basis, we may simply count the elements of each basis to find $$ \dim(W_1 \cap W_2) = n, \quad \dim(W_1) = n+i, \quad \\ \dim(W_2) = n+j, \quad \dim(W_1 + W_2) = n+i+j. $$ We can then verify the desired result by plugging these in to the desired equation.