Linear combination

Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R³. Consider the vectors e₁ = (1,0,0), e₂ = (0,1,0) and e₃ = (0,0,1). Then any vector in R³ is a linear combination of e₁, e₂, and e₃.

To see that this is so, take an arbitrary vector (a₁,a₂,a₃) in R³, and write:

${\displaystyle {\begin{aligned}(a_{1},a_{2},a_{3})&=(a_{1},0,0)+(0,a_{2},0)+(0,0,a_{3})\\[6pt]&=a_{1}(1,0,0)+a_{2}(0,1,0)+a_{3}(0,0,1)\\[6pt]&=a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+a_{3}\mathbf {e} _{3}.\end{aligned}}}$

Let K be the set C of all complex numbers, and let V be the set C_C(R) of all continuous functions from the real line R to the complex plane C. Consider the vectors (functions) f and g defined by f(t) := e^it and g(t) := e^−it. (Here, e is the base of the natural logarithm, about 2.71828..., and i is the imaginary unit, a square root of −1.) Some linear combinations of f and g are:

• ${\displaystyle \cos t={\tfrac {1}{2}}\,e^{it}+{\tfrac {1}{2}}\,e^{-it}}$
• ${\displaystyle 2\sin t=(-i)e^{it}+(i)e^{-it}.}$

On the other hand, the constant function 3 is not a linear combination of f and g. To see this, suppose that 3 could be written as a linear combination of e^it and e^−it. This means that there would exist complex scalars a and b such that ae^it + be^−it = 3 for all real numbers t. Setting t = 0 and t = π gives the equations a + b = 3 and a + b = −3, and clearly this cannot happen. See Euler's identity.

Let K be R, C, or any field, and let V be the set P of all polynomials with coefficients taken from the field K. Consider the vectors (polynomials) p₁ := 1, p₂ := x + 1, and p₃ := x² + x + 1.

Is the polynomial x² − 1 a linear combination of p₁, p₂, and p₃? To find out, consider an arbitrary linear combination of these vectors and try to see when it equals the desired vector x² − 1. Picking arbitrary coefficients a₁, a₂, and a₃, we want

${\displaystyle a_{1}(1)+a_{2}(x+1)+a_{3}(x^{2}+x+1)=x^{2}-1.}$

Multiplying the polynomials out, this means

${\displaystyle (a_{1})+(a_{2}x+a_{2})+(a_{3}x^{2}+a_{3}x+a_{3})=x^{2}-1}$

and collecting like powers of x, we get

${\displaystyle a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})=1x^{2}+0x+(-1).}$

Two polynomials are equal if and only if their corresponding coefficients are equal, so we can conclude

${\displaystyle a_{3}=1,\quad a_{2}+a_{3}=0,\quad a_{1}+a_{2}+a_{3}=-1.}$

This system of linear equations can easily be solved. First, the first equation simply says that a₃ is 1. Knowing that, we can solve the second equation for a₂, which comes out to −1. Finally, the last equation tells us that a₁ is also −1. Therefore, the only possible way to get a linear combination is with these coefficients. Indeed,

${\displaystyle x^{2}-1=-1-(x+1)+(x^{2}+x+1)=-p_{1}-p_{2}+p_{3}}$

so x² − 1 is a linear combination of p₁, p₂, and p₃.

On the other hand, what about the polynomial x³ − 1? If we try to make this vector a linear combination of p₁, p₂, and p₃, then following the same process as before, we get the equation

${\displaystyle {\begin{aligned}&0x^{3}+a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})\\[5pt]={}&1x^{3}+0x^{2}+0x+(-1).\end{aligned}}}$

However, when we set corresponding coefficients equal in this case, the equation for x³ is

${\displaystyle 0=1}$

which is always false. Therefore, there is no way for this to work, and x³ − 1 is not a linear combination of p₁, p₂, and p₃.

• Axler, Sheldon Jay (2015). Linear Algebra Done Right (3rd ed.). Springer. ISBN 978-3-319-11079-0.
• Katznelson, Yitzhak; Katznelson, Yonatan R. (2008). A (Terse) Introduction to Linear Algebra. American Mathematical Society. ISBN 978-0-8218-4419-9.
• Lay, David C.; Lay, Steven R.; McDonald, Judi J. (2016). Linear Algebra and its Applications (5th ed.). Pearson. ISBN 978-0-321-98238-4.
• Strang, Gilbert (2016). Introduction to Linear Algebra (5th ed.). Wellesley Cambridge Press. ISBN 978-0-9802327-7-6.

• "Linear Combinations". nLab. 27 October 2015. Retrieved 16 Feb 2021.