LA Chapter 7 by SGLee




Chapter 7

Dimension and Subspaces


7.1 Properties of bases and dimensions

7.2 Basic spaces of matrix

7.3 Rank-Nullity theorem

7.4 Rank theorem

7.5 Projection theorem

*7.6 Least square solution, 

7.7 Gram-Schmidt orthonomalization process

7.8 QR-Decomposition; Householder transformations

7.9 Coordinate vectors

7.10 Exercises



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 (e-book : Korean)  





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The vector space  has a basis, and it is a key concept to understand the vector space. 

In particular, a basis provides a tool to compare sizes of different vector spaces with infinitely many elements.

By understanding the size and structure of a vector space, one can visualize the space and efficiently use the data sitting contained within it.


In this chapter, we discuss bases and dimensions of vector spaces and then study their properties.

We also study fundamental vector spaces associated with a matrix such as row space, column space, and nullspace, along with their properties.

We then derive the Dimension Theorem describing the relationship between the dimensions of those spaces.

In addition, the orthogonal projection of vectors in  will be generalized to vectors in ,

and we will study a standard matrix associated with an orthogonal projection which is a linear transformation.

This matrix representation of an orthogonal projection will be used to study Gram-Schmidt Orthonomalization process and QR-Factorization.


It will be shown that there are many different bases for , but the number of elements in every basis for  is always .

We also show that every nontrivial subspace of  has a basis, and study how to compute an orthogonal basis from a given basis.

Furthermore, we show how to represent a vector as a coordinate vector relative to a given basis,

which is not necessarily a standard basis, and

find a matrix that maps a coordinate vector relative to a basis to a coordinator vector relative to another basis.




7.1 Properties of bases and dimensions

 Lecture Movie : ,

 Lab :


Having learned about standard bases, we will now discuss the concept of dimension of a vector space.

Previously, we learned that an axis representing time can be added to the 3-dimensional physical space.

We will now study the mathematical meaning of dimension.

In this section, we define a basis and dimension of using the concept of linear independence and study their properties.



Basis of a vector space









If a subset  of  satisfies the following two conditions, then  is called a basis for :


(1)  is linearly independent; and

(2) .







(1) If  is the subset of  consisting of all the points on a line going through the origin, then any nonzero vector in  forms a basis for.


(2) If a subset  of  represents a plane going through the origin,

         then any two nonzero vectors in  that are not a scalar multiple of the other form a basis for .                         





Let  where . Since  is linearly independent and spans  is a basis for .                


 In general  is a basis for , and it is called the standard basis for .




How to show linear independence of vectors in ?



 Set of vectors  in  is linear independent if




 Let  where 's are column vectors and .

If the homogeneous linear system  has the unique solution ,

then the columns of the matrix   are linearly independent.

In particular, for  implies the linear independence of the columns of .




The following  vectors in 




are linearly independent if and only if






 For ,




       This gives us the following linear system



       This linear system always has the trivial solution .

Furthermore,  if and only if .

Therefore  are linearly independent if and only if .             



By Theorem 7.1.1,  the following three vectors in 


are linearly independent because .             



 9                                                                ■

We can also use an inbuilt function of Sage to check whether sets of vectors are linearly independent or not.

Show that  with

  is a basis for .


To show that  is a basis for , we need to show that  is linearly independent and it spans .  or

 Since the computed determinant above is not zero,  is linearly independent.

We now show that  spans . Let  be a vector in .

Consider a linear system  in .

Note that if this linear system has a solution, then we can say  is spanned by  and so span()=.

The linear system can be written as




   more explicitly, we have a linear system in ,



    Hence we need to show that the linear system (1) has a solution to show that  spans .

   Indeed, the coefficient matrix  of the linear system (1) is invertible,

     so the linear system (1) has a solution. Therefore  is a basis for .          





Let  be a basis for . For , any subset  of  is linearly dependent. Therefore, if  is linearly independent, then   must be less than or equal to .

Since  is a basis for , each vector in  can be written as a linear combination of .

  That is, there are  such that


We now consider a formal equation with :




Then, from (2), we get




Since  are linearly independent,




Hence we get the following linear system


                     .                         (3)


 The homogeneous linear system (3) has  unknowns, , and  linear equations.

 Since , the linear system (3) must have a nontrivial solution. Therefore,  is linearly dependent.   




If  and  are bases for , then .


The proof of this theorem follows the theorem 7.1.2.

 There are infinitely many bases for . However, all the bases have the same number of vectors.










If  is a basis for , then the number of vectors in  is called the dimension of  and is denoted by .







 Note that . If its subspace  is the trivial subspace, , then .






For , the following holds:


 (1) If  is linearly independent, then  is a basis for .

 (2) If  spans  (i.e., ), then  is a basis for .





The determinant of the matrix having the vectors ,

  in  as its column vectors is




Hence  is linearly independent.

By Theorem 7.1.4,  is a basis for .                          






If  is a basis for a subspace  of , then every vector  in can be written as a unique linear combination of the vectors in .


 Since  spans , a vector  in  can be written as a linear combination of the vectors in . Suppose


               and  .


       By subtracting the second equation from the first one, we get



        Since  is linearly independent, .

       Therefore  for each  and  can be written as a unique linear

       combination of the vectors in .                                   


[Remark] Many a times a basis of  is defined to a set which satisfies conditions of Theorem 7.1.4 or Theorem 7.1.5.



Let . Then




However, the vector  can also be written as follows:





This is possible because  is not a basis for .                


7.2 Basic spaces of matrix


  Lecture Movie :

  Lab : 


Associated with an  matrix , there are four important vector spaces: row space, column space, nullspace, and eigenspace. 

 These vector spaces are crucial to study the algebraic and geometric properties of the matrix  

as well as the solution space of a linear system having  as its coefficient matrix.

In this section, we study the relationship between the column space and the row space of  and how to find a basis for the nullspace of .



Eigenspace and null space




 [Solution space, Null space]





The eigenspace  of an  matrix  associated to an eigenvalue  is a subspace of . The solution space of the homogeneous linear system  is also a subspace of . This is also called the null space of  and denoted by Null.









Basis and dimension of a solution space


 Let  be an  matrix. For the given augmented matrix  of a homogeneous linear system with ,

    by the Gauss-Jordan Elimination, we can get its RREF, .


 Suppose that matrix  has  nonzero rows.


  (1) If , then the only solution to  is . Hence the dimension of the solution space is zero.

  (2) If , then with permitting column exchanges, we can transform  as




  Then the linear system is equivalent to







  Here,  are  free variables. Hence, for any real numbers ,

    setting ,

 any solution can be written as a linear combination of  vectors as follows:




  Since  are arbitrary,


  are also solutions to the linear system. Hence, the previous linear combination of the  vectors can be written as 




  This implies that  spans the solution space of .

  In addition, it can be shown that  is linearly independent.

  Therefore  is a basis for the null space  of  and the dimension of the null space is .





 [Dimension of Null space]





For an  matrix , the dimension of the solution space of  is called the nullity of  and denoted by nullity(). That is, dim Null()nullity().







For the following matrix , find a basis for the null space of  and the nullity of .



 The RREF of the augmented matrix  for  is




Hence the general solution is




Therefore a basis and the dimension of the null space of  is


   ,  nullity() 2.           


Find a basis for the solution space of the following homogeneous linear system and its dimension.



Using Sage we can find the RREF of the coefficient matrix :

[1 3 0 5]

[0 0 1 2]

[0 0 0 0]

Hence the linear system is equivalent to




Since  and  are free variables, letting  for real numbers , the solution can be written




Hence we get the following basis and nullity:


        , nullity()           




① Finding a basis for a null space

Free module of degree 4 and rank 2 over Integer Ring

Echelon basis matrix:

[ 1  3  4 -2]

[ 0  5  6 –3]

② Computation of nullity



Column space and row space









For a given  matrix , the vectors obtained from the rows of  




are called row vectors and the vectors obtained from the columns of 




are called column vectorsThe subspace of  spanned by the row vectors , that is,




is called the row space of  and denoted by RowThe subspace of  spanned by the column vectors , that is,




is called the column space of , and denoted by Col. The dimension of the row space of  is called the row rank of , and the dimension of the column space of  is called the column rank of . The dimensions are denoted by   and  respectively, that is,            

                  dim Row,dim Col .











If two matrices  and  are row equivalent, then they have the same row space. 



 Note that the nonzero rows in the RREF of  form a basis for the row space of .

     The same result can be applied to the column space of .


For the following set , find a basis for  which is a subspace of :




Note that the subspace  is equal to the row space of the following matrix




 By Theorem 7.2.1, it is also equal to the row space of the RREF of 




Therefore the collection of nonzero row vectors of 




is a basis for .


Free module of degree 5 and rank 3 over Integer Ring

Echelon basis matrix:

[  1   0   7   0 -39]

[  0   1  -3   0  31]

[  0   0   0   1  -7]                                             



Find a basis for the column space of :




The column space of is equal to the row space of .

   By Theorem 7.2.1, it is also equal to the row space of the RREF of :



Therefore  is a basis for the column space of .


Free module of degree 4 and rank 3 over Integer Ring

Echelon basis matrix:

[ 1  0  0 -1]

[ 0  1  0  1]

[ 0  0  1  0]                                                     







For , the column rank and the row rank of  are equal.


For the proof of Theorem 7.2.2, see 

 The same number for the column rank and the row rank of  is called the rank of , and denoted by





Relationship between vector spaces associated with a matrix 





● Col(), Col()=Row(,

● Row()Null(), Null()Row(),

● Col()Null( ), Null( )Col().








For  is a hyperplane of .

   It is easy to see that  is a subspace of .


(1) If . Then


is a line in the plane passing through the origin perpendicular to the vector .

(2) Let . Then


is the plane in  passing through the origin and perpendicular to the vector .                          

7.3 Dimension theorem (Rank-Nullity Theorem)


 Lecture Movie:


In Section 7.2, we have studied the vector spaces associated to a matrix .

  In this section, we study the relationship between the size of matrix  and the dimensions of the associated vector spaces.  










The rank of a matrix  is defined to be the column rank (or the row rank) and denoted by .







 Let  be an  matrix. If , then  can be written as the following:




Hence rank()  and  nullity().





 7.3.1 [Rank-Nullity theorem]

For any , we have

                                           rank() nullity(


Let )  and the number of leading 1 in    U be   r.  Then.

 But the dimension of solution space of   0   is    (n-r)  which is the number of free variables in it.

  Since    0  and   0    are equivalent, the dimension of solution space of   0  is   (n-r) which is the nullity(). So



 The Rank-Nullity Theorem can be written as follows in terms of a linear transformation:

    If  is the standard matrix for a linear transformation , then


              ,     .





The RREF of  is .

Hence rank().  Since ,

  the dimension of the solution space for  is equal to nullity().                         





Compute the rank and nullity of the matrix , where




The RREF of  can be computed as follows

[1 0 3 7 0]

[0 1 1 3 0]

[0 0 0 0 1]

[0 0 0 0 0]


Hence rank(), and by Theorem 7.3.1,







2                                                               ■







A linear system  has a solution if and only if





 Let . Then the linear system  can be written as


                 .                   (1)


       Hence we have the following:


       has a solution.

                  There exist  satisfying the linear system (1).

                   is a linear combination of the columns of .




The linear system  has its matrix form as following.   



Since , Theorem 7.3.2 implies that the linear system has a solution.                                    










Let  be a nonzero vector. Then

    is called the orthogonal complement of 
This can be understood as the solution space of 

    (that is  .)






 Note that  nullity(  since has  variables and one equation.

     The orthogonal complement of  is a hyperplane of  ( dimensional subspace of ).






Let  be a  dimensional subspace of . Then  for some nonzero vector .


 Since , by the Rank-Nullity Theorem, . Thus

        for a nonzero vector . Therefore


Note: The Four Fundamental Subspaces 

7.4 Rank theorem


 Lecture Movie :

  Lab : 

In this section, we study the relationship between the rank of a matrix  and the theorems

        that is related to the dimension of subspaces associated to .




 7.4.1 [Rank theorem]

For any , dim Row()dim Col().


  We have seen that there exist an invertible  matrix  and an invertible  matrix  

    such that  has the block form  where  is an  identity matrix for some ,

     and the rest of the matrix is zero. For this matrix, it is obvious that row rank = column rank = .

    The strategy is to reduce an arbitrary matrix to this form. see the detail in the following.        ■





  For any ,    rank() ≤ min {}.


 Since dim Row()≤, dim Col()≤, and

          rank()=dim Row()dim Col(),    it follows that rank()≤min {}.                  




 7.4.3 [Rank theorem]

Given , the followings hold:


(1) dim Row( dim Null( the number of columns of 

                                (that is, rank()nullity()).


(2) dim Col( dim Null( the number of rows of 

                                 (that is, rank()nullity()).



 (1) follows from Theorem 7.3.1,

       (2) follows from the fact that RowCol and rankrank  

             along with replacing  in (1) by .                  





For a square matrix  of order  is invertible if and only if

                        rank( .



 If  is invertible, then  has the trivial solution only and hence Null(),

  giving nullity(). By the Rank-Nullity Theorem, we have .

       This can be reversed.                                                   


Find the rank and nullity of the following matrix:




Using Gaussian Elimination,




 Hence  and the Rank-Nullity Theorem gives .








For matrices  with multiplication  defined, the followings hold:


(1) Null()  Null().

(2) Null(Null().

(3) Col()  Col().

(4) Row(Row().



We prove only (1) here.


       .     This implies Null()  Null().

     Others can be shown similarly.                                        






                   rank() ≤ min{rank(), rank()}.



Follows from theorem 7.4.5.





Multiplying a matrix  by an invertible matrix  does not change the rank of  . That is, if , then





Follows from theorem 7.4.6.




Suppose  has rank(). Then


(1) Every submatrix  of  satisfies rank().

(2)  must have at least one  submatrix whose rank is equal to .



(1) Suppose the submatrix  is obtained by taking  rows and  columns of .

              Since ,


      (2) Since the rank of  is , there are  linearly independent rows of .

     Then the matrix  consisting of the  linearly independent rows has the rank equal to .

      We now form a matrix  by taking  linearly independent columns of .

      Then  is an  submatrix of  whose rank is equal to .                                  



Main Theorem of Inverse Matrices



 7.4.9 [Invertible Matrix Theorem]

For an  matrix , the following are equivalent:

 (1)  is invertible.

 (2) .

 (3)  is (row) equivalent to .

 (4)  is a product of elementary matrices.

*(5)  has a unique -factorization. That is, there exists a permutation matrix  such that  where  is a lower triangular matrix with all the diagonal entries equal to 1,  is an invertible diagonal matrix, and  is an upper triangular matrix whose main diagonal entries are all equal to 1.

 (6) For any  vector  has a unique solution.

 (7)  has the unique solution .

 (8) The column vectors of  are linearly independent.

 (9) The column vectors of  span .

*(10)  has a left inverse. That is, there exists a matrix  of order  such that .

 (11) .

 (12) The row vectors of  are linearly independent.

 (13) The row vectors of  span .

*(14)  has a right inverse. That is, there exists a matrix  of order satisfying .

 (15)  is one-to-one.

 (16)  is onto.

 (17)  is not an eigenvalue of .

 (18) .



 We first prove the following equivalence:



① (10)  (7)  (8)  (11)  (10)


(10)  (7): Suppose  has a left inverse  such that . If  satisfies , then  gives




Hence  has the unique solution .


(7)  (8): Suppose  has only the trivial solution.

        If  denotes the th column vector of  and , then




Hence the set  of the column vectors of  is linearly independent.


(8)  (11): Suppose the column vectors of  are linearly independent. Then ,

which is equal to the maximum number of linearly independent columns of , is equal to .


(11)  (10): Suppose . Then the rows of  are linearly independent.

     Let  be the th standard basis vector. Then the following linear systems



      are consistent for all , since rank()rank.

     Letting  be a solution to the linear systems,  is (a left) inverse of .


② (1)  (6)  (14)  (2)  (1)

(1)  (6): Suppose  is invertible. Then, for any  vector ,




Hence  has a solution .

  For the uniqueness of the solution, suppose  is another solution. Then




Therefore  has a unique solution.

(6)  (14): Suppose that for each  , the linear system  has a unique solution.

   If we take  to be , the th standard basis vector, then the following linear system




also has a unique solution. If  is the solution to the linear system,

         then the matrix  is a right inverse of .

(14)  (2): Suppose  has a right inverse  such that . Then




Hence .

(2)  (1): Suppose . If we let , then it can be shown that


Hence  is invertible.                                                      

7.5 Projection Theorem


 Lecture Movie : 

 Lab : 

In Chapter 1, we have studied the orthogonal projection in  where the vectors and their projections can be visualized.

    In this section, we generalize the concept of projection in . 

     We also show that the projection is a linear transformation and find its standard matrix,

    which will be crucial to study the Gram-Schmidt Orthogonalization and the QR-Decomposition*.

Orthogonal Projection in 








Projection (in 1-Dimensional subspace) on  




7.5.1 [Projection]

For any nonzero vector  in , every vector  can be expressed as follows:



where  is a scalar multiple of  and  is perpendicular to . Furthermore, the vectors  can be written as follows: 

                  ,  .


The proof of the above theorem is similar to that in case of orthogonal projection in the  and .

 In the above theorem, the vector  is called the orthogonal projection of  onto  and

      denoted by .

      The vector  is called the orthogonal complement of the vector .





 [Orthogonal projection on ]





The transformation  defined below




is called the orthogonal projection of  onto  (=span).







 It can be shown that the orthogonal projection  is a linear transformation.






Let  be a nonzero column vector in . Then the standard matrix of







Note that  is a symmetric matrix and .


For the proof of this theorem, see the website: 

Using the above theorem, find the standard matrix  of the orthogonal projection in 

                onto the line  passing through the origin.


  (Compare this with  in Chapter 6.)

This is a problem of finding the orthogonal projection of a vector  onto the subspace spanned by a vector .

     Hence we take  as a unit vector  on the line .

     Since the slope of the line is  and .


Therefore, by the previous theorem,





Find the standard matrix  for the orthogonal projection  in  onto the subspace spanned by the vector .



Hence, .                               



Projection of  on subspace   in 





Let  be a subspace of . Then every vector  in  can be uniquely expressed as follows:


                  where   and .


In this case  is called the orthogonal projection of  onto  and is denoted by.







Let  be a subspace of .

  If  is a matrix whose columns are the vectors in a basis for , then for each vector 




A rigorous proof uses facts from Sec 7.7 and *Sec 7.8.


Find the standard matrix for the orthogonal projection in   onto the plane .


 The general solution to  is




Thus  is a basis for the solution space of the plane .


Hence, by taking , the standard matrix is  


Since  and




[20/21  4/21 -2/21]

[ 4/21  5/21  8/21]

[-2/21  8/21 17/21]                                              


 The standard matrix  for an orthogonal projection is symmetric and idempotent ().



[Remark]  Simulation of the projection of two vectors














7.6 * Least square solutions

 Lecture Movie : 

 Lab : 

     Previously, we have studied how to find solve the linear system  when the linear system has a solution.

   In this section, we study how to find an optimal solution using projection when the linear system does not have any solution.

● Details can be found in the following websites: 

● Least square solutions with GeoGebra






● Least square solutions with Sage



7.7 Gram-Schmidt Orthonomalization process


 Lecture Movie: ,


Every basis of has elements, but all the bases are distinct. In this section,

we show that every nontrivial subspace of has a basis and

how to find an orthonormal basis from a given basis.







The subspaces and of are called trivial subspaces.

There are many different bases for , but all the bases have elements and the number is called the dimension of .






Orthogonal set and orthonormal set








For vectors in , let




If every pair of vectors in is orthogonal, then is called an orthogonal set. Furthermore, if every vector in the orthogonal set is a unit vector, then is called an orthonormal set.





 The above definition can be summarized as follows:

  is an orthogonal set.           ()

  is an orthonormal set.      ( Kronecker delta) 

(1) The standard basis for is orthonormal.

(2) In , let .

Then is orthogonal, but not orthonormal.

(3) In , let


Then the set is orthonormal.  


(4) If is an orthogonal set, then is an orthonormal set.                    

Orthogonality and Linear independence




Let be a set of nonzero vectors in . If is orthogonal, then is linearly independent.

 For , suppose



        Then, for each (, , , ),


       That is,


       Since, for , we have


       Since implies ,  we have


       Therefore, is linearly independent.                             

Orthogonal Basis and Orthonormal Basis



 [Orthonormal basis]





Let be a basis for . If is orthogonal, then is called an orthogonal basis. If is orthonormal, then is called an orthonormal basis.





Sets in (1) and (3) of are orthonormal bases of and the set in (2) is an orthogonal basis of .




Let be a basis for .


(1) If is orthonormal, then each vector in can be expressed as




   where .


(2) If is orthogonal, then .

 We prove (1) only. Since is a basis for ,

   each vector can be expressed as a linear combination of vectors in as follows:


       For each , we have



       Since is orthonormal, . Hence


Write as a linear combination of the vectors in


 that is the orthonormal basis for in (3)

Let . Then, by Theorem 7.7.2,

 . Hence

       , , .



 7.7.3  (General form of Theorem 1.3.1 in )

(1) Suppose is an orthonormal basis for . Then, since , the orthogonal projection onto the subspace

     in is




(2) If is an orthogonal basis, but not an orthonormal basis for , then can be written as


Let be a subspace of spanned by the two vectors  in an orthonormal set

 . Find the orthogonal projection of onto and

  the orthogonal component of perpendicular to .




The orthogonal component of perpendicular to is


Gram-Schmidt orthonormalization process




Let be a basis for . Then we can obtain an orthonormal basis  for from .

[Gram-Schmidt Orthonomalization]

We first derive an orthogonal basis for from the basis as follows:

[Step 1] Take .

[Step 2] Let be a subspace spanned by and let


[Step 3] Let be a subspace spanned by and and let


[Step 4] Repeat the same procedure to get


        where      .

It is clear that is orthogonal. By taking


we get an orthonormal basis for .                         

The above process of producing and orthonormal basis from a given basis is called the Gram-Schmidt Orthonormalization process.


Simulation for Gram-Schmidt Orthonomalization















Use the Gram-Schmidt Orthonomalization to find an orthonormal basis  for  from the two linearly independent vectors  and .


We first find orthogonal vectors  as follows:

[Step 1] 

[Step 2] 


Finally  where


   is an orthonormal basis.                                        


Let . Use the Gram- Schmidt Orthonomalization to find an orthonormal basis  for  using the basis  for .


We first find orthogonal vectors :

[Step 1] Take .

[Step 2] 


[Step 3] 



By normalizing , we get






① Computation for an orthogonal basis

[   1    1    0]

[-1/2  1/2    2]

[-2/9  2/9 –1/9]

② Normalization

[   1/2*sqrt(2)     1/2*sqrt(2)              0]

[-1/3*sqrt(1/2)   1/3*sqrt(1/2)  4/3*sqrt(1/2)]

[          -2/3            2/3           -1/3]

Therefore, we get an orthonormal basis


We can verify if  is orthonormal as follows:

③ Checking for orthonormality

[1 0 0]         [1 0 0]

[0 1 0]         [0 1 0]

[0 0 1]         [0 0 1]                                            ■


Let . Use the Gram-Schmidt Orthonomalization to find an orthonormal basis  for a subspace of  for which  is a basis. 





7.8 * QR-Decomposition; Householder Transformations


 Lecture Movie : 

 Lab : 

If an  matrix  has  linearly independent columns, then the Gram-Schmidt Orthogonalization can be used to decompose the matrix  in the form of  where the columns of  are the orthonormal vectors obtained by applying the Gram-Schmidt Orthogonalization to the columns of  and  is an upper triangular matrix.


  The -decomposition is widely used to compute numerical solutions to linear systems, least-squares problems, and eigenvalue and eigenvector problems.

 In this section, we briefly introduce the -decomposition.

● Details can be found in the following websites:





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7.9 Coordinate vectors

 Lecture Movie :, 

 Lab : 

In a finite-dimensional vector space, a basis is closely related to a coordinate system. 

  We have so far used the coordinate system associated to the standard basis of .

  In this section, we introduce coordinate systems based on non-standard bases. 

  We also study the relationship between coordinate systems associated to different bases. 

 If  is an ordered basis for ,

     then any vector  in  is uniquely expressed as a linear combination of the vectors in  as follows:




  Then  are called coordinates of the vector  relative to the     basis .




 [Coordinate vectors]





The scalars  in (1) are called the coordinates of  relative to the ordered basis . Furthermore, the column vector in 




is called the coordinate vector of  relative to the ordered basis  and denoted by .








The vector  in  can be expressed as follows relative to the standard basis  for :








Let .

   For  find the coordinate vector  relative to the basis  for .





we get the linear system .


By solving this linear system, we get .





 As described above, finding the coordinate vector relative to a basis is equivalent to solving a linear system.




Let  be a basis for . For vectors  in  and a scalar , the following holds:


(1) .

(2) .



 In general we have


Change of Basis


 Let  and  be two different ordered bases for .

   In the following, we consider a relationship between  and .




 Letting , the coordinate vector of  relative to  is


  and  the coordinate vector  of  relative to  can be expressed as



  Let   be the coordinate vector of  relative to  and matrix  be


  Then we have






  that is, .                                                (2)


 In the equation (2) matrix  transforms the coordinate vector  to another coordinate vector .

   Hence the matrix  is called a transition matrix from ordered basis  to ordered basis  and denoted by  . Therefore, .


 This transformation is called change of basis.

      Note that the change of basis does not modify the nature of a vector,

    but it changes coordinate vectors. The following example illustrates this.




Let  be the standard basis for  and .

    For the two different ordered bases :


(1) Find the transition matrix  from basis  to basis .

(2) Suppose . Find the coordinate vector .

(3) For , show that equation (2) holds.

(1) Since , we need to compute the coordinate vectors for  relative to . Since


 . Hence





(3) Since  and also ,


   It can be easily checked that . 

For  and ,

     let  and , both of which are bases for . Find .


Since , we first find the coordinate vectors for  relative to . Letting


we get the following three linear systems:



Note that all of the above linear systems have  as their coefficient matrix.

     Hence we can solve the linear systems simultaneously using the RREF of the coefficient matrix.

   That is, by converting the augmented matrix  in its RREF,

    we can find the values of  at the same time:



has the RREF


Therefore, the transition matrix from  to  is




[ 1  0  0|-1  2  1]

[ 0  1  0| 1  1  1]

[ 0  0  1| 2  1  2]                                               ■





Suppose  and  are two different ordered bases for  and  be the transition matrix from  to .

 Then  is invertible and its inverse  is the transition matrix from  to , that is, .


For the two bases  for  in , compute the following:


(1) The transition matrix  from basis  to basis .


(2) The coordinate vector  relative to basis  for given .


(1) Since the transition matrix from  to  is , by Theorem 7.9.2, we have



(2)  .                 


[   1    0    0|-1/2  3/2 -1/2]

[   0    1    0|   0    2   -1]

[   0    0    1| 1/2 -5/2  3/2]                                  ■

Ch. 7 Exercises




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