Linear Algebra review concepts1

Subspaces

Let V be a vector space. For a subset W of V , we say W is a subspace of V if W satises the following:

• closed under addition: for any vectors w1, w2 in W , w1 + w2 is also in W ,

• closed under scalar multiplication: for any scalar c, cw1 is also in W . 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 c1v1 + c2v2 + ... + cnvn = 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 c1v1 +c2v2 +...+cnvn = 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. Four fundamental subspaces of a matrix

  Let A be an m × n matrix. A = a1 a2 ... an where ai is the ith column of A.

• The null space of A is nul(A) = {x ∈ Rn : Ax = 0}.

• The column space of A is col(A) = span{a1, ..., an}. • The row space of A is col(AT ). • The left null space of A is nul(AT ).

• The rank of a matrix A is the number of pivots in A. • The dimension of nul(A) is the number of free variables in A. In other words, dim(nul(A)) = n − rank(A). • The dimension of col(A) is the rank of A. • The dimension of col(AT ) is the rank of A.

1there may be typos.

1 • The dimension of nul(AT ) is m − rank(A).

More facts about invertible matrices 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−1A = 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. • Nul(A) = {0}.

• Col(A) = Rn. • The columns of A are linearly independent.

• The rows of A are linearly independent.

• The columns of A span Rn. • The rows of A span Rn.

• The columns of A form a basis of Rn. • The rows of A form a basis of Rn. • rank(A) = n. • dim(Nul(A)) = 0.

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