MAT208 Sections 4.7 and 4.9Fall 2010THE FOUR FUNDAMENTAL SUBSPACES OF A MATRIXSome Review of Terms:SPANLINEAR INDEPENDENCEBASISDIMENSIONEXAMPLES:• Does the set of vectors form a basis for ? • The system becomes• Which row reduces to1 1 01 , 2 , 11 3 031 1 01 2 1 1 3 01 1 00 1 1 0 0 2What does this mean?• The determinant of the matrix is not zero.• The matrix has a full set of pivots.• The set of vectors spans . That is, every vector in the space may be written as a linear combination of the given three vectors.• The set of vectors is linearly independent. This means that no vector in the given set of vectors is a linear combination of the other vectors in the given set.• Since the given set of vectors is linearly independent and spans , the given set of vectors forms a BASIS for . Bases are not unique for a given space.• The dimension of is 3 because there are three basis vectors for the space.EXAMPLE:GIVEN A SPECIFIC VECTOR:2 1 1 03 1 31 1 2 12 2 23 1 3 0INTRODUCING:The Four Fundamental SubspacesA is an m x n matrixDescriptionRow SpaceColumn space of .All linear combinations of the columns of .Column SpaceAll linear combinations of the columns of A.Nullspace All solutions to Ax = 0.Left Nullspace All solutions to y = 0.TATATAFINDING THE FOUR FUNDAMENTAL SUBSPACESThe Row Space The Column Space• Using row operations to reduce a matrix means that we are taking linear combinations of the rows of a matrix A to come up with a matrix B.• We could reverse the process and get back to A.• Therefore, the row space of A equals the row space of B.• If 2 matrices are row equivalent, then their row spaces are the same.• The nonzero rows of the matrix form a BASIS for the row space.• The column space of a matrix consists of ALL linear combinations of the columns of the matrix. • Using row operations to reduce a matrix affects the columns. We cannot “reverse” what we have done to get back to the original matrix.• The columns of the original matrix corresponding to those columns containing PIVOTS form a basis for the column space of the matrix.Null Space Left Null Space• The nullspace of a matrix A consists of all solutions to Ax=0.• For this homogeneous system, either the trivial solution is the only solution or nontrivial solutions exist.• Recall what happens to Ax=0, where A is n x n, if the determinant is zero or nonzero.• The left nullspace consists of all solutions to the system • This subspace has particular applications that we will see later this semester.0TAyNOTATIONRow SpaceColumn SpaceNullspaceLeft NullspaceNotation may vary with textbooks, journals, etc., but you should be able to tell from the context what is being defined or referenced.()TCA()CA()NA()TNAEXAMPLE:Find bases for the four fundamental subspaces of:Row reduces to:1 2 3 62 5 6 121 3 3 6A1 2 3 60 1 0 00 0 0 0BNOTE:• These two matrices are row-equivalent.• The matrix A is not square, so we cannot talk about the determinant.• We do know that Ax=0 will have an infinite number of solutions.• We can see that the row-reduced form of A has two non-zero rows.• We can see that there are pivots in columns 1 and 2.• We have talked about the complete solution to a system:1 2 3 62 5 6 121 3 3 6A1 2 3 60 1 0 00 0 0 0Bhpx x xBasis for the row space of the matrix:• These vectors form a basis for • Note two basis vectors.• Dim =21 2 3 62 5 6 121 3 3 6A1 2 3 60 1 0 00 0 0 0B( 1,2,3,6),(2, 5, 6, 12)()TCA()TCABasis for the column space of the matrix:• As before,Note that columns 1 and 2 have pivots.• These vectors form a basis for • Note two basis vectors.• Dim =2 1 2 3 62 5 6 121 3 3 6A1 2 3 60 1 0 00 0 0 0B()CA( 1,2,1),(2, 5, 3)()CABasis for the nullspace of the matrix:• Once again, • Solve Ax=0 to get1 2 3 62 5 6 121 3 3 6A1 2 3 60 1 0 00 0 0 0B1 3 423344360x x xxxxxx3436001001xx3600Basis for nullspace: ,1001A New Definition• So far, we have calculated that the dimension (number of basis vectors needed) of each subspace (must meet the requirements for a subspace) has been two.• RANK of a matrix:• There are several ways to define rank.• The dimension of the row (column) space of a matrix is its rank.• The RANK of a matrix is the number of pivots in the row-reduced form of the matrix.Rank TheoremLet A be an m x n matrix and let r =its rank.SubspaceNotationSubspace ofDimensionRow SpacerColumn SpacerNullspacen-rLeft Nullspacem-r()TCA()CA()NA()TNAnnmmImportant Note:DIM ROW SPACE + DIM NULLSPACE=DIMDIM COLUMN SPACE + DIMENSION LEFT NULLSPACE=DIM nmWhat this tells us:WE NEED TO FIND ONE BASIS VECTOR FOR THE LEFT NULLSPACE OF A.Solve • Left nullspace:0TAy1 2 1 02 5 3 03 6 3 06 12 6 01 2 1 00 1 1 00 0 0 00 0 0 0Basis for the left nullspace:• Solving and simplifying:• is a basis vector for the left nullspace of the matrix.333311 1xxxx11 1For a 5 x 9 matrix with a rank =2, we know that:• The dimension of the row space is 2.• (Find 2 basis vectors)• Dim = r• The dimension of the column space is 2.• (Find 2 basis vectors)• Dim = r• The dimension of the nullspace is 7.• (Find 7 basis vectors)• Dim = n - r• The dimension of the left nullspace is 3.• (Find 3 basis vectors).• Dim = m - rEquivalent Statements• Let A be an n x n matrix:• A is nonsingular.• Ax=0 has only the trivial solution.• The determinant of A is not equal to zero.• The rank of A is n.• The nullity (the dimension of the nullspace of A) is zero.• The rows of A form a linearly set of vectors in .• The columns of A form a linearly independent set of vectors in