1. ## Subspaces take 2.

Let $\displaystyle V=\mathbb{R}^4$. Suppose further that $\displaystyle W_1$ is a subsapce of $\displaystyle V$ spanned by vectors $\displaystyle (1,2,0,1)$ and $\displaystyle (1,1,1,0)$, and $\displaystyle W_2$ is a subspace of $\displaystyle V$ spanned by $\displaystyle (2,3,1,1)$.

Determine $\displaystyle dim(W_1+W_2)$ and $\displaystyle dim(W_1 \cap W_2)$.
My other thread might be useful: http://www.mathhelpforum.com/math-he...subspaces.html

Sifting $\displaystyle (1,2,0,1)$ and $\displaystyle (1,1,1,0)$ gives those two vectors. Hence $\displaystyle dim(W_1)=2$.

Similarly $\displaystyle dim(W_2)=1$.

Since $\displaystyle (1,2,0,1)+(1,1,1,0)=(2,3,1,1)$ we know that $\displaystyle W_2$ is a subspace of $\displaystyle W_1$.

Therefore $\displaystyle dim(W_1 \cap W_2)=1$ and $\displaystyle dim(W_1+W_2)=2$.

Is this it? This question was worth 10 marks but my answer is remarkably small!

2. Originally Posted by Showcase_22
My other thread might be useful: http://www.mathhelpforum.com/math-he...subspaces.html

Sifting $\displaystyle (1,2,0,1)$ and $\displaystyle (1,1,1,0)$ gives those two vectors. Hence $\displaystyle dim(W_1)=2$.

Similarly $\displaystyle dim(W_2)=1$.

Since $\displaystyle (1,2,0,1)+(1,1,1,0)=(2,3,1,1)$ we know that $\displaystyle W_2$ is a subspace of $\displaystyle W_1$.

Therefore $\displaystyle dim(W_1 \cap W_2)=1$ and $\displaystyle dim(W_1+W_2)=2$.

Is this it? This question was worth 10 marks but my answer is remarkably small!