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- If span and one of these vectors can be written as a linear combination of the others, then those vectors span .
- Given vectors , it is possible to write one of the vectors as a linear combination of the other vectors iff there exist scalars (not all zero!) such that .
Recall that is a linear combination of .
The vectors are said to be linearly independent if implies that all the scalars must all be zero.
For two vectors such that
- If , then
- If , then
Thus the two vectors are linearly dependent
Which of the following collectios are linearly independent?
- No. The matrix of system is singular, therefore there are nontrivial solutions that will satisfy the equation. (See #Theorem)
Let be vectors in and let be the matrix formed by using the vectors as columns.
The vectors will be linearly dependent iff is singular and linearly independent iff is nonsingular
Let , then has a nontrivial solution iff is singular.
For Non-Square Matrix
For , the system can be written as , where is an matrix. Therefore if , the determinant of is not defined and #Theorem does not apply.
However, the system has nontrivial solutions (i.e. is linearly dependent) iff the row echelon form of has at least one free variable.
, , and
Therefore, since is a free variable, is linearly dependent.
Let . A vector can be written uniquely as a linear combination of iff are linearly independent.
Assume the solution is not unique, that is, where for some . This would mean that , where , would be linearly dependent.
Therefore, if are linearly dependent, then there exist (not all zero) such that
The second and third lines are two different representations for
Let , , and be in .
Therefore, since coefficients of terms must be equal,
, so the matrix is singular, and therefore the polynomials are linearly dependent.
The following determinant of a matrix of functions and derivatives
Is called the Wronskian
Let be -th differentiable functions along the interval . If there exists a point in such that , then are linearly independent.