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Vectors and Linear Combinations · Tutorial 120 of 1000

Linear Combinations with Parameterized Coefficients

Practise turning variable coefficients into explicit coordinate formulas and checking those formulas at chosen parameter values.

Beginner 8 min read

What You'll Learn

  • Treat each parameter-dependent coefficient as a scalar once parameter values are assigned.
  • Expand a linear combination coordinate by coordinate.
  • Simplify expressions involving one or more parameters without losing signs.
  • Check a coordinate formula by evaluating both the original combination and the formula.
  • Distinguish a check at one parameter value from a formula valid for every value.

From Variable Coefficients to Coordinate Formulas

A linear combination usually begins with fixed scalar coefficients. But the coefficients can instead depend on variables such as \(t\) or \(s\). For each chosen value of those variables, the coefficients are still real numbers, so the same rules for adding vectors and scaling them apply. The task is to keep the parameters visible while simplifying each coordinate into an explicit expression.

Throughout, the vectors are in \(\mathbb{R}^n\), their entries are real, and vectors are written as columns. If \(\mathbf{v}_1,\ldots,\mathbf{v}_p\) are fixed vectors and \(f_1,\ldots,f_p\) are scalar expressions in parameters, then \(f_1\mathbf{v}_1+\cdots+f_p\mathbf{v}_p\) is a parameterized linear combination. For any fixed parameter values, each \(f_i\) evaluates to a real scalar. The result is a vector whose coordinates depend on those parameter values.

Coordinate rule: If the \(j\)th coordinate of \(\mathbf{v}_i\) is \(v_{ij}\), then the \(j\)th coordinate of \(f_1\mathbf{v}_1+\cdots+f_p\mathbf{v}_p\) is \(f_1v_{1j}+\cdots+f_pv_{pj}\). Apply this rule separately to every coordinate, then simplify the scalar expression obtained.

This is the coordinate formula for a linear combination applied when the coefficients are expressions rather than fixed numbers. A useful working order is to write one coordinate at a time, multiply each vector entry by its matching coefficient, expand parentheses, collect like terms, and assemble the resulting entries into a column. Keeping the terms separated until the multiplication is written helps prevent sign errors.

A Coordinate Rule for Parameterized Combinations

Proposition: Let \(\mathbf{v}_1,\ldots,\mathbf{v}_p\in\mathbb{R}^n\), and let \(f_1,\ldots,f_p\) be real-valued expressions in the same parameters. If the \(j\)th coordinate of \(\mathbf{v}_i\) is \(v_{ij}\), then the \(j\)th coordinate of the resulting combination is \(\sum_{i=1}^{p} f_i v_{ij}\), for every \(j=1,\ldots,n\).

Proof. Fix any values of the parameters. Each \(f_i\) then has a real value, so the coordinate formula for a linear combination applies to these fixed scalar values. In coordinate \(j\), multiplying \(\mathbf{v}_i\) by \(f_i\) gives the entry \(f_i v_{ij}\), and adding the scaled vectors adds their entries in that coordinate. Thus the \(j\)th entry of the sum is \(f_1v_{1j}+\cdots+f_pv_{pj}=\sum_{i=1}^{p} f_i v_{ij}\). Since the parameter values and coordinate \(j\) were arbitrary, the formula holds for every parameter choice and every coordinate. \(\square\)

The proposition explains why the calculation can be organized coordinate by coordinate: each coordinate has its own scalar expression to simplify. The vector entries are fixed numbers, while the parameters appear only through the coefficients. If a coefficient has a minus sign or parentheses, those signs must be carried through the scalar arithmetic for each coordinate.

Worked Example: Two Coefficients Depending on One Parameter

For \(\mathbf{u}=\begin{bmatrix}1\\-2\\3\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}4\\1\\-1\end{bmatrix}\), find the coordinate formula for \(t\mathbf{u}+(2t-1)\mathbf{v}\).

In the first coordinate, multiply the first entries of the vectors by their matching coefficients and add:

$$ t(1)+(2t-1)(4)=t+8t-4=9t-4. $$

In the second coordinate, the entry of \(\mathbf{u}\) is \(-2\), while the entry of \(\mathbf{v}\) is \(1\):

$$ t(-2)+(2t-1)(1)=-2t+2t-1=-1. $$

In the third coordinate, the entries are \(3\) and \(-1\):

$$ t(3)+(2t-1)(-1)=3t-2t+1=t+1. $$

Therefore the explicit coordinate formula is

$$ t\mathbf{u}+(2t-1)\mathbf{v} = \begin{bmatrix} 9t-4\\ -1\\ t+1 \end{bmatrix}. $$

Check: Set \(t=2\). The coefficients are \(2\) and \(3\), so the original combination is \(2\mathbf{u}+3\mathbf{v}=\begin{bmatrix}2\\-4\\6\end{bmatrix}+\begin{bmatrix}12\\3\\-3\end{bmatrix}=\begin{bmatrix}14\\-1\\3\end{bmatrix}\). The formula gives \(\begin{bmatrix}9(2)-4\\-1\\2+1\end{bmatrix}=\begin{bmatrix}14\\-1\\3\end{bmatrix}\), as required.

Keeping Several Parameters Organized

With more than one parameter, the same coordinate rule applies. Simplify each coordinate as its own algebraic expression; there is no need to combine terms from different coordinates. In particular, a coefficient such as \(s-t\) multiplies every entry of its vector, including entries that are negative.

Worked Example: Three Vectors and Two Parameters

Let \(\mathbf{u}=\begin{bmatrix}2\\0\\-1\end{bmatrix}\), \(\mathbf{v}=\begin{bmatrix}-1\\3\\2\end{bmatrix}\), and \(\mathbf{w}=\begin{bmatrix}0\\-2\\4\end{bmatrix}\). Find the coordinate formula for \((s+1)\mathbf{u}+2t\mathbf{v}+(s-t)\mathbf{w}\).

For the first coordinate, the entries of the three vectors are \(2,-1,0\):

$$ (s+1)(2)+2t(-1)+(s-t)(0) =2s+2-2t =2s-2t+2. $$

For the second coordinate, the entries are \(0,3,-2\):

$$ (s+1)(0)+2t(3)+(s-t)(-2) =6t-2s+2t =-2s+8t. $$

For the third coordinate, the entries are \(-1,2,4\):

$$ (s+1)(-1)+2t(2)+(s-t)(4) =-s-1+4t+4s-4t =3s-1. $$

Hence

$$ (s+1)\mathbf{u}+2t\mathbf{v}+(s-t)\mathbf{w} = \begin{bmatrix} 2s-2t+2\\ -2s+8t\\ 3s-1 \end{bmatrix}. $$

Check: Choose \(s=1\) and \(t=2\). The coefficients are \(2,4,-1\), so the original combination is \(2\mathbf{u}+4\mathbf{v}-\mathbf{w}\). Its coordinates are \((4-4-0,\;0+12-(-2),\;-2+8-4)=(0,14,2)\). The formula gives \((2-4+2,\;-2+16,\;3-1)=(0,14,2)\). Both calculations agree.

One useful feature of the coordinate rule is that simple coefficient expressions stay simple in a predictable way. For example, if each coefficient is a constant plus terms that are constant multiples of \(s\) and \(t\), then each output coordinate has that same form after collecting terms. Some terms may cancel, as they do in the second and third coordinates of the example, but no extra products of parameters are introduced by multiplying by fixed vector entries.

Proposition: If every coefficient in a parameterized linear combination is affine in \(s\) and \(t\), meaning it has the form \(a+bs+ct\) for real constants \(a,b,c\), then every output coordinate is also affine in \(s\) and \(t\).

Proof. Fix a coordinate \(j\). By the coordinate rule, its value is \(\sum_{i=1}^{p} f_i v_{ij}\), where each \(v_{ij}\) is a fixed real number. Write \(f_i=a_i+b_i s+c_i t\). Then this coordinate equals \(\sum_{i=1}^{p}(a_i+b_i s+c_i t)v_{ij}\). Distributing the fixed factors and collecting the constant, \(s\), and \(t\) terms gives \(\sum_{i=1}^{p}a_i v_{ij}+(\sum_{i=1}^{p}b_i v_{ij})s+(\sum_{i=1}^{p}c_i v_{ij})t\), which is affine in \(s\) and \(t\). Since the argument applies to every coordinate \(j\), every output coordinate is affine. \(\square\)

This proposition is a useful expectation to check against a result: when all coefficients are affine and the vectors are fixed, a term such as \(st\) in an output coordinate signals an algebraic error. It does not say that every output coordinate must contain both parameters; coefficients can cancel, or their combined coefficient for a parameter can be zero.

Worked Example: A Quadratic Coefficient

Let \(\mathbf{a}=\begin{bmatrix}1\\2\end{bmatrix}\), \(\mathbf{b}=\begin{bmatrix}-2\\1\end{bmatrix}\), and \(\mathbf{c}=\begin{bmatrix}3\\-1\end{bmatrix}\). Find the coordinate formula for \((2t-1)\mathbf{a}+t^2\mathbf{b}+(1-t)\mathbf{c}\).

The first coordinate is

$$ (2t-1)(1)+t^2(-2)+(1-t)(3) =2t-1-2t^2+3-3t =-2t^2-t+2. $$

The second coordinate is

$$ (2t-1)(2)+t^2(1)+(1-t)(-1) =4t-2+t^2-1+t =t^2+5t-3. $$

Thus the result is

$$ (2t-1)\mathbf{a}+t^2\mathbf{b}+(1-t)\mathbf{c} = \begin{bmatrix} -2t^2-t+2\\ t^2+5t-3 \end{bmatrix}. $$

Check: At \(t=1\), the coefficients are \(1,1,0\), so the original combination is \(\mathbf{a}+\mathbf{b}=\begin{bmatrix}-1\\3\end{bmatrix}\). Substituting \(t=1\) into the formula gives \(\begin{bmatrix}-2-1+2\\1+5-3\end{bmatrix}=\begin{bmatrix}-1\\3\end{bmatrix}\), confirming the calculation at this parameter value.

What a Check Can and Cannot Establish

Substitution is an effective way to catch mistakes. Choose parameter values, evaluate the coefficients in the original combination, calculate its coordinates, and compare with the proposed formula evaluated at the same values. A disagreement proves that at least one calculation is wrong. Agreement is a useful check, but a check at one choice of parameters does not by itself prove that two formulas agree for every choice.

Equality criterion: Two parameterized vectors in \(\mathbb{R}^n\) are equal for every parameter choice if and only if, for every coordinate, their coordinate expressions have equal values for every parameter choice.

Proof. Suppose first that the two vectors are equal for every parameter choice. Fix any parameter values. Equality of vectors in \(\mathbb{R}^n\) means equality of their corresponding coordinates, so the two expressions in each coordinate have equal values at those parameter values. Since the choice was arbitrary, this holds for every parameter choice. Conversely, suppose the coordinate expressions have equal values for every parameter choice, in every coordinate. Fix any parameter values. All corresponding coordinates of the two resulting vectors are then equal. By the definition of equality of vectors in \(\mathbb{R}^n\), the vectors are equal for those values. Because the values were arbitrary, they are equal for every parameter choice. \(\square\)

The word “every” matters in this criterion. For example, the scalar expressions \(t\) and \(0\) agree when \(t=0\), but they do not agree for every real \(t\), since at \(t=1\) their values are \(1\) and \(0\). Checking a candidate coordinate formula at several values can raise confidence and locate errors, but those checks are not a substitute for carrying out and simplifying the coordinate calculation.

Common Mistake: Losing a Negative Sign

When a vector entry is negative, the entire coefficient must still be multiplied by that entry. In the second coordinate of the three-vector example, the term involving \(\mathbf{w}\) is \((s-t)(-2)\). A mistaken distribution is \(-2s-2t\), which treats the subtraction inside \(s-t\) as addition. The correct calculation is

$$ (s-t)(-2)=-2s+2t. $$

Using the incorrect sign would give \(6t-2s-2t=-2s+4t\) for that coordinate, rather than the correct \(-2s+8t\). At \(s=1,t=2\), the incorrect expression gives \(6\), while direct evaluation of the original second coordinate gives \(14\). The correction is to distribute the negative factor to both terms, preserving \((-2)(-t)=2t\).

A second useful safeguard is to keep the coordinate entries aligned with their vectors. The coefficient \(2t\) multiplies every coordinate of \(\mathbf{v}\), not just its first coordinate; likewise, a zero entry contributes zero only in that particular coordinate. Writing the separate scalar calculation for each coordinate makes these choices visible.

Check Your Understanding

For each question, write the output as a column vector of simplified coordinate expressions. Where asked, verify by substituting the stated parameter values.

  1. For \(\mathbf{u}=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\3\end{bmatrix}\), find \(t\mathbf{u}+(t+2)\mathbf{v}\).
  2. For \(\mathbf{a}=\begin{bmatrix}1\\0\\2\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}-1\\4\\1\end{bmatrix}\), find \((s-1)\mathbf{a}+3s\mathbf{b}\).
  3. For \(\mathbf{p}=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\mathbf{q}=\begin{bmatrix}0\\3\end{bmatrix}\), find \(t^2\mathbf{p}+(1-t)\mathbf{q}\), and check at \(t=2\).
  4. Explain why checking two parameter values alone does not establish that two coordinate expressions agree for every real parameter value.

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