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--- title: "NUS-MATH530 2.C Problem 17" source: https://www.jemoka.com/posts/kbhnus_math530_2_c_problem_17/ --- Claim Proof or give a counter example for the statement that: \begin{align} \dim\qty(U_1+U_2+U_3) = &\dim U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3) \end{align} Counterexample This statement is false. Take the following three subspaces of \(\mathbb{F}^{2}\): \begin{align} U_1 = \qty{\mqty(a \\ 0): a \in \mathbb{F}}\\ U_2 = \qty{\mqty(0 \\ b): b \in \mathbb{F}}\\ U_3 = \qty{\mqty(c \\ c): c \in \mathbb{F}} \end{align} subspace check All \(U_1\), \(U_2\), \(U_3\) are in \(\mathbb{F}^{2}\). zero Zero exists in all by setting free variables to \(0\) addition For \(U_1\) — \begin{equation} \mqty(a_1 \\ 0) + \mqty(a_2 \\ 0) = \mqty(a_1+a_2 \\ 0) \in \qty{\mqty(a \\ 0): a \in \mathbb{F}} \end{equation} and, by the same token, addition is closed for \(U_2\). For \(U_3\) — \begin{equation} \mqty(c_1 \\ c_1) + \mqty(c_2 \\ c_2) = \mqty(c_1+c_2 \\ c_1+c_2) \in \qty{\mqty(c \\ c): c \in \mathbb{F}} \end{equation} scalar multiplication For \(U_1\) — \begin{equation} \lambda \mqty(a \\ 0) = \mqty(\lambda a \\ 0) \in \qty{\mqty(a \\ 0): a \in \mathbb{F}} \end{equation} and, by the same token, scalar multiplication is closed for \(U_2\). For \(U_3\) — \begin{equation} \lambda \mqty(c \\ c) = \mqty(\lambda c \\ \lambda c) \in \qty{\mqty(c \\ c): c \in \mathbb{F}} \end{equation} constructing the counterexample Let us calculate the value of both sides of: \begin{align} \dim\qty(U_1+U_2+U_3) = &\dim U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3) \end{align} Recall that: \begin{align} U_1 = \qty{\mqty(a \\ 0): a \in \mathbb{F}}\\ U_2 = \qty{\mqty(0 \\ b): b \in \mathbb{F}}\\ U_3 = \qty{\mqty(c \\ c): c \in \mathbb{F}} \end{align} left side Let’s first construct: \begin{equation} U_1 + U_2 + U_3 \end{equation} By definition: \begin{equation} U_1 + U_2 + U_3 = \qty{u_1+u_2+u_3: u_j\in U_j} \end{equation} Therefore, taking a sample from each results as: \begin{equation} u_1+u_2+u_3 = \mqty(a \\ 0) + \mqty(0 \\ b) + \mqty(c \\c) = \mqty(a+c \\ b +c) \end{equation} This creates two free variables for slots, meaning: \begin{equation} U_1+U_2+U_3 = \mathbb{F}^{2} \end{equation} So: \(\dim \qty(U_1+U_2+U_3)=2\) right side dimension of the subspaces Let us construct a basis for each of these spaces to figure their dimension. For \(U_1\), \(\qty{\mqty(1 \\ 0)}\). We see that scaling the one vector in this basis will construct all vectors in \(\mathbb{F}^{2}\) for which the second coordinate will be \(0\) — spanning \(U_1\). Being a list with one non-zero vector, it is also linearly independent. So \(\dim U_1 = 1\). By almost the same token, for \(U_2\), \(\qty{\mqty(0 \\ 1)}\). This makes also \(\dim U_2=1\). For \(U_3\), we have \(\qty{\mqty(1 \\ 1)}\). Scaling this one vector will construct all vectors in \(\mathbb{F}^{2}\) for which both coordinates are the same — spanning \(U_3\). Being a list with one non-zero vector, it is also linearly independent. So \(\dim U_3 = 1\). This renders all three subspaces have dimension \(1\). dimension of the unions These subspaces were picked because of a surprising convenience. Their unions are all the zero vector! \begin{equation} U_1 \cap U_2 = \qty{\mqty(a \\ 0): a \in \mathbb{F}} \cap \qty{\mqty(0 \\ b): b \in \mathbb{F}} = \qty{\mqty(0 \\ 0)} \end{equation} This is because \(a=0\), \(b=0\) respectively in order to satisfy both generators. Similarly \begin{equation} U_1 \cap U_3 = \qty{\mqty(a \\ 0): a \in \mathbb{F}} \cap \qty{\mqty(c \\ c): c \in \mathbb{F}} = \qty{\mqty(0 \\ 0)} \end{equation} To satisfy both generators, \(a=c\) for the top coordinate, \(c=0\) for the bottom coordinate, so \(a=c=0\). By a similar token: \begin{equation} U_2 \cap U_3 = \qty{\mqty(0 \\ 0)} \end{equation} We established before that the span of \(\qty{}\) (which is declared linearly independent) to be \(\qty{0}\), so we see that the dimensions of all three required unions as \(0\) (as an empty list has length \(0\).) constructing the expression for the right side We have that: \begin{equation} \dim U_j = 1, j \in \qty{1,2,3} \end{equation} And that: \begin{equation} \dim U_{j} \cap U_{k} = 0 , j,k \in \{1,2,3\} \end{equation} from above. This makes— \begin{align} \dim &U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3)\\ =1&+1+1-0-0-0+0 \\ =3 \end{align} showing the counterexample We have now that: \begin{equation} \dim(U_1+U_2+U_3) = 2 \end{equation} But: \begin{align} \dim &U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3)\\ =3 \end{align} Yet \(2 \neq 3\). So: \begin{align} \dim(U_1+U_2+U_3) \neq \dim &U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3)\\ \end{align} Finishing the counter example. \(\blacksquare\)