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. \(\blacksquare\)