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Linear Systems and Linear Combinations
Def. Let 𝑉 be a vector space and 𝑆 a nonempty subset of 𝑉. A vector 𝑣 ∈ 𝑉 is called a
linear combination of vectors in 𝑆 if there exists a finite number of vectors
𝑣1 , 𝑣2 , … , 𝑣𝑛 ∈ 𝑆 such that:
𝑣 = 𝑎1 𝑣1 + 𝑎2 𝑣2 + ⋯ + 𝑎𝑛 𝑣𝑛 .
In this case we say 𝑣 is a linear combination of 𝑣1 , 𝑣2 , … , 𝑣𝑛 and 𝑎1 , 𝑎2 , … , 𝑎𝑛 are
called the coefficients of the linear combination.
When working with vector spaces it’s very common to ask given a vector 𝑣 is it a
linear combination of a given set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 . Thus it’s useful to be able
to find an answer to this question. We’ll demonstrate this method through an
example.
Ex. Let 𝑣1 =< 1, 3, 2 > , 𝑣2 =< −4, −12, −8 > , 𝑣3 =< −1, 0, 4 >, and
𝑣4 =< 1, −3, −10 > be vectors in ℝ3 . Is the vector in ℝ3 given by
< 3, 12, 12 > a linear combination of 𝑣1 , 𝑣2 , 𝑣3 , and 𝑣4 ? If so, find a set of
coefficients 𝑎1 , 𝑎2 , 𝑎3 , and 𝑎4 such that :
< 3, 12, 12 >= 𝑎1 𝑣1 + 𝑎2 𝑣2 + 𝑎3 𝑣3 + 𝑎4 𝑣4 .
Let’s start by assuming we can find 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ∈ ℝ such that
< 3, 12, 12 >= 𝑎1 𝑣1 + 𝑎2 𝑣2 + 𝑎3 𝑣3 + 𝑎4 𝑣4
= 𝑎1 < 1,3,2 > +𝑎2 < −4, −12, −8 > +𝑎3 < −1,0,4 > +𝑎4 < 1, −3, −10 >
=< 𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 , (3𝑎1 − 12𝑎2 − 3𝑎4 ), 2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 >.
, 2
Since the corresponding components on each side of the equation must be
equal, we have the following simultaneous equations:
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
3𝑎1 − 12𝑎2 − 3𝑎4 = 12
2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 = 12.
Our goal is to see if we can find 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ∈ ℝ that will satisfy all three equations.
We start by adding or subtracting multiples of these equations so that the coefficient
of 𝑎1 in the first equation is one and the subsequent equations do NOT contain an 𝑎1
term. (Note: if the first equation doesn’t have an 𝑎1 term you can switch the order of
the equations so that the first equation does have an 𝑎1 term. At least one of the
three equations must have an 𝑎1 term.).
Notice that our first equation already has an 𝑎1 term with a coefficient of one (if the
non-zero coeffiecient is not one, divide the equation by the coefficient or switch the
order of the equations with one that does have a coefficient of one.). However, we
must eliminate the 𝑎1 term from equations two and three. We can eliminate the 𝑎1
term from equation two by multiplying equation one by −3 and adding it to equation
two.
−3𝑎1 + 12𝑎2 + 3𝑎3 − 3𝑎4 = −9
3𝑎1 − 12𝑎2 + −3𝑎4 = 12
3𝑎3 − 6𝑎4 = 3 (new equation two).
, 3
To eliminate the 𝑎1 term from equation three, multiply equation one by −2 and add
it to equation three.
−2𝑎1 + 8𝑎2 + 2𝑎3 − 2𝑎4 = −6
2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 = 12
6𝑎3 − 12𝑎4 = 6 (new equation three).
So now we have the following three equations:
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
3𝑎3 − 6𝑎4 = 3
6𝑎3 − 12𝑎4 = 6.
Notice that 𝑎2 disappeared from equations two and three (this does not happen in
general). Now we want to get the next highest index of 𝑎, in this case 𝑎3 , to have a
coefficient of 1 in the second equation and not appear in the third equation. Once
again, we can switch the order of equations two and three if the next highest index of
𝑎 only apears in equation three.
So we start by dividing equation two by 3 to make the coefficient of 𝑎3 be 1.
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
𝑎3 − 2𝑎4 = 1
6𝑎3 − 12𝑎4 = 6.
Now multiply equation two by −6 and add it to equation three.
−6𝑎3 + 12𝑎4 = −6
6𝑎3 − 12𝑎4 = 6
0 = 0. (equation 3 now drops out).
Linear Systems and Linear Combinations
Def. Let 𝑉 be a vector space and 𝑆 a nonempty subset of 𝑉. A vector 𝑣 ∈ 𝑉 is called a
linear combination of vectors in 𝑆 if there exists a finite number of vectors
𝑣1 , 𝑣2 , … , 𝑣𝑛 ∈ 𝑆 such that:
𝑣 = 𝑎1 𝑣1 + 𝑎2 𝑣2 + ⋯ + 𝑎𝑛 𝑣𝑛 .
In this case we say 𝑣 is a linear combination of 𝑣1 , 𝑣2 , … , 𝑣𝑛 and 𝑎1 , 𝑎2 , … , 𝑎𝑛 are
called the coefficients of the linear combination.
When working with vector spaces it’s very common to ask given a vector 𝑣 is it a
linear combination of a given set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 . Thus it’s useful to be able
to find an answer to this question. We’ll demonstrate this method through an
example.
Ex. Let 𝑣1 =< 1, 3, 2 > , 𝑣2 =< −4, −12, −8 > , 𝑣3 =< −1, 0, 4 >, and
𝑣4 =< 1, −3, −10 > be vectors in ℝ3 . Is the vector in ℝ3 given by
< 3, 12, 12 > a linear combination of 𝑣1 , 𝑣2 , 𝑣3 , and 𝑣4 ? If so, find a set of
coefficients 𝑎1 , 𝑎2 , 𝑎3 , and 𝑎4 such that :
< 3, 12, 12 >= 𝑎1 𝑣1 + 𝑎2 𝑣2 + 𝑎3 𝑣3 + 𝑎4 𝑣4 .
Let’s start by assuming we can find 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ∈ ℝ such that
< 3, 12, 12 >= 𝑎1 𝑣1 + 𝑎2 𝑣2 + 𝑎3 𝑣3 + 𝑎4 𝑣4
= 𝑎1 < 1,3,2 > +𝑎2 < −4, −12, −8 > +𝑎3 < −1,0,4 > +𝑎4 < 1, −3, −10 >
=< 𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 , (3𝑎1 − 12𝑎2 − 3𝑎4 ), 2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 >.
, 2
Since the corresponding components on each side of the equation must be
equal, we have the following simultaneous equations:
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
3𝑎1 − 12𝑎2 − 3𝑎4 = 12
2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 = 12.
Our goal is to see if we can find 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ∈ ℝ that will satisfy all three equations.
We start by adding or subtracting multiples of these equations so that the coefficient
of 𝑎1 in the first equation is one and the subsequent equations do NOT contain an 𝑎1
term. (Note: if the first equation doesn’t have an 𝑎1 term you can switch the order of
the equations so that the first equation does have an 𝑎1 term. At least one of the
three equations must have an 𝑎1 term.).
Notice that our first equation already has an 𝑎1 term with a coefficient of one (if the
non-zero coeffiecient is not one, divide the equation by the coefficient or switch the
order of the equations with one that does have a coefficient of one.). However, we
must eliminate the 𝑎1 term from equations two and three. We can eliminate the 𝑎1
term from equation two by multiplying equation one by −3 and adding it to equation
two.
−3𝑎1 + 12𝑎2 + 3𝑎3 − 3𝑎4 = −9
3𝑎1 − 12𝑎2 + −3𝑎4 = 12
3𝑎3 − 6𝑎4 = 3 (new equation two).
, 3
To eliminate the 𝑎1 term from equation three, multiply equation one by −2 and add
it to equation three.
−2𝑎1 + 8𝑎2 + 2𝑎3 − 2𝑎4 = −6
2𝑎1 − 8𝑎2 + 4𝑎3 − 10𝑎4 = 12
6𝑎3 − 12𝑎4 = 6 (new equation three).
So now we have the following three equations:
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
3𝑎3 − 6𝑎4 = 3
6𝑎3 − 12𝑎4 = 6.
Notice that 𝑎2 disappeared from equations two and three (this does not happen in
general). Now we want to get the next highest index of 𝑎, in this case 𝑎3 , to have a
coefficient of 1 in the second equation and not appear in the third equation. Once
again, we can switch the order of equations two and three if the next highest index of
𝑎 only apears in equation three.
So we start by dividing equation two by 3 to make the coefficient of 𝑎3 be 1.
𝑎1 − 4𝑎2 − 𝑎3 + 𝑎4 = 3
𝑎3 − 2𝑎4 = 1
6𝑎3 − 12𝑎4 = 6.
Now multiply equation two by −6 and add it to equation three.
−6𝑎3 + 12𝑎4 = −6
6𝑎3 − 12𝑎4 = 6
0 = 0. (equation 3 now drops out).
