What is the dimension of two subspaces? Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and 
$$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\
V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$
Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$? 
I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is
$$
\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \} 
$$
and $V_2$ is
$$
\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.
$$
By solving the collection of the equations for $V_1$ and $V_2$, I obtain 
the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct?
Thank you very much.
 A: Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$:
$$\left(
\begin{array}{ccccccccc}
1&1&1&0&0&0&0&0&0\\
0&0&0&1&1&1&0&0&0\\
0&0&0&0&0&0&1&1&1\\
\hline
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1
\end{array}\right)$$
By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain
$$\left(
\begin{array}{ccccccccc}
\mathbf{1}&0&0&1&0&0&1&0&0\\
0&\mathbf{1}&0&0&1&0&0&1&0\\
0&0&\mathbf{1}&0&0&1&0&0&1\\
0&0&0&\mathbf{1}&1&1&0&0&0\\
0&0&0&0&0&0&\mathbf{1}&1&1\\
0&0&0&0&0&0&0&0&0
\end{array}\right)$$
Hence $\dim(V_1+V_2)=5$ and
$$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$
A: It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then


*

*Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$

*Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$.


Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional.
