User:Prof McCarthy/Linear independence

Evaluating linear dependence

Vectors in R²

Three vectors: Consider the set of vectors v₁= (1, 1), v₂= (-3, 2) and v₃= (2, 4), then the condition for linear dependence is a set of non-zero scalars, such that

$a_{1}{\begin{Bmatrix}1\\1\end{Bmatrix}}+a_{2}{\begin{Bmatrix}-3\\2\end{Bmatrix}}+a_{3}{\begin{Bmatrix}2\\4\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}},$

or

${\begin{bmatrix}1&-3&2\\1&2&4\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.$

Row reduce this matrix equation by subtracting the first equation from the second to obtain,

${\begin{bmatrix}1&-3&2\\0&5&2\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.$

Continue the row reduction by (i) dividing the second equation by 5, and then (ii) multiplying by 3 and adding to the first equation, that is

${\begin{bmatrix}1&0&22/5\\0&1&4/5\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.$

We can now rearrange this equation to obtain

${\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}=-a_{3}{\begin{Bmatrix}22/5\\4/5\end{Bmatrix}}.$

which shows that non-zero a_i exist so v₃= (2, 4) can be defined in terms of v₁= (1, 1), v₂= (-3, 2). Thus, the three vectors are linearly dependent.

Two vectors: Now consider the linear dependence of the two vectors v₁= (1, 1), v₂= (-3, 2), and check,

$a_{1}{\begin{Bmatrix}1\\1\end{Bmatrix}}+a_{2}{\begin{Bmatrix}-3\\2\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}},$

or

${\begin{bmatrix}1&-3\\1&2\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.$

The same row reduction presented above yields

${\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.$

which shows that non-zero a_i do not exist so v₁= (1, 1) and v₂= (-3, 2) are linearly independent.

Alternative method using determinants

An alternative method uses the fact that n vectors in $\mathbb {R} ^{n}$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero.

In this case, the matrix formed by the vectors is

$A={\begin{bmatrix}1&-3\\1&2\end{bmatrix}}.\,\!$

We may write a linear combination of the columns as

$A\Lambda ={\begin{bmatrix}1&-3\\1&2\end{bmatrix}}{\begin{bmatrix}\lambda _{1}\\\lambda _{2}\end{bmatrix}}.\,\!$

We are interested in whether AΛ = 0 for some nonzero vector Λ. This depends on the determinant of A, which is

$\det A=1\cdot 2-1\cdot (-3)=5\neq 0.\,\!$

Since the determinant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.

Otherwise, suppose we have m vectors of n coordinates, with m < n. Then A is an n×m matrix and Λ is a column vector with m entries, and we are again interested in AΛ = 0. As we saw previously, this is equivalent to a list of n equations. Consider the first m rows of A, the first m equations; any solution of the full list of equations must also be true of the reduced list. In fact, if 〈i₁,...,i_m〉 is any list of m rows, then the equation must be true for those rows.

$A_{{\langle i_{1},\dots ,i_{m}}\rangle }\Lambda =\mathbf {0} .\,\!$

Furthermore, the reverse is true. That is, we can test whether the m vectors are linearly dependent by testing whether

$\det A_{{\langle i_{1},\dots ,i_{m}}\rangle }=0\,\!$

for all possible lists of m rows. (In case m = n, this requires only one determinant, as above. If m > n, then it is a theorem that the vectors must be linearly dependent.) This fact is valuable for theory; in practical calculations more efficient methods are available.

Example II

Let V = Rⁿ and consider the following elements in V:

${\begin{matrix}\mathbf {e} _{1}&=&(1,0,0,\ldots ,0)\\\mathbf {e} _{2}&=&(0,1,0,\ldots ,0)\\&\vdots \\\mathbf {e} _{n}&=&(0,0,0,\ldots ,1).\end{matrix}}$

Then e₁, e₂, ..., e_n are linearly independent.

Proof

Suppose that a₁, a₂, ..., a_n are elements of R such that

$a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+\cdots +a_{n}\mathbf {e} _{n}=0.\,\!$

Since

$a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+\cdots +a_{n}\mathbf {e} _{n}=(a_{1},a_{2},\ldots ,a_{n}),\,\!$

then a_i = 0 for all i in {1, ..., n}.

Example III

Let V be the vector space of all functions of a real variable t. Then the functions e^t and e^2t in V are linearly independent.

Proof

Suppose a and b are two real numbers such that

ae^t + be^2t = 0

for all values of t. We need to show that a = 0 and b = 0. In order to do this, we divide through by e^t (which is never zero) and subtract to obtain

be^t = −a.

In other words, the function be^t must be independent of t, which only occurs when b = 0. It follows that a is also zero.

Example IV

The following vectors in R⁴ are linearly dependent.

${\begin{matrix}\\{\begin{bmatrix}1\\4\\2\\-3\end{bmatrix}},{\begin{bmatrix}7\\10\\-4\\-1\end{bmatrix}}\mathrm {and} {\begin{bmatrix}-2\\1\\5\\-4\end{bmatrix}}\\\\\end{matrix}}$

Proof

We need to find not-all-zero scalars $\lambda _{1}$, $\lambda _{2}$ and $\lambda _{3}$ such that

${\begin{matrix}\\\lambda _{1}{\begin{bmatrix}1\\4\\2\\-3\end{bmatrix}}+\lambda _{2}{\begin{bmatrix}7\\10\\-4\\-1\end{bmatrix}}+\lambda _{3}{\begin{bmatrix}-2\\1\\5\\-4\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}.\end{matrix}}$

Forming the simultaneous equations:

${\begin{aligned}\lambda _{1}&\;+7\lambda _{2}&&-2\lambda _{3}&=0\\4\lambda _{1}&\;+10\lambda _{2}&&+\lambda _{3}&=0\\2\lambda _{1}&\;-4\lambda _{2}&&+5\lambda _{3}&=0\\-3\lambda _{1}&\;-\lambda _{2}&&-4\lambda _{3}&=0\\\end{aligned}}$

we can solve (using, for example, Gaussian elimination) to obtain:

${\begin{aligned}\lambda _{1}&=-3\lambda _{3}/2\\\lambda _{2}&=\lambda _{3}/2\\\end{aligned}}$

where $\lambda _{3}$ can be chosen arbitrarily.

Since these are nontrivial results, the vectors are linearly dependent.