If you find any mistakes, please make a comment! Thank you.

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

Good Linear Algebra textbooks (not complete)

Linear Algebra for Machine (Deep) Learning

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This Post Has 34 Comments

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

  2. thanks

  3. Thank you man!!!

  4. Is 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.

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

    1. Okay, will do.

  6. This 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?

  7. I 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!

  8. Where can I donate to you?
    Without you I would never hat made it

  9. On 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.

    1. Oh, I thought you were the author of LADR. Never mind. LOL

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

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

  12. Wow! What a wonderful blog for those who want to explore linear algebra!

  13. Thank u

  14. I 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!

    1. Wait a minute... That i is not under the radical sign! xD

  15. First 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.

  16. The 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.

    1. Sorry, 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.

      1. Let 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?

        1. You 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.

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

          2. Thanks! Fixed.

          3. By 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."

  17. god bless you man , you are awesome

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

  19. Gracias :)

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

  21. Thanks. 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. ;)

  22. It'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.

  23. Thank for your sharing the solutions.

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