44.2© 2012 Pearson Education, Inc.Vector SpacesNULL SPACES, COLUMN SPACES, AND LINEAR TRANSFORMATIONSSlide 4.2- 2© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX Definition: The null space of an matrix A, written as Nul A, is the set of all solutions of the homogeneous equation . In set notation,. Theorem 2: The null space of an matrix A is a subspace of . Equivalently, the set of all solutions to a system of m homogeneous linear equations in n unknowns is a subspace of . Proof: Nul A is a subset of because A has ncolumns. We need to show that Nul A satisfies the three properties of a subspace. mnx0ANul {x : x is in and x 0}nAAmnnx0AnnSlide 4.2- 3© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX 0 is in Null A. Next, let u and v represent any two vectors in Nul A.  Thenand To show that is in Nul A, we must show that .  Using a property of matrix multiplication, compute Thus is in Nul A, and Nul A is closed under vector addition. u0Av0Auv(u v) 0A(u v) u v 0 0 0AAAuvSlide 4.2- 4© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX Finally, if c is any scalar, thenwhich shows that cu is in Nul A. Thus Nul A is a subspace of . An Explicit Description of Nul A There is no obvious relation between vectors in Nul Aand the entries in A. We say that Nul A is defined implicitly, because it is defined by a condition that must be checked.(u) ( u) (0) 0AccAcnSlide 4.2- 5© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX No explicit list or description of the elements in Nul A is given. Solving the equation amounts to producing an explicit description of Nul A. Example 1: Find a spanning set for the null space of the matrix.x0A361171223124584ASlide 4.2- 6© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX Solution: The first step is to find the general solution of in terms of free variables. Row reduce the augmented matrix to reduceechelon form in order to write the basic variables in terms of the free variables: ,x0A0A120130001220000000124534523022000xxxxxxxSlide 4.2- 7© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX The general solution is ,, with x2, x4, and x5free. Next, decompose the vector giving the general solution into a linear combination of vectors where the weights are the free variables. That is, 124523xxxx34522xxx 12452234524544552321310022022010001xxxxxxxxxxxxxxxx            uwvSlide 4.2- 8© 2012 Pearson Education, Inc.NULL SPACE OF A MATRIX. ----(1) Every linear combination of u, v, and w is an element of Nul A.  Thus {u, v, w} is a spanning set for Nul A.1. The spanning set produced by the method in Example (1) is automatically linearly independent because the free variables are the weights on the spanning vectors.2. When Nul A contains nonzero vectors, the number of vectors in the spanning set for Nul A equals the number of free variables in the equation .245uvwxxxx0ASlide 4.2- 9© 2012 Pearson Education, Inc.COLUMN SPACE OF A MATRIX Definition: The column space of an matrix A, written as Col A, is the set of all linear combinations of the columns of A. If , then. Theorem 3: The column space of an matrix Ais a subspace of . A typical vector in Col A can be written as Ax for some x because the notation Ax stands for a linear combination of the columns of A. That is,.mn1aanA1Col Span{a ,...,a}nAmnmCol {b:b x for some x in }nAASlide 4.2- 10© 2012 Pearson Education, Inc.COLUMN SPACE OF A MATRIX The notation Ax for vectors in Col A also shows that Col A is the range of the linear transformation . The column space of an matrix A is all of if and only if the equation has a solution for each b in . Example 2: Let ,and . xxAmnmxbAm242125733786A 32u103v13Slide 4.2- 11© 2012 Pearson Education, Inc.COLUMN SPACE OF A MATRIXa. Determine if u is in Nul A. Could u be in Col A?b. Determine if v is in Col A. Could v be in Nul A? Solution: a. An explicit description of Nul A is not needed here. Simply compute the product Au.32421 0 02u2573 3013786 3 00A   Slide 4.2- 12© 2012 Pearson Education, Inc.COLUMN SPACE OF A MATRIX u is not a solution of , so u is not in Nul A. Also, with four entries, u could not possibly be in Col A, since Col A is a subspace of .b. Reduce to an echelon form. The equation is consistent, so v is in Col A.x0A3vA24213 24213v257310154237863 000171A     xvASlide 4.2- 13© 2012 Pearson Education, Inc.KERNEL AND RANGE OF A LINEAR TRANSFORMATION With only three entries, v could not possibly be in Nul A, since Nul A is a subspace of .  Subspaces of vector spaces other than are often described in terms of a linear transformation instead of a matrix. Definition: A linear transformation T from a vector space V into a vector space W is a rule that assigns to each vector x in V a unique vector T (x) in W, such thati. for all u, v in V, and ii. for all u in V and all scalars c.4n(u v) (u) (v)TTT (u) (u)Tc cTSlide 4.2- 14© 2012 Pearson Education, Inc.KERNEL AND RANGE OF A LINEAR TRANSFORMATION The kernel (or null space) of such a T is the set of all u in V such that