# Matrix Diagonalization - Eigenvectors

I'm not sure that I've understand what I'm doing when I'm trying to diagnolize the matrice.Maybe one of the reason of this is, I can't think it geometrically or I can't understand the purpose of this. So here it's my question : What if I've found different eigenvectors from others ? Is the result acceptable or matrice has one diagonal form? (I mean there's only one way to write diagonal of matrice). The definition of diagonal matrix says:

a diagonal matrix is a matrix (usually a square matrix) in which the entries outside the main diagonal (↘) are all zero. The diagonal entries themselves may or may not be zero.

İf my answer provides this definiton,does it mean that it's true?

(please edit my question for any language mistakes).

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When you say "different eigenvalues from others" do you mean "different eigenvalues", or what is referring to that word "others"? – DonAntonio Jun 16 '13 at 9:06
You write each of the eigenvectors as column vector, and put them together (in any order you like) to form a square matrix, then applying the conjugation of this matrix you created to the original matrix gives you a diagonal matrix. – Secret Math Jun 16 '13 at 9:11
@DonAntonio; A = [0 8; -2 0] I have found P=[2i 2; 1 i] in this case P is equal our eigenvector.But on Youte where I watching the lecture about that he find P different from me? Is it possible? – Erbil Jun 16 '13 at 9:26
Yes, it is possible as you can choose any eigenvector you want to for, $\;P\;$ . Read my answer...and please: use LaTeX to write mathematics here! – DonAntonio Jun 16 '13 at 9:38

It seems to be your matrix is

$$A=\begin{pmatrix}0&8\\\!\!-2&0\end{pmatrix}\implies\det(xI-A)=\begin{vmatrix}x&\!\!-8\\2&x\end{vmatrix}=x^2+16=(x-4i)(x+4i)$$

You have two different eigenvalues $\,\pm 4i\;$, with eigenvalues:

$$\lambda=-4i:\;\;-4ix-8y=0\implies x=2iy\implies\;\;\text{for example}\;\;\binom{2i}{1}$$

$$\lambda=4i:\;\;4ix-8y=0\implies x=-2iy\implies\;\;\text{for example}\;\;\binom{\!\!-2i}{1}$$

Thus, taking

$$P:=\begin{pmatrix}2i&\!\!-2i\\1&\;1\end{pmatrix}$$

you get

$$P^{-1}AP=\begin{pmatrix}\!\!-4i&0\\0&4i\end{pmatrix}$$

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Ok thanks for solution.But what if I say y is not equal to 1. --> for example y = 2. [4i 2]. Is the solution acceptable? – Erbil Jun 16 '13 at 9:41
I honestly don't understand what "y=2.[4i 2]" is... – DonAntonio Jun 16 '13 at 9:42
Sorry for that.I just trying to say what if we take y is equal to two. You have answered the question,thank you. – Erbil Jun 16 '13 at 9:47
He's saying that taking $v_i'=2v_i$ would also be an acceptable eigenvector for $\lambda_i$, and asking if that changes the final diagonal matrix. In answer to that question, no, because, as you might have seen, the diagonal entries on the final matrix are exactly the eigenvalues, and whichever vector you choose in an eigenspace will still have the same eigenvalue! – Tim Jun 16 '13 at 9:48
Yes @Erbil, they both are acceptable. It all depends on what order you construct the columns of the matrix $\,P\,$ . Both diagonal matrix are similar. – DonAntonio Jun 16 '13 at 10:05