NUS-MATH530 2.C Problem 17
Claim Proof or give a counter example for the statement that:
\begin{align} \dim\left(U_1+U_2+U_3\right) = &\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 = \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\}\\ U_2 = \left\{\mqty(0 \\ b): b \in \mathbb{F}\right\}\\ U_3 = \left\{\mqty(c \\ c): c \in \mathbb{F}\right\} \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 \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\} \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 \left\{\mqty(c \\ c): c \in \mathbb{F}\right\} \end{equation}
scalar multiplication For U_1 —
\begin{equation} \lambda \mqty(a \\ 0) = \mqty(\lambda a \\ 0) \in \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\} \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 \left\{\mqty(c \\ c): c \in \mathbb{F}\right\} \end{equation}
constructing the counterexample Let us calculate the value of both sides of:
\begin{align} \dim\left(U_1+U_2+U_3\right) = &\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 = \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\}\\ U_2 = \left\{\mqty(0 \\ b): b \in \mathbb{F}\right\}\\ U_3 = \left\{\mqty(c \\ c): c \in \mathbb{F}\right\} \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 = \left\{u_1+u_2+u_3: u_j\in U_j\right\} \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 \left(U_1+U_2+U_3\right)=2
right side dimension of the subspaces
Let us construct a basis for each of these spaces to figure their dimension.
For U_1, \left\{\mqty(1 \\ 0)\right\}. 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, \left\{\mqty(0 \\ 1)\right\}. This makes also \dim U_2=1.
For U_3, we have \left\{\mqty(1 \\ 1)\right\}. 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 = \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\} \cap \left\{\mqty(0 \\ b): b \in \mathbb{F}\right\} = \left\{\mqty(0 \\ 0)\right\} \end{equation}
This is because a=0, b=0 respectively in order to satisfy both generators.
Similarly
\begin{equation} U_1 \cap U_3 = \left\{\mqty(a \\ 0): a \in \mathbb{F}\right\} \cap \left\{\mqty(c \\ c): c \in \mathbb{F}\right\} = \left\{\mqty(0 \\ 0)\right\} \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 = \left\{\mqty(0 \\ 0)\right\} \end{equation}
We established before that the span of \left\{\right\} (which is declared linearly independent) to be \left\{0\right\}, 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 \left\{1,2,3\right\} \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