sum of subsets — Jemoka Knowledge Base

The sum of subsets is the definition of addition upon two subsets. Apparently, the unions of subsets are almost never subspaces (they don’t produce linearity?) Therefore, we like to work with sum of subsets more. Remember this has arbitrarily many things!! as a part of the content. When defining, remember to open that possibility. constituents Sub-sets of V named U_1, U_2, \dots, U_{m} requirements The sum of subsets U_1, \dots, U_{m} is defined as:

\begin{equation} U_1, \dots, U_{m} = \{u_1+\dots+u_{m}: u_1\in U_1, \dots, u_{m} \in U_{m}\} \end{equation}

“all elements formed by taking one element from each and add it.” additional information sum of subspaces is the smallest subspace with both subspaces Suppose U_1, \dots U_{m} are subspaces of V, then U_1+\dots +U_{m} is the smallest subspace of V containing U_1, \dots, U_{m}. Proof: Is a subspace— clearly 0 is in the sum. (taking 0 from each subspace and adding) addition and scalar multiplication inherits (closed in each subspace, then, reapplying definition of sum of subsets) Smallest containing subspace— Because a subspace is closed under addition, if a subspace contains U_{1}, \dots, U_{m} you can always add each of the constituent elements manually to form every U_1+\dots+U_{m}. Conversely, the subspace U_1+\dots +U_{m} should contain U_1, \dots, U_{m} by simply setting the coefficients except for the one you are interested in to 0. Therefore, as both subsets contain each other; they are equivalent. dimension of sums Let there be two finite-dimensional subspaces: U_1 and U_2. Then:

\begin{equation} \dim(U_1+U_2)=\dim U_1+\dim U_{2} - \dim(U_1 \cap U_2) \end{equation}

Proof: let us form an basis of U_1 \cap U_{2}: u_1, \dots u_{m}; this indicates to us that \dim(U_1 \cap U_{2}) = m. Being a basis of U_1 \cap U_{2}, it is linearly independent in U_1 (which forms a part of the intersection. As any linearly independent list (in this case, in U_1) can be expanded into a basis of U_1. Let’s say by some vectors v_1 \dots v_{j}. Therefore, we have that: The new basis is u_1, \dots u_{m}, v_1, \dots v_{m}, and so:

\begin{equation} \dim U_1 = m+j \end{equation}

By the same token, let’s just say some w_1, \dots w_{k} can be used to extend u_1, \dots u_{m} into a basis of U_2 (as u_1, \dots u_{m} is also an linearly independent list in U_2). So:

\begin{equation} \dim U_{2} = m+k \end{equation}

We desire that \dim(U_1+U_2)=\dim U_1+\dim U_{2} - \dim(U_1 \cap U_2). Having constructed all three of the elements, we desire to find a list that is length (m+j)+(m+k)-m = m+j+k that forms a basis of U_1+U_2, which will complete the proof. Conveniently, u_1, \dots u_{m}, v_1, \dots v_{j}, w_1, \dots w_{k} nicely is list of length m+j+k. Therefore, we desire that that list forms a basis of U_1+U_{2}. As pairwise in this list are the basis of U_1 and U_2, this list can span both U_1 and U_2 (just zero out the “other” sublist—zero w if desiring a basis of U_1, v if U_2 —and you have a basis of each space. As U_1+U_2 requires plucking a member from each and adding, as this list spans U_1 and U_2 separately (again, it forms the basis of the each space), we can just use this list to construct individually each component of U_1+U_2 then adding it together. Hence, that long combo list spans U_1+U_2. The only thing left is to show that the giant list there is linearly independent. Let’s construct:

\begin{equation} a_1u_1+ \dots + a_{m}u_{m} + b_1v_1 + \dots + b_{j}v_{j} + c_1w_1 + \dots + c_{k}w_{k} = 0 \end{equation}

to demonstrate linearly independence, Moving the w to the right, we have that:

\begin{equation} a_1u_1+ \dots + a_{m}u_{m} + b_1v_1 + \dots + b_{j}v_{j} =-(c_1w_1 + \dots + c_{k}w_{k}) \end{equation}

Recall that u_1 \dots v_{j} are all vectors in U_1. Having written -(c_1w_1 + \dots + c_{k}w_{k}) as a linear combination thereof, we say that -(c_1w_1 + \dots + c_{k}w_{k}) \in U_1 due to closure. But also, w_1 \dots w_{k} \in U_2 as they form a basis of U_2. Hence, -(c_1w_1 + \dots + c_{k}w_{k}) \in U_2. So, -(c_1w_1 + \dots + c_{k}w_{k}) \in U_1 \cap U_2. And we said that u_1, \dots u_{m} are a basis for U_1 \cap U_{2}. Therefore, we can write the c_{i} sums as a linear combination of u:

\begin{equation} d_1u_1 \dots + \dots + d_{m}u_{m} = (c_1w_1 + \dots + c_{k}w_{k}) \end{equation}

Now, moving the right to the left again:

\begin{equation} d_1u_1 \dots + \dots + d_{m}u_{m} - (c_1w_1 + \dots + c_{k}w_{k}) = 0 \end{equation}

We have established before that u_1 \dots w_{k} is a linearly independent list (it is the basis of U_2.) So, to write 0, d_1 = \dots = c_{k} = 0. Substituting back to the original:

\begin{align} a_1u_1+ \dots + a_{m}u_{m} + b_1v_1 + \dots + b_{j}v_{j} &=-(c_1w_1 + \dots + c_{k}w_{k}) \\ &= 0 \end{align}

recall u_1 \dots v_{j} is the basis of U_1, meaning they are linearly independent. The above expression makes a_1 = \dots b_{j} = 0. Having shown that, to write 0 via u, v, \dots w requires all scalars a,b,c=0, the list is linearly independent. Having shown that the list of u_1, \dots v_1, \dots w_1 \dots w_{k} spans U_1+U_2 and is linearly independent within it, it is a basis. It does indeed have length m+j+k, completing the proof. \blacksquare