Linear Algebra/The Permutation Expansion/Solutions

SolutionsEdit

These summarize the notation used in this book for the  - and  - permutations.

 

This exercise is recommended for all readers.
Problem 1

Compute the determinant by using the permutation expansion.

  1.  
  2.  
Answer
  1. This matrix is singular.
     
  2. This matrix is nonsingular.
     
This exercise is recommended for all readers.
Problem 2

Compute these both with Gauss' method and with the permutation expansion formula.

  1.  
  2.  
Answer
  1. Gauss' method gives this
     
    and permutation expansion gives this.
     
  2. Gauss' method gives this
     
    and the permutation expansion gives this.
     
This exercise is recommended for all readers.
Problem 3

Use the permutation expansion formula to derive the formula for   determinants.

Answer

Following Example 3.6 gives this.


 
Problem 4

List all of the  -permutations.

Answer

This is all of the permutations where  


 


the ones where  


 


the ones where  


 


and the ones where  .


 
Problem 5

A permutation, regarded as a function from the set   to itself, is one-to-one and onto. Therefore, each permutation has an inverse.

  1. Find the inverse of each  -permutation.
  2. Find the inverse of each  -permutation.
Answer

Each of these is easy to check.

  1. permutation                
    inverse                
  2. permutation                                                
    inverse                                                
Problem 6

Prove that   is multilinear if and only if for all   and  , this holds.


 


Answer

For the "if" half, the first condition of Definition 3.2 follows from taking   and the second condition follows from taking  .

The "only if" half also routine. From   the first condition of Definition 3.2 gives   and the second condition, applied twice, gives the result.

Problem 7

Find the only nonzero term in the permutation expansion of this matrix.


 


Compute that determinant by finding the signum of the associated permutation.

Answer

To get a nonzero term in the permutation expansion we must use the   entry and the   entry. Having fixed on those two we must also use the   entry and the   entry. The signum of   is   because from


 


the two row swaps   and   will produce the identity matrix.

Problem 8

How would determinants change if we changed property (4) of the definition to read that  ?

Answer

They would all double.

Problem 9

Verify the second and third statements in Corollary 3.13.

Answer

For the second statement, given a matrix, transpose it, swap rows, and transpose back. The result is swapped columns, and the determinant changes by a factor of  . The third statement is similar: given a matrix, transpose it, apply multilinearity to what are now rows, and then transpose back the resulting matrices.

This exercise is recommended for all readers.
Problem 10

Show that if an   matrix has a nonzero determinant then any column vector   can be expressed as a linear combination of the columns of the matrix.

Answer

An   matrix with a nonzero determinant has rank   so its columns form a basis for  .

Problem 11

True or false: a matrix whose entries are only zeros or ones has a determinant equal to zero, one, or negative one. (Strang 1980)

Answer

False.

 
Problem 12
  1. Show that there are   terms in the permutation expansion formula of a   matrix.
  2. How many are sure to be zero if the   entry is zero?
Answer
  1. For the column index of the entry in the first row there are five choices. Then, for the column index of the entry in the second row there are four choices (the column index used in the first row cannot be used here). Continuing, we get  . (See also the next question.)
  2. Once we choose the second column in the first row, we can choose the other entries in   ways.
Problem 13

How many  -permutations are there?

Answer

 

Problem 14

A matrix   is skew-symmetric if  , as in this matrix.

 

Show that   skew-symmetric matrices with nonzero determinants exist only for even  .

Answer

In   the exponent   must be even.

This exercise is recommended for all readers.
Problem 15

What is the smallest number of zeros, and the placement of those zeros, needed to ensure that a   matrix has a determinant of zero?

Answer

Showing that no placement of three zeros suffices is routine. Four zeroes does suffice; put them all in the same row or column.

This exercise is recommended for all readers.
Problem 16

If we have   data points   and want to find a polynomial   passing through those points then we can plug in the points to get an   equation/  unknown linear system. The matrix of coefficients for that system is called the Vandermonde matrix. Prove that the determinant of the transpose of that matrix of coefficients

 

equals the product, over all indices   with  , of terms of the form  . (This shows that the determinant is zero, and the linear system has no solution, if and only if the  's in the data are not distinct.)

Answer

The   case shows what to do. The pivot operations of   and   give this.

 

Then the pivot operation of   gives the desired result.

 
Problem 17

A matrix can be divided into blocks, as here,

 

which shows four blocks, the square   and   ones in the upper left and lower right, and the zero blocks in the upper right and lower left. Show that if a matrix can be partitioned as

 

where   and   are square, and   and   are all zeroes, then  .

Answer

Let   be  , let   be  , and let   be  . Apply the permutation expansion formula

 
 

Because the upper right of   is all zeroes, if a   has at least one of   among its first   column numbers   then the term arising from   is   (e.g., if   then   is  ). So the above formula reduces to a sum over all permutations with two halves: first   are rearranged, and after that comes a permutation of  . To see this gives  , distribute.

 
This exercise is recommended for all readers.
Problem 18

Prove that for any   matrix   there are at most   distinct reals   such that the matrix   has determinant zero (we shall use this result in Chapter Five).

Answer

The   case shows what happens.

 

Each term in the permutation expansion has three factors drawn from entries in the matrix (e.g.,   and  ), and so the determinant is expressible as a polynomial in   of degree  . Such a polynomial has at most   roots.

In general, the permutation expansion shows that the determinant can be written as a sum of terms, each with   factors, giving a polynomial of degree  . A polynomial of degree   has at most   roots.

? Problem 19

The nine positive digits can be arranged into   arrays in   ways. Find the sum of the determinants of these arrays. (Trigg 1963)

Answer

This is how the answer was given in the cited source.

When two rows of a determinant are interchanged, the sign of the determinant is changed. When the rows of a three-by-three determinant are permuted,   positive and   negative determinants equal in absolute value are obtained. Hence the   determinants fall into   groups, each of which sums to zero.

Problem 20

Show that

 

(Silverman & Trigg 1963)

Answer

This is how the answer was given in the cited source.

When the elements of any column are subtracted from the elements of each of the other two, the elements in two of the columns of the derived determinant are proportional, so the determinant vanishes. That is,

 
? Problem 21

Let   be the sum of the integer elements of a magic square of order three and let   be the value of the square considered as a determinant. Show that   is an integer. (Trigg & Walker 1949)

Answer

This is how the answer was given in the cited source.

Let

 

have magic sum  . Then

 

and  . Hence, adding rows and columns,

 
? Problem 22

Show that the determinant of the   elements in the upper left corner of the Pascal triangle

 

has the value unity. (Rupp & Aude 1931)

Answer

This is how the answer was given in the cited source. Denote by   the determinant in question and by   the element in the  -th row and  -th column. Then from the law of formation of the elements we have

 

Subtract each row of   from the row following it, beginning the process with the last pair of rows. After the   subtractions the above equality shows that the element   is replaced by the element  , and all the elements in the first column, except  , become zeroes. Now subtract each column from the one following it, beginning with the last pair. After this process the element   is replaced by  , as shown in the above relation. The result of the two operations is to replace   by  , and to reduce each element in the first row and in the first column to zero. Hence   and consequently

 

ReferencesEdit

  • Silverman, D. L. (proposer); Trigg, C. W. (solver) (Jan. 1963), "Quickie 237", Mathematics Magazine (American Mathematical Society) 36 (1) .
  • Strang, Gilbert (1980), Linear Algebra and its Applications (2nd ed.), Hartcourt Brace Javanovich 
  • Trigg, C. W. (proposer) (Jan. 1963), "Quickie 307", Mathematics Magazine (American Mathematical Society) 36 (1): 77 .
  • Trigg, C. W. (proposer); Walker, R. J. (solver) (Jan. 1949), "Elementary Problem 813", American Mathematical Monthly (American Mathematical Society) 56 (1) .
  • Rupp, C. A. (proposer); Aude, H. T. R. (solver) (Jun.-July 1931), "Problem 3468", American Mathematical Monthly (American Mathematical Society) 37 (6): 355 .