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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.

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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.

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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 $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$

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