Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 163
Four.I.2.16 This one
( )
1 2
A = B =
3 4
is easy to check.
|A + B| =
∣ 2 4
6 8∣ = −8 |A| + |B| = −2 − 2 = −4
By the way, this also gives an example where scalar multiplication is not preserved |2 · A| ̸= 2 · |A|.
Four.I.2.17 No, we cannot replace it. Remark 2.2 shows that the four conditions after the replacement
would conflict — no function satisfies all four.
Four.I.2.18 A upper-triangular matrix is in echelon form.
A lower-triangular matrix is either singular or nonsingular. If it is singular then it has a zero on
its diagonal and so its determinant (namely, zero) is indeed the product down its diagonal. If it is
nonsingular then it has no zeroes on its diagonal, and can be reduced by Gauss’ method to echelon
form without changing the diagonal.
Four.I.2.19 (a) The properties in the definition of determinant show that |M i (k)| = k, |P i,j | = −1,
and |C i,j (k)| = 1.
(b) The three cases are easy to check by recalling the action of left multiplication by each type of
matrix.
(c) If T S is invertible (T S)M = I then the associative property of matrix multiplication T (SM) = I
shows that T is invertible. So if T is not invertible then neither is T S.
(d) If T is singular then apply the prior answer: |T S| = 0 and |T |·|S| = 0·|S| = 0. If T is not singular
then it can be written as a product of elementary matrices |T S| = |E r · · · E 1 S| = |E r | · · · |E 1 | · |S| =
|E r · · · E 1 ||S| = |T ||S|.
(e) 1 = |I| = |T · T −1 | = |T ||T −1 |
Four.I.2.20
(a) We must show that if
T kρi+ρj
−→
then d(T ) = |T S|/|S| = | ˆT S|/|S| = d( ˆT ). We will be done if we show that pivoting first and
then multiplying to get ˆT S gives the same result as multiplying first to get T S and then pivoting
(because the determinant |T S| is unaffected by the pivot so we’ll then have | ˆT S| = |T S|, and hence
d( ˆT ) = d(T )). That argument runs: after adding k times row i of T S to row j of T S, the j, p entry
is (kt i,1 + t j,1 )s 1,p + · · · + (kt i,r + t j,r )s r,p , which is the j, p entry of ˆT S.
(b) We need only show that swapping T ρi↔ρj
−→ ˆT and then multiplying to get ˆT S gives the same result
as multiplying T by S and then swapping (because, as the determinant |T S| changes sign on the
row swap, we’ll then have | ˆT S| = −|T S|, and so d( ˆT ) = −d(T )). That argument runs just like the
prior one.
(c) Not surprisingly by now, we need only show that multiplying a row by a nonzero scalar T kρ i
−→ ˆT
and then computing ˆT S gives the same result as first computing T S and then multiplying the row
by k (as the determinant |T S| is rescaled by k the multiplication, we’ll have | ˆT S| = k|T S|, so
d( ˆT ) = k d(T )). The argument runs just as above.
(d) Clear.
(e) Because we’ve shown that d(T ) is a determinant and that determinant functions (if they exist)
are unique, we have that so |T | = d(T ) = |T S|/|S|.
Four.I.2.21 We will first argue that a rank r matrix has a r×r submatrix with nonzero determinant.
A rank r matrix has a linearly independent set of r rows. A matrix made from those rows will have
row rank r and thus has column rank r. Conclusion: from those r rows can be extracted a linearly
independent set of r columns, and so the original matrix has a r×r submatrix of rank r.
We finish by showing that if r is the largest such integer then the rank of the matrix is r. We need
only show, by the maximality of r, that if a matrix has a k×k submatrix of nonzero determinant then
the rank of the matrix is at least k. Consider such a k×k submatrix. Its rows are parts of the rows
of the original matrix, clearly the set of whole rows is linearly independent. Thus the row rank of the
original matrix is at least k, and the row rank of a matrix equals its rank.
Four.I.2.22 A matrix with only rational entries can be reduced with Gauss’ method to an echelon
form matrix using only rational arithmetic. Thus the entries on the diagonal must be rationals, and
so the product down the diagonal is rational.
ˆT