Vectors in \(\mathbb{R}^n\) and Their Operations
The familiar rules for adding arrows and scaling them can be expressed exactly in coordinates. This makes \(\mathbb{R}^n\) a basic example of a real vector space: its elements are ordered lists of \(n\) real numbers, and its operations act on corresponding coordinates. Verifying this example connects the abstract vector space axioms to calculations that can be performed directly.
Fix \(n\in\mathbb{N}\). An element of \(\mathbb{R}^n\) is a column vector \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), where every \(x_i\in\mathbb{R}\). Two vectors in \(\mathbb{R}^n\) are equal exactly when their corresponding coordinates are equal. We use the usual addition and scalar multiplication of real numbers to define operations on these columns.
The operations preserve the number of coordinates: adding two \(n\)-component columns gives another \(n\)-component column, and multiplying an \(n\)-component column by a real number also gives an \(n\)-component column. In particular, both results belong to \(\mathbb{R}^n\). This is the closure requirement for the operations.
Worked Example: Adding and Scaling Columns
Let \(\mathbf{x}=(2,-1,0,3)^{\top}\) and \(\mathbf{y}=(-4,5,2,1)^{\top}\). Compute \(\mathbf{x}+\mathbf{y}\) and \(-2\mathbf{y}\).
Add corresponding coordinates in order:
For scalar multiplication, multiply every coordinate of \(\mathbf{y}\) by \(-2\):
Check: each answer has four real coordinates, so both are in \(\mathbb{R}^4\). For example, adding the displayed sum to \((2,-1,0,3)^{\top}\) gives \((0,3,2,7)^{\top}\), which is the coordinatewise sum required by the definition.
Verifying the Vector Space Axioms
For a real vector space, addition must be associative and commutative, there must be an additive identity and additive inverses, and scalar multiplication must satisfy the compatibility and distributive rules. The scalar \(1\) must act as an identity. We verify each condition for \(\mathbb{R}^n\). The real-number laws used in the verification hold in every coordinate.
Proof. Let \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), \(\mathbf{y}=(y_1,\dots,y_n)^{\top}\), and \(\mathbf{z}=(z_1,\dots,z_n)^{\top}\) be arbitrary vectors in \(\mathbb{R}^n\), and let \(a,b\in\mathbb{R}\). Closure follows from the definitions: each coordinate of \(\mathbf{x}+\mathbf{y}\) is a sum of real numbers, and each coordinate of \(a\mathbf{x}\) is a product of real numbers, so both results are in \(\mathbb{R}^n\).
For addition, the \(i\)-th coordinate of \((\mathbf{x}+\mathbf{y})+\mathbf{z}\) is \((x_i+y_i)+z_i\), while the \(i\)-th coordinate of \(\mathbf{x}+(\mathbf{y}+\mathbf{z})\) is \(x_i+(y_i+z_i)\). These are equal by associativity of real addition, for every \(i\), so vector addition is associative. The \(i\)-th coordinates of \(\mathbf{x}+\mathbf{y}\) and \(\mathbf{y}+\mathbf{x}\) are \(x_i+y_i\) and \(y_i+x_i\), equal by commutativity of real addition. Thus vector addition is commutative.
Define \(\mathbf{0}=(0,\dots,0)^{\top}\). The \(i\)-th coordinate of \(\mathbf{x}+\mathbf{0}\) is \(x_i+0=x_i\), so \(\mathbf{x}+\mathbf{0}=\mathbf{x}\). Define \(-\mathbf{x}=(-x_1,\dots,-x_n)^{\top}\). The \(i\)-th coordinate of \(\mathbf{x}+(-\mathbf{x})\) is \(x_i+(-x_i)=0\), hence \(\mathbf{x}+(-\mathbf{x})=\mathbf{0}\). These establish the additive identity and an additive inverse for each vector.
For scalar multiplication, the \(i\)-th coordinates of \(a(b\mathbf{x})\) and \((ab)\mathbf{x}\) are \(a(bx_i)\) and \((ab)x_i\), equal by associativity of real multiplication. Hence \(a(b\mathbf{x})=(ab)\mathbf{x}\). The \(i\)-th coordinate of \(1\mathbf{x}\) is \(1x_i=x_i\), so \(1\mathbf{x}=\mathbf{x}\).
Finally, the \(i\)-th coordinates of \(a(\mathbf{x}+\mathbf{y})\) and \(a\mathbf{x}+a\mathbf{y}\) are \(a(x_i+y_i)\) and \(ax_i+ay_i\), equal by distributivity in \(\mathbb{R}\). The \(i\)-th coordinates of \((a+b)\mathbf{x}\) and \(a\mathbf{x}+b\mathbf{x}\) are \((a+b)x_i\) and \(ax_i+bx_i\), also equal by real distributivity. Since the corresponding coordinates agree for every \(i\), the vectors agree. All vector space axioms hold. \(\square\)
The proof uses an important feature of coordinate vectors: to verify a vector identity, it is enough to compare the coordinates on both sides. The vector axioms do not require a new kind of arithmetic; each one reduces to a familiar real-number law in each coordinate.
The Zero Vector and Additive Inverses
The zero vector in \(\mathbb{R}^n\) is not a single real number. It is the column with zero in every coordinate. Likewise, the additive inverse changes the sign of every coordinate. The relevant identities can be checked directly: \(\mathbf{x}+\mathbf{0}=\mathbf{x}\) and \(\mathbf{x}+(-\mathbf{x})=\mathbf{0}\). In particular, a negative coordinate becomes positive in the inverse, and a positive coordinate becomes negative.
Worked Example: The Zero Vector and an Additive Inverse
For \(\mathbf{u}=(-3,0,5,-2)^{\top}\in\mathbb{R}^4\), write down \(\mathbf{0}\) and \(-\mathbf{u}\), then check both defining identities.
The zero vector has four zero coordinates, and the inverse changes each sign:
Check the identity operation coordinate by coordinate:
Now check the inverse:
Check: each coordinate sum in the inverse calculation is zero, including the second coordinate, where both entries were already zero.
The inverse displayed in the theorem is the unique additive inverse. Indeed, if \(\mathbf{w}\) also satisfies \(\mathbf{x}+\mathbf{w}=\mathbf{0}\), then in each coordinate \(x_i+w_i=0\), so \(w_i=-x_i\). Thus \(\mathbf{w}=(-x_1,\dots,-x_n)^{\top}=-\mathbf{x}\). Similarly, the displayed zero vector is the unique additive identity: if \(\mathbf{z}\) is an identity, then \(\mathbf{z}+\mathbf{0}=\mathbf{0}\) by the identity property for \(\mathbf{z}\), but \(\mathbf{z}+\mathbf{0}=\mathbf{z}\) by the definition of \(\mathbf{0}\), so \(\mathbf{z}=\mathbf{0}\).
Combining the Operations
Scalar multiplication can be used together with vector addition to form expressions such as \(a\mathbf{x}+b\mathbf{y}\). The definition gives a direct coordinate rule: the \(i\)-th coordinate is \(ax_i+by_i\). This rule is consistent with the vector space axioms because scalar multiplication distributes over addition, and addition is performed componentwise.
Worked Example: Solving a Vector Equation
Let \(\mathbf{a}=(1,-2,3)^{\top}\) and \(\mathbf{b}=(5,0,-1)^{\top}\). Solve \(2\mathbf{x}+\mathbf{a}=\mathbf{b}\) in \(\mathbb{R}^3\).
Write \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\). The left side is \((2x_1+1,\,2x_2-2,\,2x_3+3)^{\top}\). Equality of vectors means equality in each coordinate, so
Solving these three real equations gives \(2x_1=4\), \(x_1=2\); \(2x_2=2\), \(x_2=1\); and \(2x_3=-4\), \(x_3=-2\). Therefore \(\mathbf{x}=(2,1,-2)^{\top}\).
Check:
Common Mistake: Applying a Scalar to Only Some Coordinates
Scalar multiplication applies the scalar to every coordinate. For example, a tempting but incorrect calculation is \(-2(1,-3,2)^{\top}=(-2,-3,2)^{\top}\), as though only the first coordinate were scaled. The definition requires multiplying each entry:
The incorrect answer fails a direct check: its second coordinate should be \((-2)(-3)=6\), not \(-3\), and its third coordinate should be \((-2)2=-4\), not \(2\). A reliable procedure is to write the scalar next to each coordinate before multiplying, especially when negative signs are present.
Why This Example Matters
The standard operations on \(\mathbb{R}^n\) are concrete models for the vector space axioms. Addition combines corresponding entries; scalar multiplication scales them all at once; and the zero and inverse are obtained coordinate by coordinate. The verification also explains why an equality of vectors in \(\mathbb{R}^n\) can be checked through its coordinates: every required identity reduces to an identity of real numbers.
Check Your Understanding
Use the componentwise definitions to calculate and check each answer.
- For \(\mathbf{v}=(4,-1,2)^{\top}\) and \(\mathbf{w}=(-3,5,0)^{\top}\), find \(\mathbf{v}+\mathbf{w}\) and \(3\mathbf{v}\).
- Write the zero vector in \(\mathbb{R}^5\), and find the additive inverse of \((2,0,-4,1,-3)^{\top}\).
- Verify coordinate by coordinate that \(-(\mathbf{x}+\mathbf{y})=(-\mathbf{x})+(-\mathbf{y})\) for \(\mathbf{x}=(1,-2)^{\top}\) and \(\mathbf{y}=(3,4)^{\top}\).
- Solve \(3\mathbf{x}+(1,2)^{\top}=(10,-1)^{\top}\) in \(\mathbb{R}^2\), then substitute your answer into the equation.
- In one sentence, explain why checking the corresponding coordinates is enough to establish equality of two vectors in \(\mathbb{R}^n\).
