The Row Space of a Matrix

Transcription

1 These notes closely follow the presentation of the material given in David C Lay s textbook Linear Algebra and its Applications (rd edition) These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein The Row Space of a Matrix Definition The row space of an m n matrix A denoted by rowa is the set of all linear combinations of the row vectors of A Definition The column space of an m n matrix A denoted by cola is the set of all linear combinations of the column vectors of A Definition The null space of an m n matrix A denoted by nula is the set of all solutions x of the equation Ax m Here are some basic observations about the row space column space and null space: If A is an m n matrix then cola is a subspace of m and rowa is a subspace of n In particular cola is the span of the columns of A and rowa is the span of the rows of A If A is an m n matrix then dimcola dimnula n If A is an m n matrix then rowa cola T and cola rowa T

2 Example For the 4 matrix find bases for cola nula and rowa Solution From earlier work that we have done we know how to find bases for cola and nula Since ~ we see that the first second and fourth columns of A are the pivot columns of A We also know that the pivot columns of A form a basis for cola Thus cola is three dimensional and a basis for cola is 7 To find a basis for nula we observe that every solution of Ax 4 has the form x x x x x 4 s t s t t s s t 7 Therefore nula is two dimensional and a basis for nula is To find a basis for rowa we use the fact that rowa cola T Observing that s

3 A T ~ 7 we see that the first second and fourth columns of A T are the pivot columns of A T Therefore cola T rowa is three dimensional and a basis for cola T rowa is 8 7 7

4 How to Find a Basis for cola and rowa in One Fell Swoop If A is an m n matrix then every vector v cola is a linear combination of the columns of A This means that v Ax for some vector x n Likewise every vector w rowa is a linear combination of the rows of A This means that w A T y for some vector y m We will use this way of looking at things to show that the row spaces of any two equivalent matrices are the same! Lemma If A and B are m n matrices and A~B then there exists an invertible m m matrix C such that B CA In particular if A is any m n matrix then there exists an invertible m m matrix C such that rrefa CA Proof The fact that A~B means that if we start with A and perform a certain sequence of elementary row operations then we arrive at B This means that there is a sequence E E E k of elementary m m matrices such that E k E E B However recall that all elementary matrices are invertible and recall that any product of invertible matrices is invertible This means that the matrix C E k E E is invertible We have shown that B CA where C is an invertible m m matrix Example As an example to illustrate the above lemma let us consider the process of finding rrefa for the matrix We do this step by step as follows: rrefa 4

5 In terms of elementary matrices this procedure is written as which can be written as Thus in this example we see that rrefa CA where C is the matrix C 4

6 Theorem If A and B are m n matrices and A~B then rowa rowb In particular if A is any m n matrix then rowa rowrrefa Proof Since A~B then we know that there exists an invertible m m matrix C such that B CA We will now show that rowb rowa Suppose that w rowb This means that there exists a vector y m such that w B T y From this we obtain w CA T y w A T C T y w A T C T y Since the vector C T y is in m the last equation above shows us that w is a linear combination of the columns of A T This means that w rowa We have proved that rowb rowa We will now show that rowa rowb Suppose that w rowa This means that there exists a vector y m such that w A T y From this we obtain w C B T y w B T C T y w B T C T y Since the vector C T y is in m the last equation above shows us that w is a linear combination of the columns of B T This means that w rowb We have proved that rowa rowb Since rowb rowa and rowa rowb then rowa rowb Theorem IfAisanm n matrix then the non zero rows of rrefa form a basis for rowa (and also form a basis for rowrrefa since rowa rowrrefa) 6

7 Example Use the above theorem to find a basis for the row space of the matrix Solution Since rrefa we see that a basis for rowa is (Note that this basis for rowa is different than the one that we found when we did this problem earlier using a different method) 7

8 Example Find bases for the column space null space and row space of the matrix 4 6 Solution Since rrefa we see that a basis for cola is 4 a basis for nula is and a basis for rowa is 8

9 Remark If A and B are m n matrices and A~B then it is not necessarily true that cola colb In particular it is not necessarily true that cola colrrefa However it is always true that dimcola dimcolrrefa Theorem If A is any m n matrix then the row space and the column space of A have the same dimension Proof The dimension of the column space of A is the number of pivot columns of rrefa The dimension of the row space of A is the number of non zero rows of rrefa However the number of pivot columns of rrefa is the same as the number of non zero rows of rrefa Therefore dimcola dimrowa Example For the matrix we have seen that dimcola dimrowa For the matrix 4 6 we have seen that dimcola dimrowa Definition IfAisanm n matrix then we define the rank of A denoted by ranka tobe the dimension of the column space of A (which is the same as the dimension of the row space of A) If A is a square matrix of size n n and ranka n then we say that A has full rank Theorem An n n matrix A is invertible if and only if A has full rank Remark Any of the many other statements (for example A~I n ) that are given in the Square Matrix Theorem are equivalent to the statement that A has full rank 9