ALG1_L14

Direct Sums

External direct sums

Definition 1(Definition).

Let $V$ and $W$ be two vector spaces. Define the external direct sum like so:

$$V\oplus W\equiv \{(v,w)\mid v\in V,w\in W\}=V\times W.$$

We define the following operations, which make $V\oplus W$ a vector space.

$$\begin{array}{rlrl}\text{Addition}& & (v_1,w_1)+(v_2,w_2)& =(v_1+v_2,w_1+w_2)\\ \text{Scalar Multiplication}& & c(v,w)& =(cv,cw)\end{array}$$

Notice that $V\oplus W$ contains a copy of $V$ and $W$, i.e. there exists an injection from $V$ and $W$ into $V\oplus W$. $V$ and $W$ are NOT subspaces of $V\oplus W$.

Theorem 2(Lemma).

$\dim (V\oplus W)=\dim V+\dim W$.

Proof:
Let a basis for $V$ be $\{{\mathbf{v}}_i{\}}_{i=1}^n$ and a basis for $W$ be $\{{\mathbf{w}}_j{\}}_{j=1}^m$.
Then $\{({\mathbf{v}}_i,0){\}}_{i=1}^n\cup \{(0,{\mathbf{w}}_j){\}}_{j=1}^m$ is a basis for $V\oplus W$. ❏

Internal Direct Sum

Definition 3(Definition).

Suppose $W_1,W_2,\dots W_k$ are subspaces of $W$. Define the internal direct sum of these subspaces like so:

$$\begin{array}{rl}{W}_1+W_2+\cdots +W_k& \equiv \text{Span}\left(W_1\cup W_2\cup \dots \cup W_k\right)\\ & =\{w_1+w_2+\cdots +w_k\mid w_i\in W_i\}\end{array}.$$

The internal direct sum of $W_1,W_2,\dots W_k$ is the smallest subspace of $W$ containing every $W_i$.

Definition 4(Definition).

Call $W_1,W_2,\dots W_k$ linearly independent if

$${\mathbf{w}}_1+{\mathbf{w}}_2+\cdots +{\mathbf{w}}_k=0⟹{\mathbf{w}}_i=0.\text{\hspace{1em}(wi∈Wi)}$$

Relation between external and internal direct sums

For $W_1,W_2\subseteq V$, define

$$\begin{array}{r}S:W_1\oplus W_2\to V\\ ({\mathbf{w}}_1,{\mathbf{w}}_2)\mapsto {\mathbf{w}}_1+{\mathbf{w}}_2\end{array}\text{\hspace{1em}(w1∈W1, w2∈W2)}$$

Notice that $S$ is linear and that $\mathrm{Im}S=W_1+W_2$.

$\ker S$ is the set $\{({\mathbf{w}}_1,{\mathbf{w}}_2)\mid {\mathbf{w}}_1\in W_1,{\mathbf{w}}_2\in W_2,{\mathbf{w}}_1+{\mathbf{w}}_2=0\}$. So, $({\mathbf{w}}_1,{\mathbf{w}}_2)\in \ker S$ implies ${\mathbf{w}}_1=-{\mathbf{w}}_2=\mathbf{w}$ (say). As subspaces are closed under scalar multiplication, $\mathbf{w}\in W_1\cap W_2$. Thus,

$$\ker S=\{(\mathbf{w},-\mathbf{w})\mid \mathbf{w}\in W_1\cap W_2\}.$$

We may define $T:\ker S\to W_1\cap W_2$ such that $(-\mathbf{w},\mathbf{w})\mapsto \mathbf{w}$. We can see that $T$ is a linear map. In fact, since $T$ is also bijective, $T$ is an isomorphism!

$$\ker S\cong W_1\cap W_2.$$

Theorem 5(Theorem).

$\dim (W_1+W_2)=\dim W_1+\dim W_2-\dim (W_1\cap W_2)$.

Proof
Using the rank nullity theorem, we have

$$\begin{array}{rl}\dim \text{Domain of S}& =\dim \ker S+\dim \mathrm{Im}S\\ \dim (W_1\oplus W_2)& =\dim \ker S+\dim (W_1+W_2)\\ \dim (W_1+W_2)& =\dim W_1+\dim W_2-\dim (W_1\cap W_2).\end{array}$$

❏

Alternate Proof (Artin)
Let a basis of $W_1\cap W_2$ be $A=\{{\mathbf{\alpha }}_1,{\mathbf{\alpha }}_2,\dots ,{\mathbf{\alpha }}_k\}$.
We may extend this to a basis of $W_1$ by appending $B=\{{\mathbf{\beta }}_1,{\mathbf{\beta }}_2,\dots ,{\mathbf{\beta }}_l\}$.
We may also extend it to a basis of $W_2$ by appending $C=\{{\mathbf{\gamma }}_1,{\mathbf{\gamma }}_2,\dots ,{\mathbf{\gamma }}_m\}$.
Then, the claimed formula is

$$\dim (W_1+W_2)=(k+l)+(k+m)-k=k+l+m$$

So, we have to show that $A\cup B\cup C$ is a basis of $W_1+W_2$. Clearly, $A\cup B\cup C$ spans $W_1+W_2$ since $A\cup B$ spans $W_1$ and $A\cup C$ spans $W_2$. Also, $A\cup B$ and $A\cup C$ are linearly independent sets, since they are bases. So, it only remains to show that $B\cup C$ is linearly independent. To this end, consider a linear combination of vectors in $B\cup C$ which is $0$.

$$\begin{array}{rl}& \sum _{i=1}^l{b}_i{\mathbf{\beta }}_i+\sum _{i=1}^m{c}_i{\mathbf{\gamma }}_i=0\\ ⟹& \sum _{i=1}^l{b}_i{\mathbf{\beta }}_i=-\sum _{i=1}^m{c}_i{\mathbf{\gamma }}_i\equiv \mathbf{w}\end{array}$$

Clearly, $\mathbf{w}\in W_1$ and $\mathbf{w}\in W_2$ $⟹$ $\mathbf{w}\in W_1\cap W_2$. Thus,

$$\begin{array}{rl}& \mathbf{w}=\sum _{i=1}^k{a}_i{\mathbf{\alpha }}_i\\ ⟹& \sum _{i=1}^k{a}_i{\mathbf{\alpha }}_i-\sum _{i=1}^l{b}_i{\mathbf{\beta }}_i=0\\ ⟹& b_i=0\forall i,\text{ since }A\cup B\text{ is linearly independent.}\end{array}$$

Similarly, $c_i=0$ for all $i$. Thus, $B\cup C$ is linearly independent. ❏

!!!! this seems sus. Take 3 lines in ${\mathbb{R}}^2$.

Theorem 6(Corollary).

$$\dim \left(\sum _{i=1}^n{W}_i\right)=\sum _{i=1}^n(-1{)}^{n+1}\left(\sum _{1\le i_1<\cdots <i_k\le n}\dim (W_{i_1}\cap \cdots \cap W_{i_k})\right)$$

A more general statement is:

$$\begin{array}{rl}S:W_1\oplus W_2\oplus \cdots \oplus W_k& \to W_1+W_2+\cdots +W_k\subset V\\ (w_1,w_2,\dots ,w_k)& \mapsto w_1+w_2+\cdots +w_k\end{array}$$

We can see that

$$\begin{array}{rl}S\text{ is surjective}& ⟺\text{span}(W_1,W_2,\dots ,W_k)=V\\ S\text{ is injective}& ⟺W_1,W_2,\dots ,W_k\text{ are linearly independent}\end{array}$$

Remark

For $k=2$,

$$W_1,W_2\text{ linearly independent}⟺W_1\cap W_2=\{0\}.$$

For $k=3$,

$$W_1,W_2,W_3\text{ linearly independent}⟹W_1\cap W_2=W_2\cap W_3=W_3\cap W_1=\{0\}.$$

Note that the backwards implication is not true for $k\ge 3$.

Warning

Almost all the literature I could find has these completely different definitions for the “sum of subspaces” and “direct sums”:

Definition 7(Definition).

Suppose $U_1,\dots ,U_m$ are subspaces of $V$. The sum of $U_1,\dots ,U_m$, denoted $U_1+\cdots +U_m$, is the set of all possible sums of elements of $U_1,\dots ,U_m$.

$$U_1+\cdots +U_m=\{{\mathbf{u}}_1+\cdots +{\mathbf{u}}_m|{\mathbf{u}}_1\in U_1,\dots ,{\mathbf{u}}_m\in U_m\}.$$

Definition 8(Definition).

The sum $U_1+\cdots +U_m$ is called a direct sum is each element of $U_1+\cdots +U_m$ can be written in only one way as a sum ${\mathbf{u}}_1+\cdots +{\mathbf{u}}_m$, where each ${\mathbf{u}}_j$ is in $U_j$. If $U_1+\cdots +U_m$ is a direct sum, then it is denoted by $U_1\oplus \cdots \oplus U_m$.

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Determinants

A determinant is a function $\det :{\mathcal{M}}_{n\times n}(\mathbb{R})\to \mathbb{R}$ which satisfies the following properties:

1. A row operation ($R_i\to R_i+\lambda R_j$) on the matrix does not change the determinant
2. Scaling a row scales the determinant
3. Swapping rows negates the determinant (antisymmetry) (can be derived from 1 and 2)
4. $\det I_n=1$

Derivation of (3) from (1) and (2):

$$\begin{array}{rl}& D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_j,\dots ,{\mathbf{v}}_k,\dots ,{\mathbf{v}}_n)\\ & =D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_j,\dots ,\underset{k}{\underbrace{{\mathbf{v}}_k-{\mathbf{v}}_j}},\dots ,{\mathbf{v}}_n)\\ & =D({\mathbf{v}}_1,\dots ,\underset{j}{\underbrace{{\mathbf{v}}_j+({\mathbf{v}}_k-{\mathbf{v}}_j)}},\dots ,\underset{k}{\underbrace{{\mathbf{v}}_k-{\mathbf{v}}_j}},\dots ,{\mathbf{v}}_n)\\ & =D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_k,\dots ,\underset{k}{\underbrace{{\mathbf{v}}_k-{\mathbf{v}}_j}},\dots ,{\mathbf{v}}_n)\\ & =D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_k,\dots ,\underset{k}{\underbrace{({\mathbf{v}}_k-{\mathbf{v}}_j)-{\mathbf{v}}_k}},\dots ,{\mathbf{v}}_n)\\ & =D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_k,\dots ,-{\mathbf{v}}_j,\dots ,{\mathbf{v}}_n)\\ & =-D({\mathbf{v}}_1,\dots ,{\mathbf{v}}_k,\dots ,{\mathbf{v}}_j,\dots ,{\mathbf{v}}_n).\end{array}$$