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User:Prof McCarthy/Linear independence

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Evaluating linear dependence

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Vectors in R2

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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

or

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

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

We can now rearrange this equation to obtain

which shows that non-zero ai 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,

or

The same row reduction presented above yields

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

Alternative method using determinants

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An alternative method uses the fact that n vectors in 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

We may write a linear combination of the columns as

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

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,...,im〉 is any list of m rows, then the equation must be true for those rows.

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

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

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

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

Proof

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

Since

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

Example III

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Let V be the vector space of all functions of a real variable t. Then the functions et and e2t in V are linearly independent.

Proof

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Suppose a and b are two real numbers such that

aet + be2t = 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 et (which is never zero) and subtract to obtain

bet = −a.

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

Example IV

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The following vectors in R4 are linearly dependent.

Proof

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We need to find not-all-zero scalars , and such that

Forming the simultaneous equations:

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

where can be chosen arbitrarily.

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