Tutorials › Linear Algebra › Computing Linear Combinations from Coefficients

Vectors and Linear Combinations · Tutorial 118 of 1000

Computing Linear Combinations from Coefficients

Practise scaling each listed vector by its coefficient, adding the results coordinatewise, and checking the final vector.

Beginner 9 min read

What You'll Learn

  • Identify each coefficient-vector pair in a linear combination
  • Scale every coordinate of a vector by its coefficient
  • Add the resulting vectors one coordinate at a time
  • Handle negative coefficients and zero coordinates correctly
  • Check a computed linear combination by recomputing its coordinates
  • Combine terms that use the same vector without changing the result

A Reliable Way to Evaluate a Linear Combination

A linear combination is a weighted sum of vectors: each vector is multiplied by its assigned scalar, and the resulting vectors are added. When the vectors and coefficients are given, the task is to carry out those operations accurately. A dependable calculation keeps the coefficient attached to the correct vector, scales that entire vector, and only then adds the results coordinate by coordinate.

Throughout, vectors are columns in \(\mathbb{R}^n\), and the coefficients are real numbers. For example, in \(2\mathbf{u}-3\mathbf{v}\), the coefficient \(2\) belongs to \(\mathbf{u}\), while the coefficient \(-3\) belongs to \(\mathbf{v}\). The minus sign is part of the second coefficient. It is often helpful to rewrite the expression as \(2\mathbf{u}+(-3)\mathbf{v}\), so that every term visibly consists of one coefficient and its vector.

Definition: Given vectors \(\mathbf{v}_1,\dots,\mathbf{v}_p\in\mathbb{R}^n\) and real numbers \(c_1,\dots,c_p\), the expression \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) is a linear combination of those vectors. To evaluate it, multiply each vector by its corresponding coefficient and add the resulting vectors.

Vector addition and scalar multiplication in \(\mathbb{R}^n\) are performed coordinatewise. Thus, to multiply a vector by a scalar, multiply every coordinate by that scalar. Then add the scaled vectors: add their first coordinates to get the first coordinate of the answer, their second coordinates to get the second coordinate, and continue through the list. Keep the coordinates in their original order throughout; the first coordinate of one vector must be added to the first coordinate of every other vector, not to a different position.

The Calculation in Three Stages

1
Match each coefficient to its vector.
Write down the terms in the order given. Include any minus sign in the coefficient, and do not move a coefficient onto a different vector.
2
Scale each vector.
Multiply every coordinate in that vector by its coefficient. A negative coefficient reverses the sign of each nonzero coordinate; zero coordinates remain zero. A zero coefficient makes the whole scaled vector the zero vector.
3
Add coordinatewise and check.
Add corresponding coordinates of the scaled vectors. Recompute each coordinate from the original terms to check that the final entries have the correct signs and values.

Keeping the stages separate helps locate mistakes. If the scaled vectors are correct but the final answer is not, the error occurred during addition. If a scaled vector is wrong, there is no need to redo the whole sum before correcting it. For exact arithmetic, retain integers and fractions instead of converting them to decimals. This makes signs and equalities easier to inspect.

Worked Example: Three Vectors with a Negative and a Fractional Coefficient

Evaluate \(2\mathbf{u}-3\mathbf{v}+\tfrac12\mathbf{w}\), where \(\mathbf{u}=\begin{bmatrix}1\\-2\\3\end{bmatrix}\), \(\mathbf{v}=\begin{bmatrix}2\\1\\-1\end{bmatrix}\), and \(\mathbf{w}=\begin{bmatrix}4\\0\\2\end{bmatrix}\).

First scale each vector by its own coefficient. The coefficient of \(\mathbf{v}\) is \(-3\), and the coefficient of \(\mathbf{w}\) is \(\tfrac12\):

$$ 2\mathbf{u}=\begin{bmatrix}2\\-4\\6\end{bmatrix}, \qquad (-3)\mathbf{v}=\begin{bmatrix}-6\\-3\\3\end{bmatrix}, \qquad \tfrac12\mathbf{w}=\begin{bmatrix}2\\0\\1\end{bmatrix}. $$

Now add corresponding coordinates. The first coordinates give \(2+(-6)+2=-2\). The second coordinates give \(-4+(-3)+0=-7\). The third coordinates give \(6+3+1=10\). Therefore,

$$ 2\mathbf{u}-3\mathbf{v}+\tfrac12\mathbf{w} = \begin{bmatrix}2\\-4\\6\end{bmatrix} + \begin{bmatrix}-6\\-3\\3\end{bmatrix} + \begin{bmatrix}2\\0\\1\end{bmatrix} = \begin{bmatrix}-2\\-7\\10\end{bmatrix}. $$

Check: Recompute the coordinates from the original terms. They are \(2(1)-3(2)+\tfrac12(4)=2-6+2=-2\), \(2(-2)-3(1)+\tfrac12(0)=-4-3+0=-7\), and \(2(3)-3(-1)+\tfrac12(2)=6+3+1=10\). These agree with the result.

The check can be done using the original expression without repeating every vector addition. In each coordinate, take that coordinate from each original vector, multiply by its coefficient, and add. This is a checking procedure for a particular calculation; it does not change the order of the main method, which is to scale the vectors and then add them.

Zero and Negative Coefficients

Some coefficients need special attention. A zero coefficient contributes a zero vector, regardless of the entries in the vector it multiplies. It is still useful to keep the term visible until the other terms have been matched correctly. A negative coefficient multiplies every coordinate, not just the coordinates that were positive. For instance, multiplying \(\begin{bmatrix}0\\2\end{bmatrix}\) by \(-1\) gives \(\begin{bmatrix}0\\-2\end{bmatrix}\): the zero coordinate stays zero.

Proposition: For any \(\mathbf{v}\in\mathbb{R}^n\), \(0\mathbf{v}=\mathbf{0}\). Consequently, a term with coefficient zero does not change the value of a linear combination.

Proof. The distributive law for scalar multiplication gives \((0+0)\mathbf{v}=0\mathbf{v}+0\mathbf{v}\). The left side equals \(0\mathbf{v}\), so this equation says \(0\mathbf{v}=0\mathbf{v}+0\mathbf{v}\). Add the additive inverse of \(0\mathbf{v}\) to both sides. The left side becomes \(\mathbf{0}\), and the right side becomes \(0\mathbf{v}\). Hence \(0\mathbf{v}=\mathbf{0}\). Adding a zero vector to any vector leaves it unchanged, so a zero-coefficient term does not affect the sum. \(\square\)

Worked Example: A Zero Coefficient and a Zero Coordinate

Evaluate \(-2\mathbf{a}+0\mathbf{b}+3\mathbf{c}\), where \(\mathbf{a}=\begin{bmatrix}0\\3\\-1\end{bmatrix}\), \(\mathbf{b}=\begin{bmatrix}4\\-2\\5\end{bmatrix}\), and \(\mathbf{c}=\begin{bmatrix}-1\\0\\2\end{bmatrix}\).

Scale all three vectors. In particular, the first coordinate of \(\mathbf{a}\) is zero and remains zero when multiplied by \(-2\):

$$ (-2)\mathbf{a}=\begin{bmatrix}0\\-6\\2\end{bmatrix}, \qquad 0\mathbf{b}=\begin{bmatrix}0\\0\\0\end{bmatrix}, \qquad 3\mathbf{c}=\begin{bmatrix}-3\\0\\6\end{bmatrix}. $$

Add the first coordinates: \(0+0+(-3)=-3\). Add the second coordinates: \(-6+0+0=-6\). Add the third coordinates: \(2+0+6=8\). Thus,

$$ -2\mathbf{a}+0\mathbf{b}+3\mathbf{c} = \begin{bmatrix}0\\-6\\2\end{bmatrix} + \begin{bmatrix}0\\0\\0\end{bmatrix} + \begin{bmatrix}-3\\0\\6\end{bmatrix} = \begin{bmatrix}-3\\-6\\8\end{bmatrix}. $$

Check: The first coordinate from the original terms is \((-2)(0)+0(4)+3(-1)=0+0-3=-3\). The second is \((-2)(3)+0(-2)+3(0)=-6+0+0=-6\). The third is \((-2)(-1)+0(5)+3(2)=2+0+6=8\). Each coordinate matches the computed vector.

Combining Terms That Use the Same Vector

Sometimes the same vector appears more than once. The terms can be combined by adding their coefficients, which may shorten the arithmetic. The coefficient addition must preserve the signs: for example, \(4\mathbf{u}-\mathbf{u}\) has combined coefficient \(4+(-1)=3\), so it equals \(3\mathbf{u}\).

Proposition: For any \(\mathbf{u}\in\mathbb{R}^n\) and scalars \(a,b\in\mathbb{R}\), \(a\mathbf{u}+b\mathbf{u}=(a+b)\mathbf{u}\).

Proof. The distributive law for scalar multiplication states that \((a+b)\mathbf{u}=a\mathbf{u}+b\mathbf{u}\). Since scalar addition is commutative, the right side can be written as \(a\mathbf{u}+b\mathbf{u}\) in either order. Therefore the two expressions are equal. This argument also applies when either coefficient is zero or negative, because the distributive law holds for all real scalars. \(\square\)

Worked Example: Repeated Vectors in a Linear Combination

Evaluate \(3\mathbf{u}-2\mathbf{v}+5\mathbf{u}+\tfrac12\mathbf{w}\), where \(\mathbf{u}=\begin{bmatrix}1\\2\end{bmatrix}\), \(\mathbf{v}=\begin{bmatrix}4\\-1\end{bmatrix}\), and \(\mathbf{w}=\begin{bmatrix}2\\6\end{bmatrix}\).

First combine the two terms involving \(\mathbf{u}\). Their coefficients add to \(3+5=8\), so the expression is \(8\mathbf{u}-2\mathbf{v}+\tfrac12\mathbf{w}\). Scale each vector:

$$ 8\mathbf{u}=\begin{bmatrix}8\\16\end{bmatrix}, \qquad (-2)\mathbf{v}=\begin{bmatrix}-8\\2\end{bmatrix}, \qquad \tfrac12\mathbf{w}=\begin{bmatrix}1\\3\end{bmatrix}. $$

Add first coordinates: \(8+(-8)+1=1\). Add second coordinates: \(16+2+3=21\). Therefore,

$$ 8\mathbf{u}-2\mathbf{v}+\tfrac12\mathbf{w} = \begin{bmatrix}8\\16\end{bmatrix} + \begin{bmatrix}-8\\2\end{bmatrix} + \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}1\\21\end{bmatrix}. $$

Check: Without combining repeated terms, the first coordinate is \(3(1)-2(4)+5(1)+\tfrac12(2)=3-8+5+1=1\). The second coordinate is \(3(2)-2(-1)+5(2)+\tfrac12(6)=6+2+10+3=21\). Both agree with the answer.

Combining repeated terms is optional. It can make a calculation shorter, but it adds a step at which coefficients can be added incorrectly. When the expression is short, scaling each term as written may be safer. When combining terms, first identify which vectors are exactly the same, then add only their coefficients, including their signs.

Common Mistakes and How to Correct Them

Mistake: changing the sign of a zero coordinate. A negative scalar reverses the sign of each nonzero coordinate, but zero has no positive or negative sign to reverse. A wrong calculation would be \((-1)\begin{bmatrix}0\\2\end{bmatrix}=\begin{bmatrix}-1\\-2\end{bmatrix}\). The first entry is wrong because \((-1)(0)=0\). The correction is

$$ (-1)\begin{bmatrix}0\\2\end{bmatrix} = \begin{bmatrix}(-1)(0)\\(-1)(2)\end{bmatrix} = \begin{bmatrix}0\\-2\end{bmatrix}. $$

Mistake: applying the coefficient to only one coordinate. For example, \(-2\begin{bmatrix}1\\3\end{bmatrix}\) is not \(\begin{bmatrix}-2\\3\end{bmatrix}\). The coefficient multiplies both entries, giving \(\begin{bmatrix}-2\\-6\end{bmatrix}\). Before adding vectors, check every entry of each scaled vector.

Mistake: adding unlike coordinates. If two vectors have two coordinates, add the first entries together and the second entries together. For \(\begin{bmatrix}2\\-1\end{bmatrix}+\begin{bmatrix}5\\4\end{bmatrix}\), the result is \(\begin{bmatrix}2+5\\-1+4\end{bmatrix}=\begin{bmatrix}7\\3\end{bmatrix}\), not a vector formed by adding entries across different positions.

Mistake: dropping a minus sign before scaling. The expression \(2\mathbf{u}-3\mathbf{v}\) has coefficients \(2\) and \(-3\), not \(2\) and \(3\). Writing \((-3)\mathbf{v}\) explicitly before multiplying helps prevent this error. After the computation, check each output coordinate against the original coefficients, as in the examples.

These checks are especially useful when a coordinate contains more than one negative number. A negative coefficient times a negative coordinate gives a positive entry, while a negative coefficient times a positive coordinate gives a negative entry. For instance, \((-3)(-2)=6\), whereas \((-3)(2)=-6\). The sign of the product depends on both signs, so it is worth writing each small multiplication before summing.

Check Your Understanding

For each question, show the scaled vectors before adding them. Check your result by evaluating each output coordinate from the original expression.

  1. Evaluate \(2\begin{bmatrix}1\\3\end{bmatrix}+(-1)\begin{bmatrix}4\\-2\end{bmatrix}\).
  2. Evaluate \(-3\begin{bmatrix}0\\2\\-1\end{bmatrix}+2\begin{bmatrix}1\\0\\4\end{bmatrix}\). Which coordinates remain zero during scaling?
  3. Evaluate \(\tfrac12\begin{bmatrix}6\\-4\end{bmatrix}+3\begin{bmatrix}-1\\2\end{bmatrix}\).
  4. Rewrite \(4\mathbf{u}-\mathbf{v}+2\mathbf{u}\) by combining repeated terms, then state the coefficient of \(\mathbf{u}\).
  5. Find and correct the error in the claim \((-2)\begin{bmatrix}0\\3\end{bmatrix}=\begin{bmatrix}-2\\-6\end{bmatrix}\).

Test yourself: Linear Algebra final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon