MA2.101 - Linear Algebra | Linear Algebra Lecture 6

Direct sum

A vector space $V$ will be called a direct sum if it is the linear sum of two vector subspaces of $V$, $W_1$ and $W_2$ and for all $\alpha \in W_1$ and $\beta \in W_2$, there are no other vectors $\alpha\prime \in W_1$ and $\beta\prime \in W_2$, such that \(\alpha + \beta = \alpha\prime + \beta\prime\)

In other words, the vector space $V$ is a direct sum of the vector subspaces $W_1$ and $W_2$, if and only if each element in $V$ can be represented as the unique sum of a pair of elements, one vector from $W_1$ and the other vector from $W_2$ respectively.

Theorem: If for a vector space $V$, $W_1 + W_2 = V$ and \(W_1 \cap W_2 = \{\phi\}\), then $V$ is a direct sum of $W_1$ and $W_2$, and vice versa.

Proof:
Backward Implication: Let $V$ be the direct sum of $W_1$ and $W_2$, then it naturally follows that $V = W_1 + W_2$.

For the second condition, since the intersection of two vector subspaces also yields a vector subspace,
$W_1 \cap W_2$ is guaranteed to contain the vector additive identity, $\phi$.
Let us now assume that some other element $\alpha$ belongs to both $W_1$ and $W_2$.
Then we have:
$\alpha + \phi$ where $\alpha \in W_1$ and $\phi \in W_2$
and $\phi + \alpha$ where $\phi \in W_1$ and $\alpha \in W_2$
Clearly we have represented the same vector, $\alpha$ in the form of two different sums taking one vector from each subspace, so $V$ cannot be the direct sum of $W_1$ and $W_2$.
This poses a contradiction, and thus no element other than $\phi$ can belong to both $W_1$ and $W_2$.
Thus, \(W_1 \cap W_2 = \{\phi\}\).

Forward Implication: Given that $W_1 + W_2 = V$ and \(W_1 \cap W_2 = \{\phi\}\), let us assume that there exist pairs $\alpha, \alpha\prime \in W_1$ and $\beta, \beta\prime \in W_2$, such that $\alpha \neq \alpha\prime$, $\beta \neq \beta\prime$, and

By the property of closure, we can say that $\alpha - \alpha\prime \in W_1$, $\beta\prime - \beta \in W_2$. Since they are equal, they represent an element present in both $W_1$ and $W_2$, but we know there is only one element that the two have in common, i.e. $\phi$.
Thus we can say that $\alpha - \alpha\prime = \phi = \beta\prime - \beta$
which in turn tells us that $\alpha = \alpha\prime, \beta = \beta\prime$.

This poses a contradiction, and thus we can conclude that there must be no such pairs of elements. Finally, this means that $V$ must be a direct sum of the vector subspaces $W_1$ and $W_2$.

Theorem: (Steinitz Replacement Theorem)
If \(\{ \bar\alpha_1, \bar\alpha_2, ..., \bar\alpha_n \}\) is a basis of $V(F)$ and $\bar\beta$ is a non zero vector belonging to V, then $\bar\beta = \sum_{i = 1}^{n} c_i \bar\alpha_i$, where $c_i \in F$ and not all $c_i$’s are zero. If $c_j \neq 0$, then \(\{ \bar\alpha_1, \bar\alpha_2,..., \bar\alpha_{j-1}, \bar\beta, \bar\alpha_{j+1}, ..., \bar\alpha_n \}\)] is a basis of $V(F)$

Proof: Let $c_j$ be the non zero, then we have:

We now need to prove that the set $W = {\bar\alpha_1, \bar\alpha_2, … , \bar\alpha_{j-1}, \bar\beta, \bar\alpha_{j+1}, … , \bar\alpha_n}$ is linearly independent.

Let

where $\delta_i \in F$

Now, we replace $\bar\beta$ with the summation $\sum_{i=1}^n c_i \cdot \bar\alpha_i$ Note that this includes $\bar\alpha_j$ as a term.
Therefore, the whole expression can be written as:

That can be re-written as

Since ${\bar\alpha_1, \bar\alpha_2, … \bar\alpha_n }$ is known to be independent, all the coefficients in the above expression must turn out to be $0$ and thus $\delta_i + \delta_j c_i = 0 \forall i \not= j$ and $\delta_j c_j = 0$ but we know $c_j \not= 0$ (from the original assumption) thus $\delta_j = 0$ and since $\delta_i + \delta_j c_i = 0 \forall i \not= j$ , we can say $\delta_i = 0$. Thus, a linear combination of vectors in $W$ being $\bar\phi$ implies that all the coefficients are $0$ and thus we can say that $W$ is linearly independent.

Now, we need to prove that $W$ spans the entire space.

Let $\bar\gamma$ be any vector in $V$ and since ${\bar\alpha_1, \bar\alpha_2, … , \bar\alpha_n}$ is a basis

We have already seen that $\bar\alpha_j$ can be represented as a linear combination of the vectors of $W$ i.e. all $\bar\alpha_i$ with $i \not=j$ and $\bar\beta$

thus we make the following substitution $\bar\alpha_j = c_j^{-1} \bar\beta - c_j^{-1} (c_1 \bar\alpha_1 + c_2 \bar\alpha_2 + … + c_{j-1} \bar\alpha_{j-1} + c_{j+1} \bar\alpha_{j+1} + … + c_n \bar\alpha_n)$

and we get

So after simplifying we’ll get the $\bar\gamma$ has been represented as a linear combination of all vectors in $W$ and thus $\bar\gamma \in L(W)$ and since $\bar\gamma$ was an arbitrary vector in the vector space, we have $V(F) \subseteq L(W)$ and since $L(W) \subseteq V(F)$ we can say that $L(W) = V(F)$

Thus we have proven that $W$ is independent and spans the entire vector space, it is therefore a basis of the vector space.

Theorem: In a Vector space $V(F)$ with the basis set $B = { \bar\alpha_1, \bar\alpha_2, …, \bar\alpha_n }$ every vector $\bar \gamma \in V(F)$ is uniquely expressible in a linear combination of vectors in $B$.

Proof: Since $B$ is a basis set of $V(F)$
Consider an arbitrary vector $\bar\gamma \in V(F)$
We will prove the theorem by contradiction therefore we assume that $\bar\gamma$ can be expressed in two different linear combinations of $B$ i.e.

where $a_1, … a_n , b_1, … b_n \in F$
and $(a_1, a_2, … a_n) \not= (b_1, b_2, … b_n)$
Now we subtract the second equation from the first to get

Notice that the RHS of the above equation is a linear combination of all the vectors in $B$ , which we know to be a basis set and therefore we know it to be linearly independent. Linear independence dictates that the only way for a linear combination leads to $\bar\phi$ is when all the coeffecients are $0$ i.e. if $a_i = b_i \forall i \text{ such that } 1\leq i\leq n$ which we assumed to be untrue therefore we have a contradiction

Hence proved by contradiction.