[Math] (Ostensible Formula) Dimension of sum of three subspaces

linear algebra

I'm trying to show
$$ \mathrm{dim}(U_1 + U_2 + U_3) = \mathrm{dim}(U_1) + \mathrm{dim}(U_2) + \mathrm{dim}(U_3) -\mathrm{dim}(U_1 \cap U_2) -\mathrm{dim}(U_1\cap U_3) -\mathrm{dim}(U_2 \cap U_3) + \mathrm{dim}(U_1 \cap U_2 \cap U_3)$$

for $U_i$ subspaces of $V$. I tried to do it by extending a basis

$ u_1,…,u_n$ of $U_1 \cap U_2 \cap U_3$ to bases

$u_1,…,u_n,v_1,…,v_m$ of $U_1 \cap U_2$

$u_1,…,u_n, w_1,…,w_k$ of $u_2 \cap U_3$

$u_1,…,u_n,x_1,…,x_j$ of $U_1 \cap U_3$

and then extending these to bases

$u_1,…,u_n,v_1,…,v_m,y_1,…,y_i$ of $U_1$

$u_1,…,u_n, w_1,…,w_k,z_1,…,z_h$ of $U_2$

$u_1,…,u_n,x_1,…,x_j, \eta_1,….,\eta_l$ of $U_3$.

Then I calculated that the RHS of the equation equals $i+h+l+n$. But I can't figure out how to arrive at the same sum for $ \mathrm{dim}(U_1 + U_2 + U_3)$. How can I finish this proof?

Best Answer

he statement seems to be wrong, it is only true for two subspaces. See https://mathoverflow.net/a/23501/38249

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