Linear independence, span, basis, dimension - and their connection with linear systems

Linear independence span basis dimension - and their connection with linear systems Linear independence of a set of vectors: We say the set of vectors v v..v k is linearly independent provided c v c v..c k v k only if all coefficients are zero. Connection with linear systems: c v c v..c k v k only if all coefficients are zero same as Ac has only the solution c where A v v.. v k Span of a set of vectors: set of all linear combinations of the vectors in the set the span of a set of vectors is a subspace: it is closed under addition and scalar multiplication (a subspace is a vector space in its own right) Ex: Consider : span It s a subspace of R 4 Here s a vector in the span: 7 7 4 Another vector in the span: 4 So for instance any linear combination of these two vectors is also in the span. That property is what makes a subspace. Question: Is the vector 7 7 in the span of those two vectors?

c c Is there a solution? This is a linear system: c c no solution. Answer: No it is NOT in the span. The third equation is A key question: Is the following set of vectors linearly independent? If not find a subset that is linearly independent and spans the same subspace. We can answer all these questions by considering the question of linear independence. Independence: c If c is only solution then they are independent; otherwise they are linearly dependent. So we solve - the augmented matrix has a column of zeros on the right but since it always remains a column of zeros we don t include it and just remember it s there.

There are free variables (c and c so there are nonzero solutions: set of vectors is linearly dependent. This one row reduction also answers the other questions. Consider: Can we express the third vector as a linear combination of the first two vectors? This is also a linear system with augmented matrix at the left. This tells me exactly c c so: Now consider: can we express the fifth column in terms of the first second fourth column. This problem results in the augmented matrix below which is then reduced. Yes: col *col*col*col4 But colcolcol4 is a linearly independent set: So the answer to this example problem is that colcolcol4 are linearly independent and span the same set (because they generate col col and hence every other vector in the space) General answer to this question: Given a set of vectors find a subset that is linearly

independent with the same span. Answer: Put vectors into a matrix reduce to row echelon form (enough to determine pivot columns). Then the vectors which wind up as pivot columns (basic variables) form the set that you are looking for. Every other column can be expressed as a linear combination of these columns. If you reduce all the way to reduced row echelon form you can see the coefficients of every nonbasic column expressed as a linear combination of the basic columns. Basis of a vector space (or subspace): A set of vectors that is linearly independent and spans the space. This is the most important idea in linear algebra. Ex: From our previous calculations V span has a basis The number of vectors in a basis of a given space is the same for any basis and is called the dimension of the vector space. The dimension of V is three in this example. The null space of a matrix: The set of solutions of Ax is a subspace. This subspace is called the nullspace of the matrix: Here s why it s a subspace If Aū and Av then Aaū bv aaū bav a b. This shows that the set of solutions is closed under linear combinations. That makes it a subspace. Ex: Take a matrix A. What s a basis for the space of solutions of Ax the null space of A? Ans - reduce and solve the problem Ax 4

The free variables are x x and our fundamental solutions are: x h : These two vectors are obviously linearly independent (just look at the x and x positions) and they span the set of solutions (we discussed that before - any choice of free variables can be obtained by a linear combination of these two solutions and the basic variable values must necessarily follow because they are unique once the free variables are chosen). So for this example the dimension of the set of solutions of Ax the null space of A is two. General observation: The dimension of the null space of a matrix is the number of nonbasic columns/free variables. If we combine the two types of problems we just considered we arrive at an important theorem: Given a matrix A the dimension of the space spanned by the columns (equal to the number of pivot columns/basic variables) plus the dimension of the null space (equal to the number of nonbasic columns/free variables) is equal to n the number of columns. The rank r of a matrix: Definition: number of nonzero rows in row reduced echelon form We also have: r number of pivots in row (reduced) echelon form r number of independent columns in the matrix r the dimension of the span of the columns r number of independent rows (the nonzero rows in row reduced echelon form are linearly independent and every row in A can be generated from those rows) r the dimension of the span of the rows