Download Proof Writing Problem Set I

Proof Writing Problem Set I
Linear Algebra MATH 228-02 Fall 2015
September 26, 2015
SOLUTIONS
1
Proof Problem Set I
September 26, 2015
MATH 228-02
1.] Let A = [aij ] be an m × n matrix. Let c, d ∈ R be scalars. Use the definition of matrix addition and
scalar multiplication to show that the following distributive law holds for scalar multiplication:
(c + d)A = cA + dA.
Proof. Let A = [aij ] be an m × n matrix, such that aij ∈ R is the element of the matrix A in the
ith row and j th column where i = 1, 2, . . . , m and j = 1, 2, . . . , n. Also, let c, d ∈ R be scalars, then
c + d ∈ R is also a scalar since addition is closed in the real number system. By the definition of
scalar multiplication, it follows
(c + d)A = (c + d)[aij ] = (c + d)aij .
From here, we know (c+d)aij = caij +daij since real numbers obey the distributive property. Hence,
from the definition of matrix addition and scalar multiplication, we have
(c + d)A = [caij + daij ]
= [caij ] + [daij ]
= c[aij ] + d[aij ]
= cA + dA.
We have shown (c + d)A = cA + dA.
2.] Let A = [aij ], B = [bij ], and C = [cij ] be m × n matrices. Use the definition of matrix addition to
show that matrix addition is associative:
(A + B) + C = A + (B + C).
Proof. Let A = [aij ], B = [bij ], and C = [cij ] be m × n matrices, such that aij , bij , cij ∈ R
are the elements of the matrices A, B, and C, respectively, in the ith row and j th column where
i = 1, 2, . . . , m and j = 1, 2, . . . , n. By the definition of matrix multiplication, it follows that
(A + B) + C = [aij ] + [bij ] + [cij ]
= [aij + bij ] + [cij ]
= [(aij + bij ) + cij ].
Now, we have a single matrix whose ith , j th element is (aij + bij ) + cij . Since aij , bij , and cij are
real numbers and are associative, it follows that
(aij + bij ) + cij = aij + bij + cij = aij + (bij + cij ).
Therefore, by the definition of matrix addition, we obtain
(A + B) + C = [aij + (bij + cij )]
= [aij ] + [bij + cij ]
= [aij ] + [bij ] + [cij ]
= A + (B + C).
We have shown that (A + B) + C = A + (B + C).
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Proof Problem Set I
September 26, 2015
MATH 228-02
3.] Let A = [aij ] be an m × n matrix and B = [bjk ] be an n × p matrix. Let c ∈ R be a scalar. Using
the definition of matrix and scalar multiplication, prove the following equation:
c(AB) = A(cB).
Proof. Let A = [aij ] be an m × n matrix and B = [bjk ] be an n × p matrix, such that aij , bjk ∈ R are
the elements in the ith row and j th column of A and the j th row and k th column of B, respectively,
where i = 1, 2, . . . , m, j = 1, 2, . . . , n, and k = 1, 2, . . . , p. Also, let c ∈ R be a scalar. Then by the
definition of matrix multiplication, we obtain


n
X
c(AB) = c[aij ][bjk ] = c 
aij bjk  .
j=1
By the definition of scalar multiplication, we have


n
X
c(AB) = c
aij bjk  .
j=1
From here, we can bring the constant c into the summation since c, aij and bjk are real numbers
that obey the distributive property:


n
X
c(AB) = 
caij bjk  .
j=1
Further, since real number multiplication is associative, it follows that caij bjk = aij (cbjk ). We have


n
X
aij (cbjk ) .
c(AB) = 
j=1
Finally, by applying the definitions of matrix and scalar multiplication, we arrive at the desired
result:


n
X
c(AB) = 
aij (cbjk )
j=1
= [aij ][cbjk ]
= [aij ](c[bjk ])
= A(cB).
4.] Let A = [aij ] and B = [bij ] be m × n matrices, and C = [cjk ] be an n × p matrix. Use the definition
of matrix addition and multiplication to prove the following distributive law for matrices:
(A + B)C = AC + BC.
Proof. Let A = [aij ] and B = [bij ] be m × n matrices, and C = [cjk ] be an n × p matrix. Let aij ,
bij ∈ R be the elements in the ith row and j th column of the matrices A and B, respectively, where
i = 1, 2, . . . , m and j = 1, 2, . . . , n. Also, let cjk ∈ R be the element in the j th row and k th column
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Proof Problem Set I
September 26, 2015
MATH 228-02
of the matrix C for j = 1, 2, . . . , n and k = 1, 2, . . . , p. Then by the definition of matrix addition, we
have
(A + B)C = [aij ] + [bij ] [cjk ] = (aij + bij ) [cjk ].
By the definition of matrix multiplication, we obtain


n
X
(A + B)C =  (aij + bij )cjk  .
j=1
Since aij , bij , and cjk are real numbers, they obey the distributive law so that (aij + bij )cjk =
aij cjk +bij cjk . Furthermore, addition of the real numbers is associative, which means the summation
can be written as follows:
n
n
n
n
X
X
X
X
(aij + bij )cjk =
aij cjk + bij cjk =
aij cjk +
bij cjk .
j=1
j=1
j=1
j=1
From this, we have


n
X
(A + B)C =  (aij + bij )cjk 
j=1


n
n
X
X
=
aij cjk +
bij cjk 
j=1
j=1

 

n
n
X
X
=
aij cjk  + 
bij cjk 
j=1
j=1
= [aij ][cjk ] + [bij ][cjk ]
= AC + BC,
which follows from the definition of matrix addition and multiplication. We have successfully shown
that (A + B)C = AC + BC.
5.] An n × n matrix is said to be upper triangular if all the entries above the diagonal are zero. More
precisely, if A = [aij ], then A is upper triangular precisely when aij = 0 for all i > j. Using the
definition of matrix multiplication, show that if A and B are two n × n upper triangular matrices,
then the product C = AB is also upper triangular.
Proof. Let A = [aij ] and B = [bjk ] be a n × n matrices, such that aij , bjk ∈ R are the elements
in the ith row and j th column of A and the j th row and k th column of B, respectively, where
i, j, k = 1, 2, . . . , n. Suppose A and B are upper triangular so that if i > j, then aij = 0 and if j > k,
then bjk = 0. Then by the definition of matrix multiplication, the matrix C = AB is an n × n matrix
denoted by C = [cik ] for i, k = 1, 2, . . . , n. We wish to show cik = 0 if i > k. Consider the following
matrix product:


n
X
aij bjk  .
[cik ] = [aij ][bjk ] = 
j=1
We note that for i ≤ k, our summation can be written as


i−1
k
n
X
X
X
[cik ] = 
aij bjk +
aij bjk +
aij bjk  .
j=1
j=i
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j=k+1
Proof Problem Set I
September 26, 2015
MATH 228-02
The first summation from j = 1 to j = i − 1 is zero as aij = 0 since i > j. Also, the summation on
the right from j = k + 1 to j = n is zero as bjk = 0 since j > k. Hence, we are left with


k
X
[cik ] = 
aij bjk  .
j=i
We note that this sum has terms in it if i ≤ k. As i approaches k, the number of terms decreases,
which means if i > k, no more terms are within the sum. Therefore, for i > k we have cik = 0, as
desired. This shows that the matrix C = AB is upper triangular.
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