Vectors and vector spaces

Keep direction, magnitude and physical meaning together

A displacement describes a change of position with direction as well as magnitude. An arrow represents it: the tail marks the beginning and the head marks the end. We use A and B as names for vectors; |A| means the magnitude of A. A nonzero displacement has a direction. The zero displacement has magnitude zero and no unique direction.

Our original hypothetical case uses one chosen point in a fixed plane. The perpendicular x- and y-axes have a fixed origin; right is positive x and up is positive y. Positions and displacement components are expressed in metres. The supplied exact points are O = (0 m, 0 m), P = (2 m, 1 m) and Q = (3 m, 0 m). These are stipulated model values, not measurements.

A useful calculation must retain the quantity kind as well as units. Metrology distinguishes quantity kind from dimension: having the same dimension does not establish the same kind. Here we combine displacements in one frame. A displacement and a velocity cannot be added as though they were the same quantity; writing both as two-number pairs does not remove that distinction.

Read components in fixed axes

A pair lists the x-component first and the y-component second. For a displacement from a beginning point to an end point, subtract their corresponding coordinates: A_x = x_end − x_begin and A_y = y_end − y_begin. Thus O to P gives A = (2 m, 1 m), and P to Q gives B = (1 m, −1 m). The negative y-component of B means downward displacement in our chosen axes.

A signed component is a scalar quantity along a stated axis; its component vector points along that axis. In a perpendicular Cartesian frame, the magnitude is |A| = sqrt(A_x^2 + A_y^2). Here sqrt means the nonnegative square root. For our A, |A| = sqrt((2 m)^2 + (1 m)^2) = sqrt(5) m. Its positive components do not make its magnitude 3 m.

Dimensionless unit vectors e_x and e_y specify the positive axis directions, each with magnitude one. We can write A = (2 m)e_x + (1 m)e_y. The metre belongs to the scalar components, not to e_x or e_y. Components depend on the chosen basis; changing an axis direction changes the corresponding component signs without changing the represented displacement.

Add, subtract and scale compatible vectors

Use the same frame and compatible units, then add corresponding components. Our two successive displacements give R = A + B = (2 m + 1 m, 1 m − 1 m) = (3 m, 0 m). The head-to-tail construction places B's tail at A's head; R runs from the first tail to the final head. The arrows show net displacements, not the detailed paths or durations between the stipulated points.

A dimensionless real factor multiplies both components: 2A = (4 m, 2 m) and −A = (−2 m, −1 m). A negative factor reverses a nonzero vector's direction, while the magnitude is multiplied by the factor's absolute value. Multiplication by zero gives the zero vector. Subtraction means adding the opposite: A − B = A + (−B) = (1 m, 2 m).

Our resultant has magnitude 3 m, whereas the two magnitudes are sqrt(5) m and sqrt(2) m. Adding those magnitudes does not give |R|. Cancellation of the y-components explains the difference. Compute the vector sum first, then its magnitude; an arrow diagram must use the same physical length scale on both spatial axes to represent directions and lengths faithfully.

Build combinations and test independence

A linear combination is a sum of scalar multiples of vectors. Their span is the collection of all such results. A list is linearly independent when the only combination that gives the zero vector uses zero for every coefficient. These definitions refer to a chosen vector space and its allowed scalar numbers.

For our compatible displacement vectors, one original combination is 2A − B = (4 m − 1 m, 2 m + 1 m) = (3 m, 3 m). The factors 2 and −1 are dimensionless. If a coefficient carried physical units, the resulting quantity kind and units would need separate checking; it would not be the same operation by real numbers within the displacement space.

To study the algebra cleanly, now use numerical pairs with real entries. The pair (1, 0) can produce every pair (c, 0) by scaling, but it cannot produce (0, 1). The list (1, 0), (2, 0) is dependent because 2(1, 0) − (2, 0) = (0, 0) uses nonzero coefficients. Both stay on one line, so this list does not span all pairs in the plane.

A basis spans without redundancy

A basis is both spanning and linearly independent. Every vector then has unique coefficients in that basis. A finite-dimensional vector space has a finite basis; its dimension is the number of basis vectors. Here finite-dimensional concerns the number of independent algebraic directions, not a finite number of elements or a bounded region in physical space.

In the space R² of all ordered real pairs, define u = (1, 0) and w = (1, 1). Every pair can be written as (a, b) = (a − b)u + b w. Comparing the second components fixes the coefficient of w as b, and then the first fixes the coefficient of u as a − b. This proves existence and uniqueness directly, so u and w form a basis. They need not be perpendicular or have equal magnitudes.

The same numerical pair (2, 1) has coefficients (2, 1) in the usual basis (1, 0), (0, 1), but coefficients (1, 1) in our basis u, w. State the basis before interpreting coefficients. The vector-space dimension of R² is 2. A physical displacement represented by a pair still has the physical quantity dimension of length; the count 2 does not turn it into an area.

Vector spaces can contain objects beyond arrows

A real vector space is a set with addition and multiplication by real numbers, closed under both operations and satisfying linear rules: commutative and associative addition, a zero and additive opposites, distributive and associative scaling, and an identity factor one. Closure alone is insufficient. The word scalar here means an allowed real number, not an arbitrary physical scalar quantity with units.

For a small original abstract example, let the elements be all expressions p(z) = a + b z, where a and b are real numbers and z is a dimensionless real input. Include constant expressions and the zero expression. Add expressions by adding like coefficients, and multiply every coefficient by a real number. These operations remain in the collection and inherit the required linear rules from real arithmetic.

The expressions 1 and z form a basis: a + b z = a(1) + b(z). If c + d z = 0 for every z, evaluating at z = 0 gives c = 0 and evaluating at z = 1 then gives d = 0. Thus the basis is independent as well as spanning. This space has dimension 2 and infinitely many elements. Its vectors are expressions, not spatial arrows; the vector-space rules alone do not assign a physical direction, length or angle to them.

Worked examples

Try it yourself

1. A stipulated displacement is (−3 m, 4 m) in perpendicular axes with right and up positive. Is its magnitude negative? Find it and describe the component directions.

   Show the answer and reasoning

   Its magnitude is 5 m. The x-component points left and the y-component points up.

   The magnitude is sqrt((−3 m)^2 + (4 m)^2) = sqrt(25) m = 5 m. Signed components record axis directions; magnitude is nonnegative.

2. For the stipulated displacement A = (2 m, 1 m), find A + (−A). Does the result have a uniquely defined direction?

   Show the answer and reasoning

   The result is (0 m, 0 m), with zero magnitude and no unique direction.

   Both components cancel: (2 m − 2 m, 1 m − 1 m) = (0 m, 0 m). The zero vector remains part of the displacement algebra even though a direction cannot be inferred from it.

3. Can a displacement (1 m, 0 m) be added to a velocity (1 m/s, 0 m/s) merely because both use two-component notation? What must be checked before adding physical vectors?

   Show the answer and reasoning

   No. Check compatible quantity kinds, units and a common coordinate frame.

   Displacement and velocity have different meanings and units. The notation does not define their sum as a displacement or a velocity. For compatible displacements, convert component units consistently before adding them.

4. In R², consider u = (1, 2) and v = (2, 4). Are they independent? Can their span include (0, 1)?

   Show the answer and reasoning

   They are dependent, and their span does not include (0, 1).

   2u − v = (0, 0) gives zero with nonzero coefficients. Every combination is (c + 2d)(1, 2), whose second component is twice its first. The pair (0, 1) does not satisfy that relation.

5. With u = (1, 0) and w = (1, 1), find the coefficient pair for the vector (3, 2) in the ordered basis u, w. Is it the same as the usual numerical component pair?

   Show the answer and reasoning

   Its basis coefficient pair is (1, 2), which differs from the usual component pair (3, 2).

   1(1, 0) + 2(1, 1) = (3, 2). The basis determines how coefficients encode the vector; changing that basis does not change the vector being represented.

6. The real space of expressions a + b z has basis 1, z. Does dimension 2 mean it contains only two expressions? Does its vector-space structure alone assign a spatial direction to 2 − z?

   Show the answer and reasoning

   No to both. It has two basis vectors, infinitely many elements and no automatically assigned spatial direction.

   Real coefficients a and b give infinitely many expressions. Dimension counts the basis size. A physical arrow interpretation, or a chosen notion of length and angle, would require additional definitions.

What to carry forward

• State the physical quantity, units and coordinate frame before calculating with physical vectors.
• Signed components depend on fixed axes; a magnitude is nonnegative and zero has no unique direction.
• Add compatible components before taking a magnitude; subtract by adding the opposite.
• A span contains all linear combinations; independence excludes nontrivial combinations that give zero.
• A basis is independent and spanning, giving unique coefficients; vector-space dimension counts the basis vectors.
• An abstract vector space need not describe spatial arrows or automatically carry physical lengths and directions.