Proofs in Linear Algebra

Properties of Vectors in $3$-space

The study of Linear Algebra originates from the geometric intuition of Euclidean spaces, primarily the plane ($\mathbb{R}^2$) and 3-space ($\mathbb{R}^3$). Historically, we represent a vector as a directed line segment from the origin to a terminal point. By abstracting the properties of $\mathbb{R}^n$, we develop a framework capable of characterizing diverse objects such as functions and matrices.

In the concrete case of $\mathbb{R}^3$, let $\mathbf{u} = (a_1, b_1, c_1)$ and $\mathbf{v} = (a_2, b_2, c_2)$, where $a_i, b_i, c_i \in \mathbb{R}$. We define the fundamental operations as follows:

• Vector Addition:

$$ \mathbf{u} + \mathbf{v} = (a_1 + a_2, b_1 + b_2, c_1 + c_2) $$

• Scalar Multiplication:

$$ \alpha \mathbf{u} = (\alpha a_1, \alpha b_1, \alpha c_1) $$

for any scalar $\alpha \in \mathbb{R}$.

To illustrate the coordinate structure, we recreate the data from the foundational representations of vectors:

Space	Dimension	Exemplar Vectors	Basis Vectors
$\mathbb{R}^2$	2-Space	$\mathbf{u} = (4, 3)$	$\mathbf{i}=(1,0), \mathbf{j}=(0,1)$
$\mathbb{R}^3$	3-Space	$\mathbf{v} = (2, 3, 4)$	$\mathbf{i}=(1,0,0), \mathbf{j}=(0,1,0), \mathbf{k}=(0,0,1)$

Vectors in $\mathbb{R}^3$ must satisfy eight fundamental properties.

1. Commutative:

$$ \mathbf{u} + \mathbf{v} = (a_1+b_1, a_2+b_2, a_3+b_3) = (b_1+a_1, b_2+a_2, b_3+a_3) = \mathbf{v} + \mathbf{u} $$

relying on the commutativity of real numbers.

2. Associative:

$$ (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) $$

3. Zero Vector: There exists $\mathbf{z} = (0, 0, 0)$ such that

$$ \mathbf{u} + \mathbf{z} = \mathbf{u} $$

4. Additive Inverse: For $\mathbf{v} = (b_1, b_2, b_3)$, there exists $-\mathbf{v} = (-b_1, -b_2, -b_3)$ such that

$$ \mathbf{v} + (-\mathbf{v}) = (b_1+(-b_1), b_2+(-b_2), b_3+(-b_3)) = (0, 0, 0) = \mathbf{z} $$

5. Distributive (Scalar over Vector):

$$ \alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v} $$

6. Distributive (Scalar Sum):

$$ (\alpha + \beta)\mathbf{u} = \alpha\mathbf{u} + \beta\mathbf{u} $$

7. Scalar Associative:

$$ (\alpha\beta)\mathbf{u} = \alpha(\beta\mathbf{u}) $$

8. Scalar Identity:

$$ 1\mathbf{u} = \mathbf{u} $$

The transition from viewing vectors as “directed line segments” to “linear combinations of $\mathbf{i, j, k}$” is a shift toward computational abstraction. By expressing $\mathbf{u}$ as $a_1\mathbf{i} + b_1\mathbf{j} + c_1\mathbf{k}$, we move from a spatial domain to an algebraic one.

Vector Spaces

This framework accommodates the standard $\mathbb{R}^n$ and abstract spaces like $F_{\mathbb{R}$ (all functions $f: \mathbb{R} \to \mathbb{R}$) and $\mathbb{R}[x]$ (polynomials).

Matrices

Let $M_{mn}[\mathbb{R}]$ be the set of all $m \times n$ matrices with real entries $a_{ij}$. Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ are equal if $a_{ij} = b_{ij}$ for all $i, j$.

Addition is defined as $A + B = [a_{ij} + b_{ij}]$ and scalar multiplication as $\alpha A = [\alpha a_{ij}]$.

Note that the product $AB$ is defined by the inner product of the $i$-th row of $A$ and $j$-th column of $B$:

$$ c_{ij} = \sum_{k=1}^n a_{ik}b_{kj} $$

It is very important to recognize that while matrix multiplication is a hallmark of matrix algebra, it is a separate operation not required for $M_{mn}[\mathbb{R}]$ to qualify as a vector space. Notably, $AB$ and $BA$ are not generally equal, and $BA$ may not even be defined when $AB$ is, further distinguishing multiplication from the strictly commutative and always-defined addition required of vector spaces.

Some Properties of Vector Spaces

The integrity of an algebraic system requires the rigorous proof of “obvious” properties to prevent logical contradictions.

Proof:

Assume $\mathbf{z}$ and $\mathbf{z’}$ are both zero vectors in $V$.

Since $\mathbf{z}$ is a zero vector,

$$\mathbf{z'} + \mathbf{z} = \mathbf{z'}$$

Since $\mathbf{z’}$ is a zero vector,

$$\mathbf{z} + \mathbf{z'} = \mathbf{z}$$

By commutativity,

$$\mathbf{z} = \mathbf{z} + \mathbf{z'} = \mathbf{z'} + \mathbf{z} = \mathbf{z'}$$

Thus, $\mathbf{z} = \mathbf{z’}$.

Proof:

Assume $\mathbf{v}_1$ and $\mathbf{v}_2$ are both negatives of $\mathbf{v}$. Then

$$\mathbf{v} + \mathbf{v}_1 = \mathbf{z}$$

and

$$\mathbf{v} + \mathbf{v}_2 = \mathbf{z}$$

By Axiom 3 and substitution,

$$\mathbf{v}_1 = \mathbf{v}_1 + \mathbf{z} = \mathbf{v}_1 + (\mathbf{v} + \mathbf{v}_2)$$

By Axiom 2,

$$\mathbf{v}_1 + (\mathbf{v} + \mathbf{v}_2) = (\mathbf{v}_1 + \mathbf{v}) + \mathbf{v}_2 = \mathbf{z} + \mathbf{v}_2 = \mathbf{v}_2$$

Thus $\mathbf{v}_1 = \mathbf{v}_2$.

Proof:

Since $\mathbf{v} + (-\mathbf{v}) = \mathbf{z}$, we have

$$\mathbf{z} = \mathbf{v} + (-\mathbf{v}) = (\mathbf{v} + \mathbf{v}) + (-\mathbf{v}) = \mathbf{v} + (\mathbf{v} + (-\mathbf{v})) = \mathbf{v} + \mathbf{z} = \mathbf{v}$$

Proof:

(i)

$$0\mathbf{v} = (0+0)\mathbf{v} = 0\mathbf{v} + 0\mathbf{v}$$

By Theorem 15.3,

$$0\mathbf{v} = \mathbf{z}$$

(ii)

$$\alpha\mathbf{z} = \alpha(\mathbf{z} + \mathbf{z}) = \alpha\mathbf{z} + \alpha\mathbf{z}$$

By Theorem 15.3, $\alpha\mathbf{z} = \mathbf{z}$.

Proof:

If $\alpha = 0$, the statement holds.

If $\alpha \neq 0$, then there exists $1/\alpha \in \mathbb{R}$. Multiplying gives

$$(1/\alpha)(\alpha\mathbf{v}) = (1/\alpha)\mathbf{z}$$

Using Axioms 7, 8, and Corollary 15.4(ii), we find

$$1\mathbf{v} = \mathbf{z}$$

so $\mathbf{v} = \mathbf{z}$

Subspaces

Proof

If $W$ is a subspace, it must be closed under both.

Conversely, if $W$ is nonempty and closed, it inherits Axioms 1, 2, and 5-8 from $V$. Nonemptiness implies

$$\exists \mathbf{v} \in W$$

Closure under scaling implies

$$0\mathbf{v} = \mathbf{z} \in W$$

that is, Axiom 3, and

$$(-1)\mathbf{v} = -\mathbf{v} \in W$$

that is, Axiom 4. Thus, $W$ is a vector space.

Spans of Vectors

A span describes the total “reach” of a set of vectors through the operations of addition and scaling. It is the formal method by which we define a subspace $W$ within a larger vector space $V$.

To visualize this, consider the subspace $W$ of $2 \times 2$ matrices $M_2(\mathbb{R})$ defined by the condition that the upper-right entry is zero. Any matrix $A \in W$ can be decomposed into a linear combination of three basis-like elements:

$$ A = \begin{bmatrix} a & 0 \\ b & c \end{bmatrix} = a \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + b \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$

Thus,

$$W = \langle \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \rangle$$

Proof

We verify this using the formal two-step Subspace Test:

1. Non-emptiness (Zero Vector Inclusion):

By setting all scalars $\alpha_i = 0$, we see that

$$\mathbf{z} = 0v_1 + 0v_2 + \dots + 0v_n$$

Thus, $\mathbf{z} \in W$.

2. Closure under Addition and Scalar Multiplication:

Let $\mathbf{u}, \mathbf{w} \in W$ and $\alpha \in \mathbb{R}$. Then

$$\mathbf{u} = \sum_{i=1}^n \alpha_i v_i$$

and

$$\mathbf{w} = \sum_{i=1}^n \beta_i v_i$$

Addition:

$$\mathbf{u} + \mathbf{w} = (\alpha_1 + \beta_1)v_1 + (\alpha_2 + \beta_2)v_2 + \dots + (\alpha_n + \beta_n)v_n$$

Since $\alpha_i + \beta_i \in \mathbb{R}$, $\mathbf{u} + \mathbf{w} \in W$.

Scalar Multiplication:

$$\alpha \mathbf{u} = (\alpha \alpha_1)v_1 + (\alpha \alpha_2)v_2 + \dots + (\alpha \alpha_n)v_n$$

Since $\alpha \alpha_i \in \mathbb{R}$, $\alpha \mathbf{u} \in W$.

Different sets can generate identical subspaces. For example, in the polynomial space $\mathbb{R}[x]$, the sets $S_1 = {1, 1+x^2, 1+x^2+x^4}$ and $S_2 = {1, x^2, x^4}$ span the same space. This highlights the pedagogical “So What?”: we often have a choice in how we describe a space, and we generally seek the most efficient description.

Furthermore, $\langle v_1, \dots, v_n \rangle$ is the smallest (most restrictive) subspace containing those vectors. If any other subspace $W’$ contains $S$, it must also contain $\text{span}(S)$. This ensures that a span contains no elements other than those strictly required by the axioms of a vector space.

While the span describes the “reach” of a set, it does not account for redundancy. To determine the efficiency of our building blocks, we must turn to linear independence.

Linear Dependence and Independence

The following comparison illustrates how system solutions dictate set classification:

Example 15.18: Independence in $\mathbb{R}^3$ Let the set: $\{(1,1,1), (1,1,0), (0,1,1)\}$ We build the equation system:

$$a + b = 0$$$$a + b + c = 0$$$$a + c = 0$$

Subtracting (1) from (2) yields $c=0$, which implies $a=0$ and $b=0$. Only the trivial solution exists thus the set is linearly independent.

Note that independence is “hereditary”, removing elements from an independent set cannot create dependence. With these structural relationships established, we can now examine how linear transformations move these structures between spaces.

Linear Transformations

In $\mathbb{R}^n$, transformations are often expressed as matrix multiplications

$$ T(\mathbf{v}) = A\mathbf{v} $$

For example,

$$ T(a, b, c) = (2a+c, 3c-b) $$

corresponds to a $2 \times 3$ matrix.

$$ T(v) = \begin{bmatrix} 2 & 0 & 1 \\ 0 & -1 & 3 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 2a + c \\ 3c - b \end{bmatrix} $$

Proof:

1. Preservation of Addtition:

$$(T_2 \circ T_1)(\mathbf{u} + \mathbf{v}) = T_2(T_1(\mathbf{u} + \mathbf{v})) = T_2(T_1(\mathbf{u}) + T_1(\mathbf{v}))$$$$= T_2(T_1(\mathbf{u})) + T_2(T_1(\mathbf{v})) = (T_2 \circ T_1)(\mathbf{u}) + (T_2 \circ T_1)(\mathbf{v})$$

2. Preservation of Scalar Multiplication:

$$(T_2 \circ T_1)(\alpha \mathbf{u}) = T_2(T_1(\alpha \mathbf{u})) $$$$= T_2(\alpha T_1(\mathbf{u})) = \alpha T_2(T_1(\mathbf{u})) = \alpha (T_2 \circ T_1)(\mathbf{u})$$

This property is the reason matrix multiplication is defined as it is; the product $A_2A_1$ represents the composition $T_2 \circ T_1$.

Properties of Linear Transformations

Linearity imposes strict geometric constraints:

1. A linear map must fix the origin

$$ T(\mathbf{z}) = \mathbf{z}': T(0\mathbf{z}) = 0T(\mathbf{z}) = \mathbf{z}' $$

2. Inverting a vector in the domain results in an inverted image in the codomain

$$ T(-\mathbf{v}) = -T(\mathbf{v}) $$

The Kernel and Image

Proof

Since $T(\mathbf{z}) = \mathbf{z}’, \mathbf{z} \in \text{ker}(T)$

If $\mathbf{u}, \mathbf{v} \in \text{ker}(T)$, then

$$T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \mathbf{z}' + \mathbf{z}' = \mathbf{z}'$$

Similarly,

$$T(\alpha \mathbf{u}) = \alpha T(\mathbf{u}) = \alpha \mathbf{z}' = \mathbf{z}'$$

Example

Consider $T(a, b, c) = (2a+c, 3c-b)$. To find the kernel, solve $T(a, b, c) = (0, 0)$:

$$2a + c = 0 \implies a = -c/2$$$$3c - b = 0 \implies b = 3c$$

This yields vectors of the form $(-c/2, 3c, c)$, or $c(-1/2, 3, 1)$. The kernel is thus the subspace $\langle (-1/2, 3, 1) \rangle$. This identifies the specific dimension of the domain that is “lost” or collapsed to zero by the transformation.