Linear Algebra Done Right 2nd Edition – Solutions to Selected Problems

Today, I’m starting a series of posts on my solutions to selected problems in the popular mathematics text Linear Algebra Done Right by Sheldon Axler. This is the second time I have worked through the exercises. The first time, I was in grad school and really enjoyed how this book concentrates on the math instead of the simple mechanics of linear algebra. In fact, it was of such interest that I took one of my qualifiers on linear algebra. Some of the first exercises are deceptively simple, but challenges come quickly.

I’ve always been somewhat hesitant to post solutions to problems from well know textbooks. However, solutions for this book are readily available and examples like these must be understood to be of use to the student. Simply copying the solutions won’t make the grade. In addition, my solutions may not be correct. Please use them at your own risk. Hopefully, I will remember to notate on each post whether or not the particular solution was validated by one of my professors, peers or Mr. Axler’s solution manual.

Let’s get started…

Chapter 1

1. Suppose a and b are real number, not both 0. Find real numbers c and d such that

1/(a+bi) = c+di

Solution:

1 = (a+bi)(c+di) = (ac+bd) + (ad+bc)i

\Rightarrow

ca-db=1
cb+da=0
\Rightarrow
-bca + bdb = -b
acb+ada = 0
\Rightarrow
0 + (b^2+a^2)d = -b
\Rightarrow
d=\frac{-b}{(b^2+a^2)}
ca-db=1
cb+da=0
\Rightarrow
aca-adb=a
bcb+bda=0
\Rightarrow
(a^2+b^2)c = a
\Rightarrow
c=\frac{a}{(a^2+b^2)}

Check:

\frac{a}{(a^2+b^2)} + \frac{-b}{(b^2+a^2)}i  = \frac{a-bi}{(a^2+b^2)}

and

= \frac{1}{(a+bi)}  = \frac{a-bi}{(a-bi)} \frac{1}{(a+bi)}
= \frac{a-bi}{(a-bi)(a+bi)} = \frac{a-bi}{[aa-(-b)(b)]+[ab+(-ba)]} = \frac{a-bi}{(a^2+b^2)}

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Posted in Linear Algebra Done Right, Mathematics
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