Vectors in R^n

Vectors as Coordinate Tuples

In the previous tutorial, $\mathbb{R}^n$ was treated as a metric space: its points are ordered $n$-tuples, and the Euclidean metric measures the distance between them. The same tuples also have algebraic structure. We can add them and multiply them by real numbers, coordinate by coordinate. These operations are what allow vectors to encode directions, displacements, and combinations of quantities in multivariable analysis.

Throughout, $n$ is a positive integer. A vector in $\mathbb{R}^n$ is an ordered tuple $x=(x_1,\ldots,x_n)$ of real numbers. Two vectors are equal exactly when their corresponding coordinates are equal. Define addition and scalar multiplication by

x+y=(x_1+y_1,\ldots,x_n+y_n), \qquad \alpha x=(\alpha x_1,\ldots,\alpha x_n),

where $x,y\in\mathbb{R}^n$ and $\alpha\in\mathbb{R}$. The word “scalar” here means a real number multiplying a vector. These operations do not change the number of coordinates: the result of adding two vectors in $\mathbb{R}^n$, or multiplying one by a real scalar, is again in $\mathbb{R}^n$.

Definition: A vector in $\mathbb{R}^n$ is an ordered $n$-tuple of real numbers. Vector addition and scalar multiplication are defined coordinate by coordinate as above. The zero vector is $0=(0,\ldots,0)$, and the additive inverse of $x=(x_1,\ldots,x_n)$ is $-x=(-x_1,\ldots,-x_n)$.

The vector $0$ is distinguished from the real number $0$ by context: it is an $n$-tuple, while the scalar $0$ is a real number. The same notation is standard because the intended object is usually clear. The coordinate definition gives $x+0=x$ and $x+(-x)=0$. In particular, subtraction is defined by $x-y=x+(-y)$, so

x-y=(x_1-y_1,\ldots,x_n-y_n).

This subtraction agrees with the coordinate difference used to define Euclidean distance in the previous tutorial. Thus $d_2(x,y)=\|x-y\|_2$ measures the Euclidean length of the vector from $y$ to $x$. The algebraic operations and the metric therefore describe related aspects of the same coordinate space.

The Vector-Space Laws

The coordinate formulas inherit their basic algebraic properties from the real numbers. In particular, addition is associative and commutative, scalar multiplication distributes over vector addition, and multiplying by the scalar $1$ leaves a vector unchanged. These laws make $\mathbb{R}^n$, with the stated operations, a real vector space.

Theorem (Vector-Space Structure of $\mathbb{R}^n$): Coordinatewise addition and scalar multiplication make $\mathbb{R}^n$ a real vector space. In particular, for $x,y,z\in\mathbb{R}^n$ and $\alpha,\beta\in\mathbb{R}$, $$ x+y=y+x,\quad (x+y)+z=x+(y+z),\quad x+0=x,\quad x+(-x)=0, $$ $$ \alpha(x+y)=\alpha x+\alpha y,\quad (\alpha+\beta)x=\alpha x+\beta x,\quad \alpha(\beta x)=(\alpha\beta)x,\quad 1x=x. $$

Proof. Write $x=(x_i)_{i=1}^n$, $y=(y_i)_{i=1}^n$, and $z=(z_i)_{i=1}^n$. Equality of vectors is coordinatewise, so each vector identity follows by checking its $i$-th coordinate. For commutativity, the $i$-th coordinates of $x+y$ and $y+x$ are $x_i+y_i$ and $y_i+x_i$, which are equal by commutativity in $\mathbb{R}$. For associativity, the corresponding coordinates are $(x_i+y_i)+z_i$ and $x_i+(y_i+z_i)$, equal by associativity in $\mathbb{R}$. The coordinates of $x+0$ are $x_i+0=x_i$, and the coordinates of $x+(-x)$ are $x_i+(-x_i)=0$.

For distributivity over vector addition, the $i$-th coordinate of $\alpha(x+y)$ is $\alpha(x_i+y_i)$, which equals $\alpha x_i+\alpha y_i$. This is the $i$-th coordinate of $\alpha x+\alpha y$. The $i$-th coordinates of $(\alpha+\beta)x$ and $\alpha x+\beta x$ are $(\alpha+\beta)x_i$ and $\alpha x_i+\beta x_i$, which are equal by distributivity in $\mathbb{R}$. The coordinates of $\alpha(\beta x)$ and $(\alpha\beta)x$ are both $(\alpha\beta)x_i$, and the coordinates of $1x$ and $x$ are both $x_i$. This verifies all the displayed laws, and hence the vector-space structure. $\square$

Worked Example: Adding and Scaling Vectors

Let $u=(2,-1,4)$ and $v=(-3,5,1)$ in $\mathbb{R}^3$. Addition and scalar multiplication are computed in matching coordinates:

u+v=(2+(-3),-1+5,4+1)=(-1,4,5), $$ $$ 2u=(2\cdot2,2\cdot(-1),2\cdot4)=(4,-2,8).

For example, the second coordinate of $u+v$ is $-1+5=4$, while the second coordinate of $2u$ is $2(-1)=-2$. These operations do not combine coordinates in different positions; each output coordinate uses only the corresponding input coordinates.

Linear Combinations and Coordinate Vectors

Repeated addition and scalar multiplication can be written as one expression. Given vectors $v_1,\ldots,v_m\in\mathbb{R}^n$ and scalars $a_1,\ldots,a_m\in\mathbb{R}$, the vector $a_1v_1+\cdots+a_mv_m$ is called a linear combination of $v_1,\ldots,v_m$. Its coordinates are obtained by applying the same combination to each coordinate position.

Definition: If $v_1,\ldots,v_m\in\mathbb{R}^n$ and $a_1,\ldots,a_m\in\mathbb{R}$, then $a_1v_1+\cdots+a_mv_m$ is a linear combination of those vectors. If $v_j=(v_{j1},\ldots,v_{jn})$, its $i$-th coordinate is $\sum_{j=1}^{m}a_jv_{ji}$.

The most important coordinate vectors are the standard basis vectors. For each $i\in\{1,\ldots,n\}$, let $e_i$ be the vector whose $i$-th coordinate is $1$ and whose other coordinates are $0$. For instance, in $\mathbb{R}^3$,

e_1=(1,0,0),\qquad e_2=(0,1,0),\qquad e_3=(0,0,1).

Multiplying $e_i$ by a scalar places that scalar in the $i$-th coordinate and leaves all other coordinates zero. Adding these separate contributions reconstructs any vector. This is more than a convenient notation: the coefficients in this representation are forced by the coordinates of the vector.

Theorem (Unique Standard-Coordinate Representation): Every $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$ has the representation $$ x=x_1e_1+\cdots+x_ne_n, $$ and this representation is unique. That is, if $x=a_1e_1+\cdots+a_ne_n$, then $a_i=x_i$ for every $i$.

Proof. In the sum $x_1e_1+\cdots+x_ne_n$, the $i$-th coordinate receives a contribution $x_i$ from $x_ie_i$. Every other term $x_je_j$, with $j\ne i$, has $i$-th coordinate zero. Thus the $i$-th coordinate of the sum is $x_i$, for each $i$, and the sum equals $x$.

For uniqueness, suppose $x=a_1e_1+\cdots+a_ne_n$. The $i$-th coordinate of the right-hand side is $a_i$, because $a_ie_i$ contributes $a_i$ there and every $a_je_j$ with $j\ne i$ contributes zero. The $i$-th coordinate of the left-hand side is $x_i$. Equality of the vectors therefore gives $a_i=x_i$. This holds for each $i$, proving uniqueness. $\square$

Worked Example: Writing a Vector with Standard Coordinates

In $\mathbb{R}^4$, let $x=(3,-2,0,5)$. The standard coordinate vectors are $e_1=(1,0,0,0)$, $e_2=(0,1,0,0)$, $e_3=(0,0,1,0)$, and $e_4=(0,0,0,1)$. Therefore,

$$ 3e_1-2e_2+0e_3+5e_4 =(3,0,0,0)+(0,-2,0,0)+(0,0,0,0)+(0,0,0,5) =(3,-2,0,5). $$

The zero coefficient is included because the third coordinate of $x$ is zero. The uniqueness theorem says that no different list of four coefficients can produce this same vector using these standard coordinate vectors.

Using Linear Combinations

Linear combinations let us describe vectors formed from specified directions. The scalar coefficients determine how much of each direction is used, and negative coefficients reverse the corresponding contribution. The coordinate calculation is systematic: multiply each vector by its scalar, then add corresponding coordinates.

Worked Example: Combining Two Directions

In $\mathbb{R}^2$, take $u=(1,2)$, $v=(3,-1)$, and scalars $a=2$, $b=-1$. The linear combination $2u-v$ is

$$ 2u-v =2(1,2)-(3,-1) =(2,4)-(3,-1) =(2-3,4-(-1)) =(-1,5). $$

Checking by coordinates gives first coordinate $2\cdot1-3=-1$, and second coordinate $2\cdot2-(-1)=5$. The minus sign applies to the entire vector $v$, so its second coordinate contributes $+1$, not $-1$.

Worked Example: Recovering Coefficients

Suppose $x=(4,1,-2)\in\mathbb{R}^3$ is to be written as $a e_1+b e_2+c e_3$. Since

$$ a e_1+b e_2+c e_3=(a,b,c), $$

equality with $x=(4,1,-2)$ requires equality in all three coordinates. Hence $a=4$, $b=1$, and $c=-2$, so $x=4e_1+e_2-2e_3$. Substituting these values verifies the result:

$$ 4(1,0,0)+1(0,1,0)-2(0,0,1) =(4,0,0)+(0,1,0)+(0,0,-2) =(4,1,-2). $$

This coordinate check illustrates both existence and uniqueness: the coefficients produce the desired vector, and each coefficient is prescribed by its corresponding coordinate.

Why the Vector Viewpoint Matters

A point of $\mathbb{R}^n$ can be viewed as a location, while a vector can be viewed as a displacement or as a list of components to be combined. The tuple notation is the same in both cases, but addition and scalar multiplication make the vector interpretation explicit. For example, if $p,q\in\mathbb{R}^n$ are locations, then $q-p$ is the coordinate displacement from $p$ to $q$. Adding that displacement to $p$ gives $p+(q-p)=q$.

The standard coordinate vectors provide a reference direction for each coordinate. The unique representation theorem says that every vector has exactly one list of coefficients relative to these particular vectors. It does not say that a vector has a unique expression using every possible collection of vectors; uniqueness here follows from the special coordinate structure of $e_1,\ldots,e_n$. Keeping that qualification clear is useful when later studying other systems of vectors.

The dot product and Euclidean length introduced in the previous tutorial can now be applied to vectors built by these operations. For example, their coordinate definitions imply that the dot product distributes over vector addition and that scalar multiplication can be taken out of either argument. Such compatibility connects algebraic calculations with lengths, angles, and geometric analysis. The next tutorial develops norms, which generalize the idea of length while using the vector operations established here.

Check Your Understanding

Use the coordinate definitions and results in this tutorial to answer each question.

Compute $(2,-3,1)+(4,1,-5)$ and $3(2,-3,1)$.
What is the additive inverse of $(0,6,-2)$ in $\mathbb{R}^3$?
Write $(-1,4,2)\in\mathbb{R}^3$ as a linear combination of $e_1,e_2,e_3$.
If $a e_1+b e_2+c e_3=(5,0,-3)$, what must $a,b,c$ be?
Why does equality of two vectors in $\mathbb{R}^n$ reduce the proof of a vector identity to checking its coordinates?

What You'll Learn