Theorem: The set of non-zero vectors ${\bar\alpha_1\, \bar\alpha_2\ … \bar\alpha_k}$ is linearly dependent, if and only if there exists an $\bar\alpha_r, 2 \leq r \leq k$, such that it is the linear combination of preceeding vectors.
Proof:
Note: this proof is taken primarily from the book “Finite-Dimensional Vector Space by P.R. Halmos”
(Note that prof uses the terms sufficient and necessity condition terminology,
it’s the same proof, just a different way of looking at it. I am more comfortable with this.)
RTP:
Let $W = {\bar\alpha_1\, \bar\alpha_2\ .… \bar\alpha_r…. \bar\alpha_k}$
The first part:
We assume that $r$ is the first integer for which the set ${\bar\alpha_1, \bar\alpha_2 … \bar\alpha_r}$ is linearly dependent (notice that is a subset of $W$ ).
Note that this is not an unfair assumption, because we can always say that $r =k$ and the set will become $W$
which is linearly dependent as per the left side of implication condition in the first part of the RTP.
Thus, now we can say
$a_1\cdot \bar\alpha_1 + … + a_r \cdot \bar\alpha_r = 0 \ \ … (i)$
such that not all $a_1, … a_r$ are 0
Note that $a_r$ cannot be $0$ because that would imply that the set ${\bar\alpha_1, \bar\alpha_2 … \bar\alpha_{r-1}}$ is linearly dependent, which it can’t be by our definition of the $r$ (we assumed that $\bar\alpha_r$ is the first element such that everything before it and including it gives us a linearly dependent set). Thus we can re-write $(i)$
$- a_1\cdot \bar\alpha_1 - …. - a_{r-1}\bar\alpha_{r-1} = a_r \cdot \bar\alpha_r$
and since $a_r$ is non-zero, it has a multiplicative inverse, thus
$(-a_1 a_r^{-1}) \cdot \bar\alpha_1 + (-a_2 a_r^{-1}) \cdot \bar\alpha_2 + … + (-a_{r-1} a_r^{-1}) \cdot \bar\alpha_{r-1} = \bar\alpha_r$
The second part: This is trivial because if there exists such an $\bar\alpha_r$ then the set ${\bar\alpha_1, \bar\alpha_2 .. \bar\alpha_r}$ is linearly dependent and the super set of any linearly dependent set is itself linearly dependent.
Note to the reader: the professor here goes into a mini discussion about why $r$’s range goes from $2$ to $k$ and why it can’t be $1$. I haven’t included that entire discussion because the theorem talks about _preceding vectors_ in the set and there is nothing preceding $\bar\alpha_1$ and therefore $r$ can’t be $1$.
Also note that we’re strictly concerned with a finite set.
Theorem: If the vectors space $V(F)$ is spanned by a linearly dependent set of vectors say \(W = \{\bar\alpha_1, \bar\alpha_2, ... \bar\alpha_k\}\) then there exists a proper subset of $W$ that can span the entire vector space $V(F)$
Proof: Let $W$ be linearly dependent. We proved earlier that there exists an $\bar\alpha_r$ in $W$ such that it can be expressed as a linear combination of the vectors preceding it. We can trivially say that this implies that $\bar\alpha_r$ can be represented as a linear combination of all the elements of $W$.
therefore, let
\[\bar\alpha_r = c_1 \cdot \bar\alpha_1 + c_2 \cdot \bar\alpha_2 + ... + c_k \cdot \bar\alpha_k\]where $c_1, c_2, … c_k \in F$
consider an arbitrary vector $\bar\beta$ we know that $W$ spans the entire $V(F)$ and therefore $V(F) = L(W)$ thus $\bar\beta \in L(W)$ thus
\[\bar\beta = b_1 \cdot \bar\alpha_1 + b_2 \cdot \bar\alpha_2 + ... + b_k \cdot \bar\alpha_k \\ \text{where, } b_1 .. b_k \text{ are scalars} \\ \bar\beta = b_1 \cdot \bar\alpha_1 + b_2 \cdot \bar\alpha_2 + ... + b_r \cdot \bar\alpha_r + ... b_k \cdot \bar\alpha_k\]replacing $\bar\alpha_r$
\[\bar\beta = b_1 \cdot \bar\alpha_1 + b_2 \cdot \bar\alpha_2 + ... + b_r \cdot (c_1 \cdot \bar\alpha_1 + c_2 \cdot \bar\alpha_2 + ... + c_k \cdot \bar\alpha_k) + ... b_k \cdot \bar\alpha_k \\ \\ \bar\beta = (b_1 + b_r c_1)\cdot\bar\alpha_1 + (b_2 + b_r c_2)\cdot\bar\alpha_2 + ... + (b_{r-1} + b_r c_{r-1})\cdot\bar\alpha_{r-1} + ... + (b_k + b_r c_k)\cdot\bar\alpha_k\]So $\bar\beta$ has been represented by the set of vectors $W’ = {\bar\alpha_1, \bar\alpha_2, … \bar\alpha_{r-1}, … \bar\alpha_k}$ which clearly does not have $\bar\alpha_r$ thus $W’$ proper subset of $W$ . Since $\bar\beta$ was an arbitrary vector, we can say that $W’$ spans the entire vector space.
Hence proved.
Definition: A linear basis (pl. bases) (or a coordinate system) in a vector space, is a set $X$ such that:
Definition: The number of vectors in a basis of the vector space $V(F)$ is called the dimension pf $V(F)$ and is defined by $dimm(V)$
Note: A vector space is finite dimensional if it has a finite basis. Note: A vector space may have more than one basis.
Note: number of elements of all bases are equal (proof pending)
To prove that note, I first need to introduce another theorem:
Theorem
If $V$ has a basis with $n$ elements then every set of vectors in $V$ which has more than $n$ elements is linearly dependent
Proof
consider $B = {v_1, v_2, … v_n}$ to be a basis (with $n$ elements)
consider a set $W = {w_1, w_2, … w_k}$ and $k > m$
To prove that $W$ is linearly dependent we need to prove that the equation
\[x_1 \cdot w_1 + x_2 \cdot w_2 + .. + x_k \cdot w_k = 0\]has a solution in which not all $x_i$ are $0$.
since $B$ is a basis, then it is an independent set that spans the whole vector space $V$ and therefore we can represent any vector as a unique linear combination of all the vectors in $B$.
This leads us to:
\[w_1 = a_{11} \cdot v_1 + a_{12} \cdot v_2 + ... + a_{1n} \cdot v_n \\ w_2 = a_{21} \cdot v_1 + a_{22} \cdot v_2 + ... + a_{2n} \cdot v_n \\ . . \\ .. \\ .. \\ w_k = a_{k1} \cdot v_1 + a_{k2} \cdot v_2 + ... + a_{kn} \cdot v_n\]where $a_{ij} \in F$
this allows us to rewrite $x_1 \cdot w_1 + x_2 \cdot w_2 + .. + x_k \cdot w_k = 0$ as
\[x_1(a_{11} \cdot v_1 + a_{12} \cdot v_2 + ... + a_{1n} \cdot v_n) + x_2(a_{21} \cdot v_1 + a_{22} \cdot v_2 + ... + a_{2n} \cdot v_n) + .. + x_k(a_{k1} \cdot v_1 + a_{k2} \cdot v_2 + ... + a_{kn} \cdot v_n) = 0 \\ \Rightarrow (a_{11}x_1 + a_{21}x_2 + ... + a_{k1}x_k) \cdot v_1 + ... + (a_{1n}x_1 + a_{21}x_2 + ... + a_{kn}x_k) \cdot v_n = 0\]as the $B$ is independent, the only way the final version of the equation works is if all the coefficients have to be $0$.
\[a_{11}x_1 + a_{21}x_2 + ... + a_{k1}x_k = 0 \\ .. \\ .. \\ a_{1n}x_1 + a_{21}x_2 + ... + a_{kn}x_k = 0\]these are $n$ homogenous equations with $k$ variables (all the $x_i$) with more variables than equations (as $n < k$), thus we know that there need to be infinitely many solutions, in which there has to be a solution in which not all $x_i$ are $0$ Hence proved that $W$ is dependent
Now, to prove that all bases have the same number of elements
Let there exist two bases $B$ and $W$, such that $B$ has $m$ elements and $W$ has $n$ elements and $m \not = n$. Without loss of generality, I assume that $n > m$ .
Since, they’re bases we know that they’re linearly independent sets.
As per the last theorem, we know that any set with more elements than a basis has to be linearly independent this is a contradiction to our original assumption that both $B$ and $W$ are bases.
Hence proved by contradiction that $m=n$ and consequently that all bases have the same number of elements.
Question: Show that the vectors $(1, 1, 1, 1), (0, 1, 1, 1), (0, 0, 1, 1), \text{and } (0, 0, 0, 1)$ form a basis of $\mathbb{R^4(R)}$.
If $W_1$ and $W_2$ are two subspaces of $V(F)$ then the linear sum of two subspaces is denoted by $W_1 + W_2$ and is the set of the sums $\bar\alpha_1 + \bar\alpha_2$ such that $\bar\alpha_1 \in W_1$ and $\bar\alpha_2 \in W_2$. More formally: \(W_1 + W_2 = \{\bar\alpha_1 + \bar\alpha_2: \bar\alpha_1 \in W_1, \bar\alpha_2 \in W_2\}\)
Theorem: If $W_1$ and $W_2$ are subspaces of the vector space $V(F)$, then $W_1 + W_2$ is a subspace of $V$
Proof: Let $\bar\alpha, \bar\beta \in W_1 + W_2$ therefore
\(\bar\alpha = \bar\alpha_1 + \bar\alpha_2\), where $\bar\alpha_1 \in W_1$ and $\bar\alpha_2 \in W_2$
\(\bar\beta = \bar\beta_1 + \bar\beta_2\), where $\bar\beta_1 \in W_1$ and $\bar\beta_2 \in W_2$
Now, for $a, b \in F$
\(a \cdot \bar\alpha + b \cdot \bar\beta = a \cdot(\bar\alpha_1 + \bar\alpha_2) + b \cdot (\bar\beta_1 + \bar\beta_2) \\
\Rightarrow a \cdot \bar\alpha + b \cdot \bar\beta = (a \cdot \bar\alpha_1 + b \cdot \bar \beta) + (a \cdot \bar\beta_2 + b \cdot \bar\beta_2)\)
as $\bar\alpha_1, \bar\beta_1 \in W_1$ we can say (due to closure of scalar multiplication and vector addition) that $(a \cdot \bar\alpha_1 + b \cdot \bar \beta) \in W_1$ and similarly $(a \cdot \bar\beta_2 + b \cdot \bar\beta_2) \in W_2$ and thus their summation must belong to $W_1 + W_2$ by definition.
Thus for arbitrary $\bar\alpha, \bar\beta \in W_1 + W_2$, $(a \cdot \bar\alpha + b \cdot \bar\beta) \in W_1 + W_2$.
Thus, $W_1 + W_2$ must be subspace of $V(F)$ .