Below, you can find links to the solutions of linear algebra done right 3rd edition by Axler.

Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

### My favorite Linear Algebra textbooks

- Linear Algebra Done Right 3rd ed. 2015 Edition by Sheldon Axler (errata | videos)
- Linear Algebra 2nd Edition by Kenneth M Hoffman, Ray Kunze (see solutions here)

### Good Linear Algebra textbooks (not complete)

- Introduction to Linear Algebra, Fifth Edition by Gilbert Strang, Solution Manual
- Linear Algebra and Its Applications (5th Edition) by David C. Lay, Steven R. Lay, Judi J. McDonald
- Linear Algebra with Applications 9th Edition by Steven J. Leon
- Linear Algebra 3rd Edition by Serge Lang, Solution Manual
- Linear Algebra Done Wrong by Sergei Treil

### Linear Algebra for Machine (Deep) Learning

- Introduction to Applied Linear Algebra: Vectors, Matrices, and Least Squares by Stephen Boyd, Lieven Vandenberghe
- Linear Algebra and Learning from Data by Gilbert Strang

### Table of content

- Exercises 1.A | Exercises 1.B | Exercises 1.C
- Exercises 2.A | Exercises 2.B | Exercises 2.C
- Exercises 3.A | Exercises 3.B | Exercises 3.C | Exercises 3.D | Exercises 3.E | Exercises 3.F
- Exercises 4
- Exercises 5.A | Exercises 5.B | Exercises 5.C
- Exercises 6.A | Exercises 6.B | Exercises 6.C
- Exercises 7.A | Exercises 7.B | Exercises 7.C | Exercises 7.D
- Exercises 8.A | Exercises 8.B | Exercises 8.C | Exercises 8.D
- Exercises 9.A | Exercises 9.B
- Exercises 10.A | Exercises 10.B

## anbu

2 Sep 2021Hello

Could you please suggest the best books easy to understand the fundamentals for calculus & probability Statistics for starters/beginners.

Thanks

Anbu

## motabhai

2 Aug 2021Thank you so much for this!!

## John

12 Jun 2021Can we recommend texts that should be provided solutions? Doing some self-study over the summer and it would be helpful for some of these. Considering Baby Rudin and Papa Rudin for the mean time

## Vicente Jimenez

22 Mar 2021I just wanted to say this is great. As someone who has been self-studying Axler’s book, I really appreciate your work.

## lttytrty

22 Oct 2020thanks

## jopaso

16 Oct 2020Thank you man!!!

## Tim

14 Sep 2020Is it possible to contribute solutions? I am working through Axler's Measure, Integration and Real Analysis. This site has been helpful and I wish to contribute something back.

## quanity

18 Jun 2020Please complete all the solutions of Hoffman Kunze Linear algebra specially chapter 7 8 9 10.

## Linearity

19 Jun 2020Okay, will do.

## anon

16 May 2020This site has been invaluable to me learning (by self-study) Axler's linear algebra. Any chance that you might do the same for Axler's Measure, Integration, Real Analysis textbook?

## Anonymous

12 May 2020I just wanted to let you know that this blog means so much to me; it helped me go through Hoffman-Kunze (which is a great book, I definitely recommend it.) Thank you so much!

## John

12 May 2020Where can I donate to you?

Without you I would never hat made it

## Linearity

12 May 2020Thank you very much. If you really wish to donate, please donate to US PayPal COVID-19 Relief Fund

## Qinyang Tan

3 May 2020On page 156 of LADR, the proof of 5.38 has a small typo:

where each $u_j$ is in $E(\lambda_j,T)$.

You dropped the second j.

Have a good day.

## Qinyang Tan

3 May 2020Oh, I thought you were the author of LADR. Never mind. LOL

## Linearity

3 May 2020Here is the official site of this book. http://linear.axler.net

## sicheng

15 Oct 2019waiting for help:I cannot find the list of mistakes in Linear Algebra Done Right（3E）.

## K.K.Misra

9 Oct 2019What a wonderful platform to learn Linear Algebra and Real Analysis from. Good job. Keep going!

## Joe

26 Aug 2019Wow! What a wonderful blog for those who want to explore linear algebra!

## Joe Smith

20 May 2019Thank u

## Fred

27 Nov 2017I must be very tired, but how does 1.A:2 work? Not even Wolfram Alpha agrees. Nor to the square computation or cube=1...

What am I missing?

https://www.wolframalpha.com/input/?i=((-1%2Bsqrt(3i))%2F2)%5E2%3D(-1-sqrt(3i))%2F2

I got of to a horrible start trying to dust off my linear algebra skills, haha!

## Fred

28 Nov 2017Wait a minute... That i is not under the radical sign! xD

## Robert Browning

26 Sep 2017First of all, thank you. What a wonderful resource.

Second, I am studying for my qualifying exam and I am using Axler's book. Do you have a TeX file or pdf of all the solutions compiled?

Thanks again.

## Wu Jinyang

25 Aug 2017The proof of 7.36 is awful. Took me quite a while to get the point.

In fact, It can be easier once you show that the eigenvectors of R (the square root) is also the eigenvectors of T (the eigenvalue of T is the square of the eigenvalue of R with respect to the same eigenvector), which is quite obvious.

## Mohammad Rashidi

26 Aug 2017Sorry, I don't quite understand your logic. The problem asks to show the square root is unique. Hence there might have many different choices of square roots. For different square roots, the eigenvectors may also be different.

The logic of the proof from the book is quite clear to me.

## Wu Jinyang

26 Aug 2017Let R be the positive square root of T.

Then from the definition of positive operators, R is self-adjoint. Therefore, V can have an orthonormal basis consisting of eigenvectors of R. let them be e1, ...., en. let a1,..., an be the according eigenvalues.

Thus Rei = aiei for all i = 1,... n

then RRei = ai^2 * ei. since R is a square root of T.

T ei = ai^2 * ei. which shows that e1,... en are also eigenvectors of T.

Then T have e1,...,en as eigenvectors (orthonormal and serve as a basis at the same time), with a1^2,..., an^2 as eigenvalues.

Assume that there exist another positive square root, let's call it R1. let v1,....,vn be a orthonormal basis consisting of eigenvectors. and b1,...bn be the according eigenvalues. By the same logic b1^n,....,bn^2 are eigenvalues of T.

Thus the eigenspaces of R and R1 with respect to same eigenvalue is identical. (remember that positive operators have non negative eigenvalues, if b1^2 = a1^2 then we know for sure b1 = a1). and the eigenspace is orthogonal to each other (since they are spanned by distinct orthonormal vectors). Since the direct sum of eigenspace equal V. we can apply that trick like the "orthogonal projection" to show that Rv = R1v for all v of V.

Thus we know that R and R1 are identical, only their presentation in the matrix form may be slightly different.

What do you think?

## Mohammad Rashidi

26 Aug 2017You are correct. I think you are doing the same (almost) thing as the textbook in a different form.

It is good to have other approach. Everyone's understand to a certain problem might be different. It is hard to say which is good or bad. Take the one which suits you.

## Wu Jinyang

26 Aug 2017By the way, 7B.1 should be $Te_1 = e_1$, $Te_2 = 2e_2+e_1$.

## Mohammad Rashidi

26 Aug 2017Thanks! Fixed.

## Wu Jinyang

10 Sep 2017By the way,One question about baby rudin, P.23 Chapter 1, Exercise 10. If it is possible, can you explain how did they reach the conclusion that "every complex number (with one exception) has two complex square roots."

## mhmd fthy

4 Aug 2017god bless you man , you are awesome

## Ross

27 Mar 2017This is immensely helpful to those of us who can't afford school and choose to self-study. Thank you SO MUCH.

## Josué Nazul

11 Feb 2017Gracias :)

## jackkinsella

20 Oct 2016By releasing these solutions, you've greatly enhanced an already strong book's educational value. Thanks so much :)

## Nuno Alvares Pereira

4 Aug 2016Thanks. I'm going to use this for self-study. I'm no longer having classes at univ, so it will be nice to have some help for learning this out of curiosity. Thanks once again. ;)

## essay writer reviews

12 Jun 2016It's really wonderful that this kind of exercises was being shared that it can be able to help a lot of people in testing their knowledge of algebra. Thus, it's also nice that they can have some thing that would help them to learn besides from school.

## Nadim Farhat

30 Mar 2016Thank for your sharing the solutions.