Linear Combinations as Weighted Sums

Reading a Linear Combination

Suppose several vectors are available and each is assigned a scalar amount. Scaling each vector by its assigned amount and adding the resulting vectors produces a new vector. This is a linear combination. The phrase “weighted sum” gives a useful mental picture: the scalars are the weights, the vectors are what is being weighted, and the sum is the result. The picture is only an analogy, however. Coefficients may be negative or zero, and they do not have to add to one.

Throughout this tutorial, the scalars are real numbers and the vectors belong to \(\mathbb{R}^n\). Vectors are columns. We will use the vector addition and scalar multiplication rules for \(\mathbb{R}^n\): scaling a vector produces another vector in the same space, and vectors in that space can be added.

The order of the list makes the matching explicit: \(c_i\) is attached to the vector \(\mathbf{v}_i\). A list may include the same vector more than once, and each position still has its own coefficient. For example, in the list \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3)\), a coefficient \(c_2\) multiplies the second vector, not whichever vector happens to have a particular value.

The definition describes an expression and its parts, not a special kind of vector. Once the scalar multiples have been added, the result is simply another vector in \(\mathbb{R}^n\). When reading an expression, it helps to mark four things: the specified vectors, the scalar attached to each one, the individual scaled terms, and the final sum.

What the Coefficients Do

A coefficient changes the contribution of its associated vector. A coefficient of \(1\) leaves the vector unchanged. A coefficient of \(0\) makes that term the zero vector, so it contributes nothing to the sum. A negative coefficient reverses direction as well as scaling magnitude: for example, \(-2\mathbf{v}\) is the opposite of \(2\mathbf{v}\). These statements follow from the ordinary rules for scaling and adding vectors.

Proof. For any vector \(\mathbf{v}\), scalar multiplication gives \(0\mathbf{v}=\mathbf{0}\), \(1\mathbf{v}=\mathbf{v}\), and \((-1)\mathbf{v}=-\mathbf{v}\), where \(-\mathbf{v}\) denotes the additive inverse of \(\mathbf{v}\). Substituting any of these equalities for its term in a sum gives the stated contribution. \(\square\)

Coefficients are not required to be positive, and their sum is not required to be \(1\). A weighted average is a more restricted kind of weighted sum: it uses nonnegative weights whose total is \(1\). A linear combination has no such restriction. This distinction matters because a negative coefficient is perfectly valid in a linear combination.

Zero and Unit Coefficients

Some coefficient choices make the meaning of a combination especially direct. If all coefficients are zero, every term is \(\mathbf{0}\), so the result is \(\mathbf{0}\). If one coefficient is \(1\) and all the others are zero, the result is just the vector in the position with coefficient \(1\). These are valid combinations of the specified list, even when they are not the only possible choices of coefficients.

Proof. If every coefficient is zero, then every term \(c_i\mathbf{v}_i\) equals \(\mathbf{0}\). Adding these zero vectors gives \(\mathbf{0}\). Now fix a position \(k\), set \(c_k=1\), and set \(c_i=0\) for each \(i\ne k\). The \(k\)th term is \(1\mathbf{v}_k=\mathbf{v}_k\), while every other term is \(\mathbf{0}\). Adding \(\mathbf{v}_k\) to zero vectors leaves \(\mathbf{v}_k\). Both claims follow for any positive integer \(p\), including a list with just one vector. \(\square\)

Negative Weights and the Resulting Vector

A negative coefficient does not make an expression cease to be a linear combination. Its effect is handled within the scalar multiple: first form the term, then add that term to the others. This order of interpretation helps prevent a common sign error. The sign of a coefficient belongs to its vector’s term, not to a different term later in the sum.

Common Mistakes in Reading the Expression

Do not attach a coefficient to the wrong vector. In \(2\mathbf{v}_1-3\mathbf{v}_2\), the coefficient of \(\mathbf{v}_1\) is \(2\), and the coefficient of \(\mathbf{v}_2\) is \(-3\). Reversing them changes the expression.

Do not drop a negative sign before scaling. For example, if \(\mathbf{v}_1=\begin{bmatrix}1\\2\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}3\\1\end{bmatrix}\), then a wrong calculation of \(\mathbf{v}_1-2\mathbf{v}_2\) might use \(2\mathbf{v}_2\) instead of \(-2\mathbf{v}_2\), producing \(\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}6\\2\end{bmatrix}=\begin{bmatrix}7\\4\end{bmatrix}\). The correction is to scale by the full coefficient: \(-2\mathbf{v}_2=\begin{bmatrix}-6\\-2\end{bmatrix}\). Therefore the correct sum is \(\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}-6\\-2\end{bmatrix}=\begin{bmatrix}-5\\0\end{bmatrix}\).

Do not assume that “weighted sum” means “weighted average.” A linear combination allows any real coefficients. In the example \(\tfrac12\mathbf{a}-\tfrac12\mathbf{b}\), the second coefficient is negative and the two coefficients total \(0\), yet the expression is a valid linear combination.

When identifying a combination, keep the roles separate. The vectors are the listed inputs; the coefficients say how those inputs are scaled; the terms are the scaled vectors; and the result is their sum. This vocabulary applies whether the coefficients are positive, negative, zero, whole numbers or fractions.