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Linear Algebra/Mechanics of Matrix Multiplication/Solutions
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Solutions
Problem 1
Predict the result of each multiplication by an elementary
reduction matrix, and then check by multiplying it out.
<li>
<li>
<li>
<li>
<li>
Answer
<li> The second matrix has its first row multiplied
by and
its second row multiplied by .
<li> The second matrix has its first row multiplied
by and
its second row multiplied by .
<li> The second matrix undergoes the pivot operation
of replacing the second row with times the first row added
to the second.
<li> The first matrix undergoes the column operation of: the
second column is replaced by times the first column plus the
second.
<li> The first matrix has its columns swapped.
Problem 2
The need to take linear combinations of rows and columns in tables of
numbers arises often in practice.
For instance, this is a map of part of Vermont and New York.
<TR>
<TD>In part because of Lake Champlain, there are no roads directly connecting some pairs of towns. For instance, there is no way to go from Winooski to Grand Isle without going through Colchester. (Of course, many other roads and towns have been left off to simplify the graph. From top to bottom of this map is about forty miles.)
<TD>x400px
<li> def:incidence matrix The incidence matrix
of a map is the square matrix
whose entry is the number of roads from city
to city .
Produce the incidence matrix of this map (take the cities in
alphabetical order).
<li> def:symmetric A matrix is symmetric if it equals its transpose.
Show that an incidence matrix is symmetric.
(These are all two-way streets.
Vermont doesn't have many one-way streets.)
<li> What is the significance of the square of the
incidence matrix?
The cube?
Answer
<li> The incidence matrix is this
(e.g, the first row shows that there is only one connection
including Burlington, the road to Winooski).
<li> Because these are two-way roads,
any road connecting city
to city gives a connection between city and
city .
<li> The square of the incidence matrix tells how cities
are connected
by trips involving two roads.
Problem 3
This table gives the number of hours of each
type done by each worker, and the associated pay rates.
Use matrices to compute the wages due.
<TR>
<TD>
<TR>
<TD>
<TD style="border-bottom:2px solid black;"> regular
<TD style="border-bottom:2px solid black;"> overtime
<TR>
<TD style="border-right:2px solid black;">Alan
<TD> 40
<TD> 12
<TR>
<TD style="border-right:2px solid black;">Betty
<TD> 35
<TD> 6
<TR>
<TD style="border-right:2px solid black;">Catherine
<TD> 40
<TD> 18
<TR>
<TD style="border-right:2px solid black;">Donald
<TD> 28
<TD> 0
<TD>
<TR>
<TD>
<TD style="border-bottom:2px solid black;"> wage
<TR>
<TD style="border-right:2px solid black;">regular
<TD> $25.00
<TR>
<TD style="border-right:2px solid black;">overtime
<TD> $45.00
(Remark.
This illustrates, as did the prior problem, that in practice
we often want to compute linear combinations of rows and
columns in a context where we really aren't interested in any
associated linear maps.)
Answer
The pay due each person appears in the matrix product of the two arrays.
Problem 4
Find the product of this matrix with its transpose.
Answer
The product is the identity matrix (recall that ). An explanation is that the given matrix represents, with respect to the standard bases, a rotation in of radians while the transpose represents a rotation of radians. The two cancel.
Problem 5
Prove that the diagonal matrices form a subspace of
.
What is its dimension?
Answer
The set of diagonal matrices is nonempty as the zero matrix is
diagonal.
Clearly it is closed under scalar multiples and sums.
Therefore it is a subspace.
The dimension is ; here is a basis.
Problem 6
Does the identity matrix represent the identity map if the bases are
unequal?
Answer
No.
In , with respect to the unequal bases
and ,
the identity transformation is represented by by this matrix.
Problem 7
Show that every multiple of the identity commutes with every
square matrix.
Are there other matrices that commute with all square matrices?
Answer
For any scalar and square matrix we have
.
There are no other such matrices; here is an argument for
matrices that is easily extended to .
If a matrix commutes with all others then it commutes with this
unit matrix.
From this we first conclude that the upper left entry must equal its lower right entry . We also conclude that the lower left entry is zero. The argument for the upper right entry is similar.
Problem 8
Prove or disprove: nonsingular matrices commute.
Answer
It is false; these two don't commute.
Problem 9exer:PermTimesTransEqId
Show that the product of a permutation matrix and its transpose
is an identity matrix.
Answer
A permutation matrix has a single one in each row and column, and
all its other entries are zeroes.
Fix such a matrix.
Suppose that the -th row has its one in its -th column.
Then no other row has its one in the -th column; every other
row has a zero in the -th column.
Thus the dot product of the -th row and any other row is zero.
The -th row of the product is made up of the dot products of the -th row of the matrix and the columns of the transpose. By the last paragraph, all such dot products are zero except for the -th one, which is one.
Problem 10
Show that if the first and second rows of are equal then so
are the first and second rows of .
Generalize.
Answer
The generalization is to go from the first and second rows to the -th and -th rows. Row of is made up of the dot products of row of and the columns of . Thus if rows and of are equal then so are rows and of .
Problem 11
Describe the product of two diagonal matrices.
Answer
If the product of two diagonal matrices is defined— if both are — then the product of the diagonals is the diagonal of the products: where are equal-sized diagonal matrices, is all zeros except each that entry is .
Problem 12
Write
as the product of two elementary reduction matrices.
Answer
One way to produce this matrix from the identity is to use
the column operations
of first multiplying the second column by three, and then adding the
negative of the resulting second column to the first.
Column operations, in contrast with row operations) are written from
left to right, so doing the above two operations
is expressed with this matrix product.
Remark. Alternatively, we could get the required matrix with row operations. Starting with the identity, first adding the negative of the first row to the second, and then multiplying the second row by three will work. Because successive row operations are written as matrix products from right to left, doing these two row operations is expressed with: the same matrix product.
Problem 13
Show that if has a row of zeros then
(if defined) has a row of zeros.
Does that work for columns?
Answer
The -th row of is made up of the dot products of
the -th row of with the columns of .
The dot product of a zero row with a column is zero.
It works for columns if stated correctly: if has a column of zeros then (if defined) has a column of zeros. The proof is easy.
Problem 14
Show that the set of unit matrices forms a basis for
.
Answer
Perhaps the easiest way is to show that each matrix
is a linear combination of unit matrices in one and only one way:
has the unique solution , , etc.
Problem 15
Find the formula for the -th power of this matrix.
Answer
Call that matrix .
We have
In general,
where is the -th Fibonacci number
and , ,
which is verified by induction, based on this equation.
Problem 16def:trace
The trace
of a square matrix is the sum of the
entries on its diagonal (its significance appears in
Chapter Five).
Show that .
Answer
Chapter Five gives a less computational reason— the trace of a matrix is the second coefficient in its characteristic polynomial— but for now we can use indices.
We have
while
and the two are equal.
Problem 17def:traiangular
A square matrix is
upper triangular
if its only nonzero
entries lie above, or on, the diagonal.
Show that the product of two upper triangular matrices is upper triangular.
Does this hold for lower triangular also?
Answer
A matrix is upper triangular if and only if its entry is zero whenever . Thus, if are upper triangular then and are zero when . An entry in the product is zero unless at least some of the terms are nonzero, that is, unless for at least some of the summands both and . Of course, if this cannot happen and so the product of two upper triangular matrices is upper triangular. (A similar argument works for lower triangular matrices.)
Problem 18def:Markov matrix
A square matrix is a Markov matrix
if each entry is between zero
and one and the sum along each row is one.
Prove that a product of Markov matrices is Markov.
Answer
The sum along the -th row of the product is this.
Problem 19
Give an example of two matrices of the same rank with squares of
differing rank.
Answer
Matrices representing (say, with respect to
) the maps that send
and
will do.
Problem 20
Combine the two generalizations of the identity matrix,
the one allowing entires to be other than ones, and the one allowing the
single one in each row and column to be off the diagonal.
What is the action of this type of matrix?
Answer
The combination is to have all entries of the matrix be zero
except for one (possibly) nonzero entry in each row and column.
Such a matrix can be written as the product of a permutation matrix and
a diagonal matrix, e.g.,
and its action is thus to rescale the rows and permute them.
Problem 21
On a computer multiplications are more costly than additions, so
people are interested in reducing the number of multiplications used to
compute a matrix product.
<li> How many real number multiplications are needed in formula
we gave for the product of a
matrix and a matrix?
<li>
Matrix multiplication is associative, so all associations yield
the same result.
The cost in number of multiplications, however, varies.
Find the association requiring the fewest real number multiplications
to compute the matrix product of
a matrix,
a matrix,
a matrix, and
a matrix.
<li> (Very hard.)
Find a way to multiply two matrices using only seven
multiplications instead of the eight suggested by the naive approach.
Answer
<li> Each entry
takes multiplications and there are entries.
Thus there are multiplications.
<li> Let be ,
let be ,
let be ,
let be .
Then, using the formula from the prior part,
shows which is cheapest.
<li> This is reported by Knuth
as an improvement by S. Winograd of a formula due to
V. Strassen:
where ,
takes seven multiplications and fifteen additions (save the
intermediate results).
? Problem 22
If and are square matrices of the
same size such that , does it follow that ?
(Putnam Exam 1990)
Answer
This is how the answer was given in the cited source.
No, it does not.
Let and represent, with respect to the standard bases,
these transformations of .
Observe that
Problem 23
Demonstrate these four assertions to get an
alternate proof that column rank equals row rank. Liebeck 1966
<li> iff .
<li> iff .
<li> .
<li> .
Answer
This is how the answer was given in the cited source.
<li> Obvious.
<li> If then
where .
Hence by (a).
The converse is obvious.
<li> By (b), , ... ,
are linearly
independent iff , ... ,
are linearly independent.
<li> We have
.
Thus also
and so we have .
Problem 24
Prove
(where is an matrix and so defines
a transformation of any -dimensional space
with respect to where is a basis) that
.
Conclude
<li>
iff
;
<li>
iff
;
<li>
iff
and ;
<li>
iff
;
<li> (Requires the Direct Sum subsection, which is optional.)
iff
.Ackerson 1955
Answer
This is how the answer was given in the cited source.
Let be a basis for
( might be ).
Let be such that
.
Note is linearly independent,
and extend to a basis for :
where .
Now take .
Write
and so
But , so
and we now know
spans .
To see
is linearly independent, write
and, since
we get a contradiction unless it is (clearly it is in
, but is a basis
for ).
Hence .
References
R. H. Ackerson A Note on Vector Spaces American Mathematical Monthly American Mathematical Society Dec. 1955 62 10 721.
Hans. Liebeck A Proof of the Equality of Column Rank and Row Rank of a Matrix American Mathematical Monthly American Mathematical Society Dec. 1966 73 10 1114.
PutnamWilliam Lowell Putnam Mathematical Competition, Problem A-5, 1990.