First of all, the notion of linearity does not depend on which coordinates are used. Recall that a map T : R n R m is linear if

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1 5 Matrices in Different Coordinates In this section we discuss finding matrices of linear maps in different coordinates Earlier in the class was the matrix that multiplied by x to give ( x) in standard coordinates ut what if we want to use different coordinates for the input and output? Change coordinates can make the matrix used simpler In fact if one can find a basis of eigenvectors the matrix of the linear map in those coordinates is diagonal First of all the notion of linearity does not depend on which coordinates are used Recall that a map : R n R m is linear if (a) ( x + y) ( x) + ( y) (b) (c x) c ( x) for all x y R n are scalars c We can then think of addition and scalar multiplication of vectors in a coordinate-free way using tip-to-tail addition and scaling hus properties (a) and (b) mean the same thing no matter what coordinates are used in R n and R m Recall also that every linear map : R n R m has a matrix so that ( x) x ut what if we used different coordinate for the input and output? ( ) Example 5 Let be a basis for R and : R R be the linear map with 3 4 ( ) (a) Find and express your answer in -coordinates ( (b) Find ) and express your answer in -coordinates Solution: We find ( ( ) ( 3 ) ) ( ) We strongly recommend keeping track of the maps and conversions using the following diagram: -basis: 3 4 E-basis:

2 he top row is in -coordinates and the bottom row is in standard coordinates E ( e e n ) Now it follows by linearity that ( ) ( ) x x x + x ( ) ( ) x + x 3 4 x + x 3x 4x x + x 3 4 x We put a subscript on the matrix here for reasons to indicate the coordinates used for the input and output In this case we say that 3 4 using Definition/heorem 5 Definition/heorem 5 For any linear map : R n R n and basis for R n there is a unique matrix so that ( ( x) in ) ( x in ) In addition () Proof In Definition/heorem 8 we showed where () x in ( x) x in in heorem 8 y exactly the proof as in Definition/heorem 8 but replacing all instances of vectors in standard coordinates with -coordinates instead ( ( x) in ) ( x in ) where is given by () and such a is unique What if : R n R m is a linear map and we put the input in -coordinates and the output in coordinates for some bases of R n R m respectively? his gives a matrix we call as in 53 Definition/Proposition 53 For any linear map : R n R m and bases of R n R m respectively there is a unique matrix so that ( ( x) in ) ( x in )

3 3 In addition (3) in in Proof Use exactly the proof as in Definition/heorem 8 but replacing all instances of vectors in R n in standard coordinates with -coordinates instead and all instances of vectors in R m in standard coordinates with -coordinates instead here are advantages to having the input and output expressed in the same set of coordinates Firstly one does not have to change an output into input coordinates in order to do the map again and secondly Proposition 57 We will focus on linear maps : R n R n are expressing inputs and outputs in the same coordinates for the remainder of the section Problems 6 - deal with different coordinates for the input and output here is another method we can use to change the coordinates of linear maps Proposition 54 Change of Coordinates Formula Suppose be a basis for R n with coordinate matrix C and : R n R n is a linear map hen (4) Proof We have the following diagram: C C C C -basis: x in ( x) in C C E-basis: x in E ( x) in E Each arrow is labeled by the matrix by which we multiply y looking at the two different paths from ( x in ) to ( ( x) in ) we have ( ( x) in ) ( x in ) ( ( x) in ) C C( x in ) ecause this holds for all x R n we conclude that by heorem 8 that Now multiplying on the left by C and right by C Example 55 Let C C C C C CCC C C ( be a basis for R ) and : R R be the linear map given by 3 4 Find using () and (4) and check that your answers agree Solution: Using ()

4 4 -basis: 5 S-basis: 5 using the conversions his means so by () On the other hand we have So by (4) ( C ) ( in 5 C ) 5 /3 /3 /3 /3 C C /3 /3 3 /3 /3 4 /3 /3 5 /3 /3 5 Notice that the rightmost matrix at each step keeps track of vectors in each diagram We start in lower left corners with our inputs in standard coordinates: hen multiplying by gives our outputs in 5 standard coordinates in lower right corners: Finally multiplying by C converts the outputs back to -coordinates in upper right corners: 5 In Example 55 is a simpler matrix than in that it has more zeros In fact Example 55 is an example of a more general phenomenon that we will discuss in the next section What does changing coordinates do the eigenvalues and eigenspaces of a matrix? Let us explore this question through Example 56 Example 56 Let ( 6 3 and 3 4 )

5 5 (a) Find the eigenvalues and eigenspaces of (b) Find (c) Find the eigenvalues and eigenspaces of (d) Explain why this is not a coincidence (You should realize what this is after doing (a) - (c) ) Solution: We leave it to the reader to work out the calculations Here is the answer: ( (a) Eigenvalues of 3 with eigenspace span and 7 with eigenspace span 3) 4 (b) 7 4 () (c) Eigenvalues of 3 with eigenspace span and 7 with eigenspace span 7 eigenvalues in -coordinates since was expressed in -coordinates ( ) () Notice we express (d) Why do and have the same eigenvalues? Although the eigenspaces are different they become the same if we view them in the right coordinates Viewing the eigenspaces of in -coordinates the eigenspaces are the same as well: span ( 7 ) span ( 3) span ( ) ( span ) his is not a coincidence since 3 7 are the eigenvalues of where is a linear map that does not yet reference coordinates! is the same linear map no matter what coordinates we express it in We can talk about eigenvalues of as the solutions to ( v) λ v for some v which means the same thing in any set of coordinates his also means the eigenspaces are the same spaces but just expressed in different coordinates y ( v) λ v v x Here is a summary: asis ( ) Eigenvalues Eigenvectors (Resp) 6 E 3 7 ( ) All of the properties we discussed in this class work the same way in other coordinates In fact everything we have done until now could have been done in -coordinates instead of standard coordinates! For example range( ) range( ) ker( ) ker( )

6 6 where now our answer is expressed in -ccordinates In addition operations on linear maps correspond to operations on matrices in the same way in other coordinates Proposition 57 Proposition 57 Suppose U : R n R n are linear maps is a basis for R n and c is a scalar hen (a) + U + U (b) c b c (c) U U (d) (e) k k for all positive integers k Proof We have seen that these facts hold without the subscript o see that they still hold in -coordinates take the proofs of their standard coordinate versions and replace them with -coordinates Alternatively one can prove them with the Change of Coordinates Formula Problem Exercises: ( For each of the following linear maps : R R and bases of R find expressing your answer in -coordinates hen find { 6 3 (a) 3 4 } { } 5 (b) 3 { (c) 4} { } 3 3 (d) Suppose : R 3 R 3 with and Find 3 ) ( and ) in -coordinates hen find 3 Suppose : R 3 R 3 with and 3 Find in -coordinates hen find 6 (Hint: Your answer should have a lot of zeroes) ( ( ( 4 Suppose : R R has with Find and in ) ) ) standard coordinates hen find

7 5 Suppose : R 3 R 3 has 3 in standard coordinates hen find 6 Let Problems: with 3 4 Find ( 5 which is a basis for R ) Find a linear map : R R so that () ( ) 3 3 () Prove Proposition 57 using (4) the Change of Coordinates Formula () wo n n matrices P Q are called similar denoted P Q if there exists an invertible n n matrix C so that P C QC Show the following for any matrices P Q R (a) P P (b) If P Q then Q P (c) If P Q and Q R then P R 3 () Show that (U AU) k (U A k U) for all k 4 () Show that for any polynomial p linear map : R n R n and basis of R n p( ) p( ) 5 () Suppose : R n R n is given by ( x) λ x for some scalar λ and is a basis for R n Show that λ λ λi n λ Note that this matrix is the same no matter what is! You may wonder if there are other linear maps : R n R n with the property that is the same matrix for all bases of R n It turns out that there are not Section 54 Problem 6 (3) Suppose : R n R n is a linear map and is a basis for R n with coordinate matrix C Show that finding ker( ) is the same as finding ker() and then converting the answer to -ccordinates 7 (3) Suppose : R n R n is a linear map and is a basis for R n with coordinate matrix C Show that finding range( ) is the same as finding range() and then converting the answer to -ccordinates 8 (+) Suppose U : R n R n are linear maps and is a basis of R n so that U Show that there exists another basis of R n so that U 9 (3) Suppose U V : R n R n are linear maps and are bases of R n so that U V Show that there exists another basis 3 of R n so that U 3 V 7

8 8 (3) Suppose S S R n are complementary subspaces Show that there exists a basis of R n so that π SS where the matrix is n n and the number of s is dim(s ) (3) Suppose S R n is a subspace dim(s) k and : R n R n is a linear map satisfying ( x) S for all x S Show that there exists a basis of R n so that O n kk (3) Suppose S S R n are complementary subspaces dim(s ) k and : R n R n is a linear map satisfying ( x) S for all x S ( y) S for all y S Show that there exists a basis of R n so that where A is k k and A is (n k) (n k) O n kk O kn k 3 (4) Suppose : R n R n satisfies n O but n O Show that there exists a basis of R n so that 4 (4) Suppose : R n R n be a linear map { v v v n } is a basis for R n where (λ v ) is an eigenpair of for some scalar λ Let S span { v v n } and { v v n } Let π : R n S be the projection map given by π x x x n x x n and let U π S : S S Show that λ O n U 5 (3) Show that for every linear map : R n R n there exists a linearly independent set ( v v v n ) that may have complex entries so that is upper triangular Here may also have complex entries

9 9 6 (3) Suppose : R n R m is a linear map and let k rank( ) Show that there exists bases of R n R m respectively so that where the matrix is m n and the number of s is k 7 () Suppose : R n R m is a linear map are bases of R n R m respectively with coordinate matrices C C Show that C C 8 () Suppose U : R n R m are linear maps are bases of R n R m respectively and c is a scalar Show that (a) + U + U (b) c c 9 (3) Suppose : R n R k and U : R k R m are linear maps 3 are bases of R n R k R m respectively Show that U 3 U 3 (3) Suppose : R n R n is an invertible linear map and are bases of R n Show that

Dimension. Eigenvalue and eigenvector

Dimension. Eigenvalue and eigenvector Math 112, week 9 Goals: Bases, dimension, rank-nullity theorem. Eigenvalue and eigenvector. Suggested Textbook Readings: Sections 4.5, 4.6, 5.1, 5.2 Week 9: Dimension,