# If $\dim(V) = 5$ and $\dim(W_1)=3$, and $\dim(W_2)=4$. $W_1\nsubseteq W_2$. Find $\dim(W_1+W_2)$ and $\dim(W_1\cap W_2)$.

Is my solution correct?

Let $V$ be a vector space. Let $W_1$ and $W_2$ be subspace of $V$. Then $W_1+W_2 = \{a+b\mid a\in W_1 \text{ and } b\in W_2\}$. If $\dim(V) = 5$ and $\dim(W_1)=3$, and $\dim(W_2)=4$. $W_1\nsubseteq W_2$. Find $\dim(W_1+W_2)$ and $\dim(W_1\cap W_2)$.

Solution: First, we try to construct the basis of $V$. Since $\dim(V)=5$, then any $5$ linearly independent vectors in V forms a basis. Let $B_v$ be a basis of $V$ such that $B_v=\{v_1,v_2,v_3,v_4,v_5\}$.The basis of $W_1$ be $A=\{a_1,a_2,a_3\}$.and let the basis of $W_2$ be $B=\{b_1,b_2,b_3,b_4\}$. By definition of basis, the $b_i's$ are linearly independent to each other. since $W_1\nsubseteq W_2$, then $\exists a_1\in W_1$ such that $a_1$ is linearly independent to every vector in $W_2$. Now we have $5$ linearly independent vectors in $V$ and thus they form a basis.Thus the basis of V is $\{b_1,b_2,b_3,b_4,a_1\}$.

Since $a_2,a_3\in V$, then they can be written as $a_2=c_1b_1+c_2b_2+c_3b_3+c_4b_4$ and $a_3=d_1b_1+d_2b_2+d_3b_3+d_4b_4$. hence, $a_2,a_3\in W_2$. Thus the $\dim(W_1 \cap W_2) = 2$. and therefore $\dim(W_1+W_2) = 4+3-2=5$.

• This is correct, just one comment: Once you get $\{b_1,b_2,b_3,b_3,a_1\}$ to be a basis for $V$, it automatically follows that $W_1 + W_2 = V$ and so $\dim(W_1+W_2) = 5$ – Prahlad Vaidyanathan Apr 27 '15 at 11:45
• In sort (for one part), using heavily that all relevant spaces are finite-dimensional: Since $W_1\not\subseteq W_2$, certainly $W_1+W_2$ is larger than $W_2$ hence must have dimension $>4$, hence same dimension as $V$, hence must equal $V$. – Hagen von Eitzen Apr 27 '15 at 11:47
• Moreover, to obtain the dimension of $W_1\cap W_2$, you can use the dimensions formula: $$\dim(W_1+W_2)+\dim(W_1\cap W_2)=\dim W_1+\dim W_2.$$ – Bernard Apr 27 '15 at 11:50