Question

In R

4

let L

1

​

be the subspace spaned by the vectors a=(1;2;5;1), b=(4;3;3;0), c=(7;4;1;−1), and let L

2

​

be the subspace spaned by the vectors d=(1;1;1;1), f=(−1;0;3;−1) and g=(5;2;−1;−3). Find the dimension of L

1

​

∩L

2

​

.

Answer 1

$dim (L_1\cap L_2)=dimL_1+dimL_2-dim(L_1\cup L_2).$ We can see that $dimL_1=2$ , because the rank of the matrix composed of the coordinates of the vectors $a,b,c$ equals to 2:

$rank \begin{pmatrix} 1 & 2 &5& 1\\ 4 & 3 &3&0\\ 7& 4& 1& -1\\ \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -5 &-17&-4\\ 0& -10& -34& -8 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -5 &-17&-4 \end{pmatrix}=2$ .

We can see that $dimL_2=3$ , because

$rank \begin{pmatrix} 1 & 1 &1& 1\\ -1 & 0 &3&-1\\ 5& 2& -1& -3 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 1 &1& 1\\ 0 & 1 &4&0\\ 0& -3& -6& -8 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 1 &1& 1\\ 0 & 1 &4&0\\ 0& 0& 6& -8 \end{pmatrix}=3$

If we want find the $dim(L_1\cup L_2)$ , we can find the rank of a matrix composed of vectors $a,b,c,d,f,g$ :

$rank \begin{pmatrix} 1 & 2 &5& 1\\ 4 & 3 &3&0\\ 7& 4& 1& -1\\ 1 & 1 &1& 1\\ -1 & 0 &3&-1\\ 5& 2& -1& -3 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -5 &-17&-4\\ 0& -10& -34& -8 \\ 0 & -1 &-4& 0\\ 0 &2 &8&0\\ 0& -8& -26& -8 \end{pmatrix}=$  $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -1 &-4& 0\\ 0 & -5 &-17&-4\\ 0& -8& -26& -8 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -1 &-4& 0\\ 0 & 5 &17&4\\ 0& 4& 13& 4 \end{pmatrix}=$ $rank \begin{pmatrix} 1 & 2 &5& 1\\ 0 & -1 &-4& 0\\ 0 & 0 &-3&4 \end{pmatrix}=3$

So $dim(L_1\cup L_2)=3$. It follows that  $dim (L_1\cap L_2)=2+3-3=2$ .