Learn Linear Algebra

Theorem - Parallelogram Rule for Addition

If u⃗,v⃗∈R2\vec{u}, \vec{v} \in \mathbb{R}^2 are represented by two vectors, the parallelogram formed by these two vectors has a diagonal that represents the sum of the vectors.
Proof: Let u⃗\vec{u} and v⃗\vec{v} be vectors in R2\mathbb{R}^2. Lets denote u⃗\vec{u} as [u1u2]\begin{bmatrix} u_1 \\ u_2 \end{bmatrix} and similarly, lets denote v⃗\vec{v} as [v1v2]\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}. Recall that we can add two vectors in the same dimensional space, and the sum of the two vectors is the sum of their corresponding entries. Thus, u⃗+v⃗=[u1u2]+[v1v2]=[u1+v1u2+v2]. \vec{u} + \vec{v} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix}. In the plane, both vectors u⃗\vec{u} and v⃗\vec{v} can be represented as points with coordinates P(u1,u2) P(u_1, u_2) and Q(v1,v2) Q(v_1, v_2) , both starting from the origin O(0,0) O(0, 0) . Consider the parallelogram with vertices at O(0,0) O(0, 0) , P(u1,u2) P(u_1, u_2) , Q(v1,v2) Q(v_1, v_2) , and R(x,y) R(x, y) . We aim to show that the point R R is the sum of both vectors, i.e., R=u⃗+v⃗ R = \vec{u} + \vec{v} . To prove this, we will demonstrate that the diagonals of the parallelogram bisect each other, which is a defining property of parallelograms. Let the midpoint of the diagonal PQ PQ be denoted by M1 M_1 , where: M1=(u1+v12,u2+v22). M_1 = \left( \frac{u_1 + v_1}{2}, \frac{u_2 + v_2}{2} \right). Similarly, let the midpoint of the diagonal OR OR be denoted by M2 M_2 , where: M2=(0+x2,0+y2)=(x2,y2). M_2 = \left( \frac{0 + x}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right). Since the diagonals of a parallelogram bisect each other, we must have M1=M2 M_1 = M_2 . Therefore, by equating the components of M1 M_1 and M2 M_2 , we obtain: u1+v12=x2andu2+v22=y2. \frac{u_1 + v_1}{2} = \frac{x}{2} \quad \text{and} \quad \frac{u_2 + v_2}{2} = \frac{y}{2}. Solving these equations, we find: x=u1+v1andy=u2+v2. x = u_1 + v_1 \quad \text{and} \quad y = u_2 + v_2. Thus, the coordinates of the point R R are R(x,y)=(u1+v1,u2+v2) R(x, y) = (u_1 + v_1, u_2 + v_2) , which confirms that R=u⃗+v⃗ R = \vec{u} + \vec{v} . This proves that the fourth vertex R R of the parallelogram corresponds to the sum of the two vectors u⃗ \vec{u} and v⃗ \vec{v} .

Theorem - Algebraic Properties of Vectors in Rn \mathbb{R}^n

Let u⃗,v⃗,w⃗∈Rn \vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n and scalars a,b∈R a, b \in \mathbb{R} . The following vector properties are extensions of the axiomatic basis for the real number system.

1. Commutative Property of Addition for Vectors
u⃗+v⃗=v⃗+u⃗ \vec{u} + \vec{v} = \vec{v} + \vec{u}

2. Associative Property of Addition for Vectors
(u⃗+v⃗)+w⃗=u⃗+(v⃗+w⃗) (\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})

3. Additive Identity of Vectors
u⃗+0⃗=u⃗ \vec{u} + \vec{0} = \vec{u}

4. Additive Inverse of Vectors
u⃗+(−u⃗)=0⃗ \vec{u} + (-\vec{u}) = \vec{0}

5. Scalar Distributive Property onto Vectors
a(u⃗+v⃗)=au⃗+av⃗ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v}

6. Scalar Associative Property of Multiplication with a Vector
(ab)u⃗=a(bu⃗) (ab)\vec{u} = a(b\vec{u})

7. Multiplicative Identity of Vectors
1u⃗=u⃗ 1\vec{u} = \vec{u}

Proof - Commutative Property of Addition for Vectors:

Let u⃗=(u1,u2,…,un) \vec{u} = (u_{1}, u_{2}, \dots, u_{n}) and let v⃗=(v1,v2,…,vn) \vec{v} = (v_{1}, v_{2}, \dots, v_{n}) .

u⃗+v⃗=(u1+v1,u2+v2,…,un+vn) \vec{u} + \vec{v} = (u_{1} + v_{1}, u_{2} + v_{2}, \dots, u_{n} + v_{n}) (Definition of vector addition)

=(v1+u1,v2+u2,…,vn+un) = (v_{1} + u_{1}, v_{2} + u_{2}, \dots, v_{n} + u_{n}) (Commutativity of addition in R \mathbb{R} )

=v⃗+u⃗ = \vec{v} + \vec{u} (Definition of vector addition)

Proof - Associative Property of Addition for Vectors:

Let u⃗=(u1,u2,…,un) \vec{u} = (u_1, u_2, \dots, u_n) , v⃗=(v1,v2,…,vn) \vec{v} = (v_1, v_2, \dots, v_n) , and w⃗=(w1,w2,…,wn) \vec{w} = (w_1, w_2, \dots, w_n) .

(u⃗+v⃗)+w⃗=((u1+v1)+w1,(u2+v2)+w2,…,(un+vn)+wn)(\vec{u} + \vec{v}) + \vec{w} = ((u_1 + v_1) + w_1, (u_2 + v_2) + w_2, \dots, (u_n + v_n) + w_n) (Definition of vector addition)

 =(u1+(v1+w1),u2+(v2+w2),…,un+(vn+wn))\ = (u_1 + (v_1 + w_1), u_2 + (v_2 + w_2), \dots, u_n + (v_n + w_n)) (Commutativity of addition in R\mathbb{R})

 =u⃗+(v⃗+w⃗)\ = \vec{u} + (\vec{v} + \vec{w})

Proof - Additive Identity of Vectors:

Let u⃗=(u1,u2,…,un) \vec{u} = (u_1, u_2, \dots, u_n) and we know that the 0⃗=(0,0,…,0)\vec{0} = (0 ,0 ,\dots, 0) (For this proof, the zero vector is of the same size as u⃗\vec{u}).

u⃗+0⃗=(u1+0,u2+0,…,un+0)\vec{u} + \vec{0}= (u_1 + 0, u_2 +0, \dots, u_n + 0) (Definition of vector addition)

 =(u1,u2,…,un)\ = (u_1, u_2, \dots, u_n)

=u⃗= \vec{u}

Proof - Additive Inverse of Vectors:

Let u⃗=(u1,u2,…,un) \vec{u} = (u_1, u_2, \dots, u_n) and let −u⃗=(−u1,−u2,…,−un) \vec{-u} = (-u_1, -u_2, \dots, -u_n)

u⃗+(−u⃗) =(u1+(−u1),u2+(−u2),…,(un+(−un))\vec{u} + (\vec{-u})\ = (u_1 + (-u_1), u_2 + (-u_2),\dots, (u_n + (-u_n)) (Definition of vector addition)

=(0,0,…,0)= (0,0,\dots,0) =0⃗=\vec{0}

Proof - Scalar Distributive Property onto Vectors:

Let u⃗=(u1,u2,…,un) \vec{u} = (u_1, u_2, \dots, u_n) , w⃗=(w1,w2,…,wn) \vec{w} = (w_1, w_2, \dots, w_n) and let some scalar a∈Ra \in \mathbb{R}

a(u⃗+w⃗)=a(u1+w1,u2+w2,…,un+wn)a(\vec{u} + \vec{w}) = a(u_{1} + w_{1}, u_{2} + w_{2}, \dots, u_{n} + w_{n}) (Definition of vector addition)

=(au1+aw1,au2+aw2,…,aun+awn)= (au_1 + aw_1, au_2 + aw_2,\dots,au_n + aw_n)

Proof - Scalar Associative Property of Multiplication with a Vector:

Let a,b∈Rna, b \in \mathbb{R}^n and let u⃗=(u1,u2,…,un)\vec{u} = (u_1, u_2, \dots, u_n)

Then we have that

ab(u⃗)=ab(u1,u2,…,un)ab(\vec{u})= ab(u_1, u_2, \dots, u_n)

=(abu1,abu2,…,abun)= (abu_1, abu_2, \dots, abu_n)

=(a(bu1),a(bu2),…,a(bun))= (a(bu_1), a(bu_2), \dots, a(bu_n))

=a(bu1,bu2,…,bun)= a(bu_1, bu_2, \dots, bu_n)

=a(bu⃗)=a(b\vec{u})

Proof - Multiplicative Identity of Vectors:

Let u⃗=(u1,u2,…,un)\vec{u} = (u_1, u_2, \dots, u_n)

If we multiply u⃗\vec{u} by 1 we get 1 ⋅\cdot u⃗\vec{u}

=1(u1,u2,…,un)= 1(u_1, u_2,\dots,u_n)

=(1u1,1u2,…,1un)=(1u_1, 1u_2,\dots, 1u_n)

=(u1,u2,…,un)=(u_1, u_2,\dots,u_n)

=u⃗= \vec{u}

Theorem

A vector equation c1v⃗1+c2v⃗2+⋯+cnv⃗n =b⃗c_1\vec{v}_{1} + c_2\vec{v}_{2} + \dots + c_n\vec{v}_{n}\ = \vec{b} has the same solution set as the linear system whose augmented matrix is [v⃗1,v⃗2,…,v⃗n,b⃗][\vec{v}_{1}, \vec{v}_{2},\dots,\vec{v}_{n}, \vec{b}]
Proof: Given a valid solution c1,c2,…,cp∈Rc_1, c_2,\dots,c_p \in \mathbb{R} exists for b⃗=c1v⃗1+c2v⃗2+⋯+cpv⃗p\vec{b} = c_1\vec{v}_{1} + c_2\vec{v}_{2} + \dots + c_p\vec{v}_{p} where v⃗1,v⃗2,…,v⃗p∈Rn\vec{v}_{1},\vec{v}_{2},\dots,\vec{v}_{p} \in \mathbb{R}^{n}. Recall that we can write b⃗=c1v⃗1+c2v⃗2+⋯+cpv⃗p\vec{b} = c_1\vec{v}_{1} + c_2\vec{v}_{2} + \dots + c_p\vec{v}_{p} as

c1(v11v12⋮v1n)+c2(v21v22⋮v2n)+⋯+cp(vp1vp2⋮vpn)=(b1b2⋮bn)c_1 \left( \begin{array}{c} v_{11} \\ v_{12} \\ \vdots \\ v_{1n} \end{array} \right) + c_2\left( \begin{array}{c} v_{21} \\ v_{22} \\ \vdots \\ v_{2n} \end{array} \right) + \cdots + c_p\left( \begin{array}{c} v_{p1} \\ v_{p2} \\ \vdots \\ v_{pn} \end{array} \right) = \left( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \end{array} \right)

We can express this as a system of linear equations now:

{c1v11+c2v21+⋯+cpvp1=b1c2v12+c2v22+⋯+cpvp2=b2…cpv1n+c2v2n+⋯+cpvpn=bn \left\{ \begin{array}{l} c_{1}v_{11} + c_{2}v_{21} + \dots + c_{p}v_{p1} = b_1 \\\\ c_{2}v_{12} + c_{2}v_{22} + \dots + c_{p}v_{p2} = b_2 \\\\ \dots \\\\ c_{p}v_{1n} + c_{2}v_{2n} + \dots + c_{p}v_{pn} = b_n \\\\ \end{array} \right.

Since we have a system of equations now we can create an augmented matrix for this system:

[v11v21⋯vp1b1v12v22⋯vp2b2⋮⋮⋱⋮⋮v1nv2n⋯vpnbn] \left[ \begin{array}{cccc|c} v_{11} & v_{21} & \cdots & v_{p1} & b_1 \\\\ v_{12} & v_{22} & \cdots & v_{p2} & b_2 \\\\ \vdots & \vdots & \ddots & \vdots & \vdots \\\\ v_{1n} & v_{2n} & \cdots & v_{pn} & b_n \end{array} \right]

This in turn can now be written in the following form: [v⃗1v⃗2⋯v⃗pbn] \left[ \begin{array}{cccc|c} \vec{v}_{1} & \vec{v}_{2} & \cdots & \vec{v}_{p} & b_n \end{array} \right]

Since the solution c1,c2,…,cpc_1, c_2,\dots,c_p holds true for b⃗=c1v⃗1+c2v⃗2+⋯+cpv⃗p\vec{b} = c_1\vec{v}_{1} + c_2\vec{v}_{2} + \dots + c_p\vec{v}_{p} then it must also follow that the solution holds true for [v⃗1v⃗2⋯v⃗pbn\begin{array}{cccc|c} \vec{v}_{1} & \vec{v}_{2} & \cdots & \vec{v}_{p} & b_n \end{array}] since we showed that they are both equivelant.

Theorem

b⃗\vec{b} can be generated by a linear combination of the vectors {v⃗1,v⃗2,…,v⃗p\vec{v}_1,\vec{v}_2,\dots,\vec{v}_p} if and only if there is a solution to the vector equation x1v⃗1+x2v⃗2+⋯+xpv⃗p=b⃗x_1\vec{v}_1 + x_2\vec{v}_2+\cdots+x_p\vec{v}_p = \vec{b}

⇔\Leftrightarrow b⃗\vec{b} can be generated by a linear combination of the vectors {v⃗1,v⃗2,⋯ ,v⃗p\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_{p}}.

⇔\Leftrightarrow There exist some weights (c1,c2,⋯ ,cp)(c_1, c_2,\cdots, c_p) such that c1v⃗1+c2v⃗2+⋯+cpv⃗p=b⃗c_1\vec{v}_1 + c_2\vec{v}_2 +\cdots+c_p\vec{v}_p = \vec{b}.

⇔\Leftrightarrow There is a solution to the vector equation x1v⃗1+x2v⃗2+⋯+xpv⃗p=b⃗x_1\vec{v}_1 + x_2\vec{v}_2 + \cdots + x_p\vec{v}_p = \vec{b} where x1=c1,x2=c2,⋯ ,xp=cpx_1 = c_1, x_2 = c_2, \cdots, x_p = c_p