# How to find $\dim( \operatorname{span}(M)+ \operatorname{span}(N))$ and $\dim(\operatorname{span}(M\cap N))$?

$$M=\{(1,2,1),(3,1,5),(3,-4,7)\}$$, $$N=\{(1,0,-1),(1,2,1)\}$$

Let $$A= \operatorname{span}(M)$$ and $$B= \operatorname{span}(N)$$

I want to know how to find $$\;(a)\;\dim(A+B)$$ and $$(b)\;\dim(A\cap B)$$

$$\begin{bmatrix} 1 & 3 & 3 \\ 2 & 1 & -4 \\ 1 & 5 & 7 \end{bmatrix}$$$$\sim$$ $$\begin{bmatrix} 1 & 3 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$ Since, only the first two vectors are linearly independent, we have $$\dim(A)=2$$

Clearly, $$\dim(B)=2$$ since it is a linearly independent set(They are not multiple of each other).

I want to make use of:

$$\dim(A+B)= \dim(A)+\dim(B)-\dim(A\cap B)$$

So how do I find either of $$\dim(A+B)$$ or $$\dim(A\cap B)$$ to get the other ?

Does making a single matrix with the vectors in $$M$$ and $$N$$ and calculating its rank help?

$$\begin{bmatrix} 1 & 3 & 3 & 1 &1\\ 2 & 1 & -4&0 &2 \\ 1 & 5 & 7 &-1&1 \end{bmatrix}$$ $$\sim$$ $$\begin{bmatrix} 1 & 3 & 3 & 1 &1\\ 0 & -5 & -10&-2 &0 \\ 0 & 0 & 0 &14&0 \end{bmatrix}$$ So this gives a rank of 3

• @DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2? Commented Jan 27, 2019 at 11:40
• Yes, it does. But where is this written? "So, $\dim(A)=2$" is a very poor proof. Commented Jan 27, 2019 at 11:40
• I have written the matrices in my post Commented Jan 27, 2019 at 11:41
• You have written the matrices, yes, but no word about why the rank is not full. Commented Jan 27, 2019 at 11:42
• To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before. Commented Jan 27, 2019 at 11:42