Linear Algebra Exercises-n-Answers.pdf

114 Linear Algebra, by Hefferon
Three.IV.3.30 It is false; these two don’t
(
commute.
) ( )
1 0 0 0
0 0 1 0
Three.IV.3.31 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 i-th row has its one in its j-th column. Then no
other row has its one in the j-th column; every other row has a zero in the j-th column. Thus the dot
product of the i-th row and any other row is zero.
The i-th row of the product is made up of the dot products of the i-th row of the matrix and the
columns of the transpose. By the last paragraph, all such dot products are zero except for the i-th
one, which is one.
Three.IV.3.32 The generalization is to go from the first and second rows to the i 1 -th and i 2 -th rows.
Row i of GH is made up of the dot products of row i of G and the columns of H. Thus if rows i 1 and
i 2 of G are equal then so are rows i 1 and i 2 of GH.
Three.IV.3.33 If the product of two diagonal matrices is defined — if both are n×n — then the product
of the diagonals is the diagonal of the products: where G, H are equal-sized diagonal matrices, GH is
all zeros except each that i, i entry is g i,i h i,i .
Three.IV.3.34 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. ( ) ( ) ( )
1 0 1 0 1 0
−→ −→
0 1 0 3 −3 3
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.
) ( )
1 0 1 0
0 3 −1 1
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.
Three.IV.3.35 The i-th row of GH is made up of the dot products of the i-th row of G with the
columns of H. The dot product of a zero row with a column is zero.
It works for columns if stated correctly: if H has a column of zeros then GH (if defined) has a
column of zeros. The proof is easy.
Three.IV.3.36 Perhaps the easiest way is to show that each n×m matrix is a linear combination of
unit matrices in
⎛
one and only
⎞
one way:
⎛
⎞ ⎛
⎞
1 0 . . .
0 0 . . . a 1,1 a 1,2 . . .
⎜
c 1 ⎝
0 0 ⎟
⎜
⎠ + · · · + c n,m ⎝
⎟ ⎜
.
.
⎠ = ⎝
⎟
.
⎠
.
0 . . . 1 a n,1 . . . a n,m
has the unique solution c 1 = a 1,1 , c 2 = a 1,2 , etc.
Three.IV.3.37 Call that matrix F (. We)
have ( ) ( )
F 2 2 1
= F 3 3 2
= F 4 5 3
=
1 1
2 1
3 2
In general,
( )
F n fn+1 f
=
n
f n f n−1
where f i is the i-th Fibonacci number f i = f i−1 + f i−2 and f 0 = 0, f 1 = 1, which is verified by
induction, based on this equation.
( ) ( ) ( )
fi−1 f i−2 1 1 fi f
=
i−1
f i−2 f i−3 1 0 f i−1 f i−2
Three.IV.3.38 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
trace(GH) = (g 1,1 h 1,1 + g 1,2 h 2,1 + · · · + g 1,n h n,1 )
+ (g 2,1 h 1,2 + g 2,2 h 2,2 + · · · + g 2,n h n,2 )
+ · · · + (g n,1 h 1,n + g n,2 h 2,n + · · · + g n,n h n,n )