Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some online tool for diagonalizing 2x2 matrices or at least finding the eigenvectors and eigenvalues of it. I don't like to download any stuf because I'm not able to, some online tool will do the job. Thanks.

share|cite|improve this question
Wolfram Alpha works. Try this. – J. M. Nov 4 '10 at 13:26
@Guesswhoitis. Of course, WolframAlpha pretty much can do everything, and it's free so it's awesome. – Derek 朕會功夫 May 10 '15 at 3:58

Wolframalpha has an option. Try this:

share|cite|improve this answer

Let $A$ be your $2$ by $2$ diagonalizable matrix, let $\lambda$ and $\mu$ be its eigenvalues, and let $I$ be the $2$ by $2$ identity matrix.

If $\lambda=\mu$, then $A=\lambda I$ and there is nothing to do.

If $\lambda\not=\mu$, then the nonzero columns of $A-\mu I$ (such always exist) are $\lambda$-eigenvectors.

[Recall that the eigenvalues of $$\begin{pmatrix}a&b\\ c&d\end{pmatrix}$$ are the roots of $X^2-(a+d)\,X+ad-bc$.]

[There is an obvious generalization to $n$ be $n$ matrices: in the above recipe to get a $\lambda$-eigenvector, replace $A-\mu I$ by the product of the $A-\mu I$, where $\mu$ runs over the eigenvalues not equal to $\lambda$.]

To prove this in the $2$ by $2$ case, it suffices to check $$A^2-(a+d)\,A+(ad-bc)\,I=0,$$ which is straightforward. This is (a particular case of) the Cayley-Hamilton Theorem.

share|cite|improve this answer
The OP is asking for online tools. – Rasmus Aug 18 '11 at 10:56
Dear @Rasmus: Thanks. I agree. [The OP doesn’t seem to be around anyway, but this changes nothing.] I just wanted to make sure the (virtual) OP was aware that s/he was asking for online tools for solving a quadratic equation. I suspect that s/he was not aware of that. Some people believe that to find the eigenvectors in this case you must solve linear systems. [There are probably more online tools that solve quadratic equations than online tools that diagonalize matrices. This tells the OP which kind of online tools are needed.] ... But I'm open to dialogue... – Pierre-Yves Gaillard Aug 18 '11 at 11:26
I see your point now. Thank you for the explanation. +1 – Rasmus Aug 18 '11 at 14:42

Try the Online Matrix Calculator.

share|cite|improve this answer
And which option does the $S \Lambda S^{-1}$ decomposition? I do not see it. – Val Jul 30 '13 at 12:07
@Val, check Eigenvalues/eigenvectors. – lhf Jul 30 '13 at 13:53
It produces you the list of eigenvalues/eigenvectors. It does not put them into the $S \Lambda S^{-1}$ form, which is what I call "diagonalization". does. – Val Jul 31 '13 at 10:31 decomposes the matrix into $S \Lambda S^{-1}$

share|cite|improve this answer
It requires the eigenvalues to be rational, which is quite stupid. Otherwise it displays "Not Enough Rational Eigenvalues". – Alex M. Jan 12 at 10:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.