Let v1 , ..., vk , b be vectors of Rn .
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The following are equivalent:

•  v1 v2 . . . vk b  is consistent.
• There exists scalars x1 , ..., xk ∈ R such that x1 v1 + ... + xk vk = b.
• b is a linear combination of {v1 , ..., vk }.
• b is in span{v1 , ..., vk }.
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• b is in col(A) where A =  v1 v2 . . . vk .
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Let A be an n × n matrix.
The following are equivalent
:
• A is invertible.
• There exists a matrix A−1 such that AA−1 = A−1 A = I .
• A has n pivots in echelon form (every row and every column has a pivot).
• The reduced row echelon form of A is I .
• A is a product of elementary matrices.
• There is a unique solution to Ax = 0 (x = 0).
• For any b ∈ Rn , Ax = b has a unique solution.
• N ul(A) = {0}.
• Col(A) = Rn .
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Linear independence, dependence, and bases
Let {v1 , v2 , ..., vn } be a set of vectors of a vector space V .
• {v1 , v2 , ..., vn } are linearly independent if c1 v1 + c2 v2 + ... + cn vn = 0 has only the trivial
solution c1 = c2 = ... = cn = 0.
• {v1 , v2 , ..., vn } are linearly dependent if they are not linearly independent. I.e., if c1 v1 + c2 v2 +
... + cn vn = 0 has a nontrivial solution. Note that c1 = c2 = ... = cn = 0 is still a solution; therefore
we have at least 2 solutions. It follows that there must be innitely many solutions to the system.
• If {v1 , v2 , ..., vn } are linearly dependent, there must exist a vi that is a linear combination of the
other vectors. I.e., vi ∈ span{v1 , ..., vi−1 , vi+1 , ..., vn }.
• If vi ∈ span{v1 , ..., vi−1 , vi+1 , ..., vn }, then
span{v1 , ..., vi−1 , vi , vi+1 , ..., vn } = span{v1 , ..., vi−1 , vi+1 , ..., vn }.
• {v1 , v2 , ..., vn } form a basis of V if the following hold:
span{v1 , v2 , ..., vn } = V ,
{v1 , v2 , ..., vn } are linearly independent.
• Given a vector space, there are innitely many choices for a basis. However every basis of the vector
space will have the same number of vectors.
• The dimension of V, denoted dim(V ), is the number of vectors in a basis of V .
• Suppose V has dimension k .
Any set of k linearly independent vectors of V form a basis of V (the k vectors therefore span
V ).
Any set of k vectors that span V form a basis of V (the k vectors are therefore linearly
independent).
• Given a basis {v1 , ..., vn } of V , any vector u ∈ V can be written as a UNIQUE linear combination
of the basis vectors.
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