R b. x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 1 1, x h. , x p. x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9
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1 The full solution of Ax b x x p x h : The general solution is the sum of any particular solution of the system Ax b plus the general solution of the corresponding homogeneous system Ax. ) Reduce A b to reduced row echelon form A b ~ R b ) Along the way check if there are solutions. ) If there are solutions consider the solutions of Ax which are the same as the solutions of Rx. This set of solutions is denoted by x h the general solution of the homogeneous system. Locate the basic variables corresponding to the locations of the "leading s" or the "pivots" in R. The remaining variables are the nonbasic variables. Given any choice of values of the nonbasic variables there is a unique solution for the basic variables. For each nonbasic variable generate a solution of Ax by setting that nonbasic variable equal to and the other nonbasic variables to zero. The basic variables are then completely and uniquely determined and can be read off of R. Together these "fundamental" solutions of Ax generate all solutions of Ax by linear combination; they are referred to as a "basis" of the solution space of Ax. 4) Read off a particular solution x p of Ax b by solving Rx b with all nonbasic variables set to zero. ) Now the general solution: x x p x h Ex: x x x x 4 x x 6 x 7 x 8 x 9 R b Basic variables: x x x 6 x 8 Nonbasic variables: x x 4 x x 7 x 9 Fundamental solutions (the nonbasic choices are in boldbace): x h x x x x 4 x x 6 x 7 x 8 x 9 x p
2 x c c c c 4 c This can be written in matrix form as: c x c c c 4 c Subspaces of vector spaces; subspaces of R n A subspace is a vector space that is a subset of a another vector space. To be a subspace a subset of a vector space just needs to be closed under addition and scalar multiplication. In R n subspaces are generally defined in two different ways: Implicitly: If A is mn the set of solutions of Ax is a subspace of R n. We just need to check that the sum of two vectors in the set is also in the set and similarly for a scalar multiple. In turn this is simply the argument: If Aū and Av then Aū v (this shows that ifū v are in the set then so isū v ) If Aū then Acū caū (this shows that ifūis in the set then so is cū If a subspace is specified implicitly then it is easy to test if a vector is in the subspace but not so easy to generate vectors in the subspace. Explicitly: If v.. v k are in R n then the set of all linear combinations c v c v...c k v k is a subspace of R n. This subspace is called the span of the vectors v.. v k or the space spanned by the vectors v.. v k.
3 Let s check: Ifū v are in the set then ū a v a v...a k v k v b v b v...b k v k and so ū v a b v a b v...a k b k v k showing thatū v is in the set. Similarly if ū a v a v...a k v k then cū ca v ca v...ca k v k which shows that ifūis in the set so is cū. If a subspace is specified explicitly then it is easy to generate vectors in the subspace but not so easy to test is a vector in the subspace. Linearly independent vectors: The idea: A set of vectors is linearly independent if no one vector in the set can be generated from the other vectors by a linear combination. The mathematical definition: A set v.. v k of vectors in a vector space is linearly independent if c v...c k v k only when c c.. c n. Test for linear independence in R n : Check if the homogeneous system Ax with A v.. v k has a nonzero solution. The span of a set of vectors: The span of a set v.. v k of vectors in a vector space is the subspace of all linear combinations of those vectors i.e. spanv.. v k c v c v...c k v k. Is a given vector in the span of a set of vectors? To check if v c v c v...c k v k is possible we solve the system Ax v with A v.. v k. We say a set of vectors in a vector space (or subspace) spans the space if every vector in the space is in the span of the set. Example: In solving Ax we write the solution as a linear combination of fundamental solutions that we generate through Gaussian elimination to reduced row echelon form. Those fundamental solutions span the solution space Ax a subspace of R n. The basis of a vector space: A basis of a vector space (or subspace) is a set of vectors that spans the space and is linearly independent. For the subspace Ax the set of fundamental solutions that we generate is a basis. For a subspace of R n that is defined explicitly as the span of a set of vectors it is always possible to choose a subset of the set that is linearly independent and spans the same space. So if we consider the subspace spanv.. v k c v c v...c k v k we can identify a subset of the v i that are independent and span the same space. To do that we
4 form the matrix A v.. v k and identify those vectors which contain pivots or leading s in the row echelon form. Those span the same space and are independent and form a basis. Ex: Consider the vectors Their span is a subspace of the space consisting of all vectors in R 4 whose components sum to zero: x. How many independent vectors can be found among those six? Is the span of these vectors the same as the entire space x? These questions can be answered by row operations: ~ The first three vectors in the original set are independent and span the same space. Now do these vectors - x? span the entire space The answer is yes. This makes use of an important result: The number of vectors in any basis of a vector space is always the same. It is called the dimension of the space. The space x has dimension three and because of that any three independent vectors in that subspace must span the space. Let s repeat this important result: The number of vectors in any basis of a given vector space is always the same. It is called the dimension of the space. One consequence of this is the following: If the dimension of a space is known to be say4 and we have a set of 4 independent vectors in that space then they must span the space and hence constitute a basis. (If they did not span the space we could find a th independent vector which is not possible if the dimension is 4). Similarly if I have 4 vectors that span the space then they must be linearly independent and hence constitute a basis of the space. (If they were not independent we could drop one out and still span the space but you cannot span a space of dimension 4 with three vectors.) We saw that in discussing the span of a set of vectors we were led to the matrix formed by the vectors and could investigate various questions about that span using reduction to row 4
5 echelon form (for determining linear independence and dimensions) and to reduced row echelon form (for expressing "extraneous" vectors as linear combinations of the independent basis vectors). Now we go in the reverse direction: Given a matrix A we define its column space as the span of its columns. We define its row space as the span of its rows (you can think of rows as column vectors on their sides or as vectors in their own right using the usual matrix operations of addition and scalar multiplication). We define the null space of A as the subspace of solutions of Ax. Now if A is mn we can say the following: dimension of the row space of A dimension of column space of A number of independent columns number of leading entries/pivots in its row echelon form number of not-all-zero rows in the row echelon form of A This requires the following observation: the not-all-zero rows of the reduced row echelon form R of A are linearly independent and are a basis of the row space. This is easy to see: Every row of R came from A so are rows in the column space; every row of A can be obtained from R so the rows of R span the row space. The linear independence is easy to see. So we have the following picture: ~ The rows of R are a basis of the row space. The columns of A that have leading entries in R are a basis of the column space. This number is the same and is called the rank r of the matrix A. The rank r of a matrix is the number of nonzero rows in its row echelon form. r dimension of row spacedimension of column spacenumber of independent columnsnumber of independent rows (note that just as we could identify a subset of the columns as a basis of the column space we could with a bit more computation identify a subset of the rows as a basis of the row space. However the basis of the row space that we have is not of that form and usually cannot be directly ascertained directly from the row echelon form.) The dimension of the null space of a matrix A n r. This is how many free/nonbasic variables there are. The rank is a well known concept so you should know what it means and how it relates to subspaces of R m and R n related to a given matrix. *************************************************************************************** Here s an important observation: The rows of A and the solutions of Ax can both be considered vectors in R n. If we take a basis of the row space consisting of r vectors and a basis of the null space consisting of n r vectors that makes n vectors the dimension of R n. Is that collection a basis of R n? YES.!
6 We have seen that any subspace of R n can be defined as the span of a set of vectors. Some subspaces can be represented as the nullspace of a matrix. Can every subspace of R n be represented as the null space of a matrix? YES! Here s how: Take a basis of the subspace of R n under consideration (or simply the set of vectors that span the subspace in question). Put them in as rows of a matrix A. Now find a basis for the the null space of A the fundamental solutions of Ax. Take these basis vectors and put them in as the rows of another matrix call it B. Now every row of A as a vector in R n is a solution of Bx. And it is not hard to show that in fact there are "no other" solutions: the span of the rows of A the original subspace in question is exactly the same as the null space of B. In this way any subspace can be expressed (we say represented) as the null space of a matrix. Ex: Consider the span of. Denote this subspace by the letter V. Find a matrix B such that V is the null space of B. First use these vectors as rows of a matrix A and find a basis for the null space of A :: ~ 4 R / / /4 is a basis for the null space of A. Note that the dimension of V is two and a basis of V are the nonzero rows of R. Now we form B from the basis of vectors of the null space of A computed above put in as rows: B 4 The null space of B is the vector space V! 6
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