Why a Coordinate Formula Helps
A linear combination is built by scaling vectors and adding the results. Once a vector has several coordinates, it is useful to know exactly how those operations affect each coordinate. The rule is that the \(i\)-th coordinate of the result comes from combining the \(i\)-th coordinates of the input vectors, using the same coefficients. This is not a separate operation or a shortcut with different rules: it follows directly from how addition and scalar multiplication are defined in \(\mathbb{R}^n\).
Throughout, the scalars are real numbers, and vectors in \(\mathbb{R}^n\) are written as columns. For a positive integer \(n\), write a vector \(\mathbf{v}\in\mathbb{R}^n\) as \(\mathbf{v}=(v_1,\dots,v_n)^{\top}\), where \(v_i\) denotes its \(i\)-th coordinate. Addition and scalar multiplication are defined coordinate by coordinate: if \(\mathbf{u}=(u_1,\dots,u_n)^{\top}\) and \(c\in\mathbb{R}\), then \(\mathbf{u}+\mathbf{v}=(u_1+v_1,\dots,u_n+v_n)^{\top}\) and \(c\mathbf{u}=(cu_1,\dots,cu_n)^{\top}\).
Deriving the Formula
The indices in \(v_{ki}\) keep track of two different things: \(k\) identifies which vector is being used, and \(i\) identifies the coordinate within that vector. To find coordinate \(i\) of the combination, hold \(i\) fixed and read that coordinate from every vector. The coefficient \(c_k\) stays paired with \(\mathbf{v}_k\) throughout.
Proof. First scale each vector \(\mathbf{v}_k\) by its coefficient \(c_k\). By the definition of scalar multiplication, coordinate \(i\) of \(c_k\mathbf{v}_k\) is \(c_kv_{ki}\). Now add the \(p\) scaled vectors. By the definition of vector addition, coordinate \(i\) of their sum is the sum of their coordinate-\(i\) entries, namely \(c_1v_{1i}+\cdots+c_pv_{pi}=\sum_{k=1}^{p}c_kv_{ki}\). This argument holds for each \(i=1,\dots,n\), so listing those coordinates in order gives exactly the vector displayed in the theorem. \(\square\)
The formula has two useful features. First, it preserves the order of the coordinates: the first output coordinate uses only first coordinates, the second uses only second coordinates, and so on. Second, the coefficients are not changed when moving from one coordinate to another. A coefficient belongs to a whole vector, so it multiplies every coordinate of that vector.
Worked Example: A Combination of Two Vectors in R³
Calculate \(3\mathbf{v}_1-2\mathbf{v}_2\) when \(\mathbf{v}_1=(1,-2,4)^{\top}\) and \(\mathbf{v}_2=(2,1,-1)^{\top}\). Then verify the result by scaling and adding the vectors.
Here the coefficients are \(c_1=3\) and \(c_2=-2\). Apply the coordinate formula, pairing first coordinates with first coordinates, second with second, and third with third:
For a check using vector operations, scale each vector separately: \(3\mathbf{v}_1=(3,-6,12)^{\top}\) and \(-2\mathbf{v}_2=(-4,-2,2)^{\top}\). Adding coordinate by coordinate gives \((3-4,-6-2,12+2)^{\top}=(-1,-8,14)^{\top}\), agreeing with the formula.
Applying the Rule with More Vectors
With more vectors, the principle is unchanged: for every output coordinate, multiply each vector’s entry in that position by its coefficient, then add those products. Writing the coordinate calculations separately makes it easier to keep each coefficient attached to the correct vector, especially when a coefficient is negative or fractional.
Worked Example: Three Vectors in R²
Calculate \(2\mathbf{u}_1-\mathbf{u}_2+\tfrac12\mathbf{u}_3\) for \(\mathbf{u}_1=(1,3)^{\top}\), \(\mathbf{u}_2=(-2,4)^{\top}\), and \(\mathbf{u}_3=(6,-2)^{\top}\). Verify the answer by first forming the scaled vectors.
The first coordinate is \(2(1)-(-2)+\tfrac12(6)=2+2+3=7\). The second coordinate is \(2(3)-4+\tfrac12(-2)=6-4-1=1\). Therefore the coordinate formula gives
To check, compute each scaled vector: \(2\mathbf{u}_1=(2,6)^{\top}\), \(-\mathbf{u}_2=(2,-4)^{\top}\), and \(\tfrac12\mathbf{u}_3=(3,-1)^{\top}\). Their sum is \((2+2+3,6-4-1)^{\top}=(7,1)^{\top}\), as claimed.
A coefficient of zero fits the same rule. Its vector contributes zero in every coordinate, because \(0v_{ki}=0\) for each \(i\). This is why a term with zero coefficient can be omitted from the calculation, but it is still useful to retain it while setting up the expression so that every coefficient remains matched to its original vector.
Worked Example: A Zero Coefficient in R⁴
Calculate \(-2\mathbf{w}_1+0\mathbf{w}_2+3\mathbf{w}_3\), where \(\mathbf{w}_1=(1,0,-2,3)^{\top}\), \(\mathbf{w}_2=(4,-1,2,0)^{\top}\), and \(\mathbf{w}_3=(-1,2,1,-2)^{\top}\).
Using the formula, calculate all four coordinates in order:
Thus the result is \((-5,6,7,-12)^{\top}\). Check it by scaling the vectors: \(-2\mathbf{w}_1=(-2,0,4,-6)^{\top}\), \(0\mathbf{w}_2=(0,0,0,0)^{\top}\), and \(3\mathbf{w}_3=(-3,6,3,-6)^{\top}\). Their sum is \((-2+0-3,0+0+6,4+0+3,-6+0-6)^{\top}=(-5,6,7,-12)^{\top}\).
Using Coordinates to Check Equality
The coordinate formula also gives a precise test for whether two proposed linear combinations are equal. Since vectors in \(\mathbb{R}^n\) are ordered lists, two such vectors are equal exactly when their coordinates match in every position. For combinations of the same vectors, this means comparing the scalar sums separately in each coordinate; the coefficients in the two combinations do not have to be the same.
Proof. Suppose the two combinations are equal. By the coordinate formula, their \(i\)-th coordinates are respectively \(\sum_{k=1}^{p}c_kv_{ki}\) and \(\sum_{k=1}^{p}d_kv_{ki}\). Equal vectors have equal coordinates in every position, so these scalar sums are equal for every \(i\). Conversely, suppose the scalar sums are equal for every \(i=1,\dots,n\). The coordinate formula says these sums are exactly the coordinates of the two combinations. Thus the combinations have equal coordinates in every position, so they are equal as vectors. \(\square\)
This test checks equality of the resulting vectors, not equality of the coefficients. Different coefficients can produce the same vector when the vectors in the combination allow it. The coordinate test establishes equality by comparing the actual outputs, coordinate by coordinate; it does not require making an assumption about whether the input vectors are independent.
Common Mistake: Mixing Coordinate Positions
A frequent error is to combine entries from different positions, or to use a coefficient with the wrong vector. For example, take \(\mathbf{a}=(1,2)^{\top}\), \(\mathbf{b}=(3,4)^{\top}\), and calculate \(2\mathbf{a}-\mathbf{b}\). A wrong calculation might use the first coordinate of \(\mathbf{a}\) with the second coordinate of \(\mathbf{b}\) to claim that the first output coordinate is \(2(1)-4=-2\). That calculation does not follow the coordinate formula: an output coordinate uses the same position from every vector.
The correct first coordinate uses the first entries \(1\) and \(3\), giving \(2(1)-3=-1\). The correct second coordinate uses \(2\) and \(4\), giving \(2(2)-4=0\). Hence \(2\mathbf{a}-\mathbf{b}=(-1,0)^{\top}\). Checking by scaling gives \(2\mathbf{a}=(2,4)^{\top}\) and \(-\mathbf{b}=(-3,-4)^{\top}\), whose sum is \((-1,0)^{\top}\).
To avoid this mistake, write the vectors with coordinates visible, keep their positions aligned, and calculate one output coordinate at a time. The number of output coordinates must equal the number of coordinates in each input vector. In the example above, there are two entries in each vector, so the result also has two entries.
Why the Formula Is Useful
The coordinate formula lets a vector calculation be translated into ordinary scalar arithmetic without changing the meaning of the expression. It is especially useful when checking a computed combination: one can either scale each whole vector and add, or compute the corresponding scalar combination in each coordinate. The two methods agree because both use the coordinate definitions of scaling and addition.
For a final check, confirm three things before accepting a result: each input vector has the same number of coordinates; each output coordinate uses only entries in that same position; and the coefficient multiplying an entry is the coefficient of that entry’s vector. Then, when the arithmetic is manageable, recompute by scaling the vectors separately and adding them. Agreement between the two calculations is a direct check that both the coordinate pairing and the arithmetic are correct.
Check Your Understanding
Use the coordinate formula in each question, and check at least one result by scaling and adding the vectors.
- For \(\mathbf{v}_1=(2,-1,0)^{\top}\) and \(\mathbf{v}_2=(-3,4,5)^{\top}\), calculate \(2\mathbf{v}_1+\mathbf{v}_2\) coordinate by coordinate.
- Calculate \(-\mathbf{u}_1+\tfrac12\mathbf{u}_2\) for \(\mathbf{u}_1=(2,6)^{\top}\) and \(\mathbf{u}_2=(8,-4)^{\top}\).
- Write the coordinate formula for \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3\) when each vector is in \(\mathbb{R}^3\).
- Explain why a vector with coefficient zero contributes zero to every coordinate of a linear combination.
- State the coordinate-by-coordinate condition that must hold for two linear combinations of the same vectors to be equal.