Obtain orthogonal "eigenvectors" for non-symmetric 2x2 matrix

Question

[Please excuse ignorance, I'm not a mathematician, though I do data analysis.]

The question is strictly for $2\times2$ real non-symmetric matrix $\bf A$. Let us assume, having real eigenvalues.

I've learned how to compute the eigenvalues and the eigenvectors for such a matrix, so that $\bf V D V^{-1} = A$, where $\bf D$ is the diagonal matrix of eigenvalues $\lambda$. Note that $\bf V$ are usually non-orthogonal eigenvectors.

But instead of this I want to find orthogonal "eigenvectors" $\bf U$ and the upper triangular $\bf T$ ($\lambda$ being its diagonal) so that $\bf U T U' = A$.

I know this is what eigen functions usually do via QR iterations of some sort. (I've read a bit about it and can program such iterations.) It is called Schur form, if I'm correct.

My question: Is it possible to obtain $\bf T$ and $\bf U$ for a $2\times2$ case in a closed form - without iterations? Is that easy to do "by hand"? How?

Illustration

A 
 -5  4 
  3  6

Eivenvalues are -6 an 7

Eigenvectors:
V 
  -.9701425001   .3162277660 
   .2425356250   .9486832981

V'V
   1.000000000   -.076696499 
   -.076696499   1.000000000
(not orthogonal)

Restore A:
V  *   -6   0   *   inv(V)
        0   7
yields A

-------------------------------------
But, 

U  *   -6   1   *   t(U)
        0   7
(T is in the middle) also yields A

where U
   .9701425001   .2425356250 
  -.2425356250   .9701425001

is orthogonal:
U'U
   1.000000000    .000000000 
    .000000000   1.000000000

Need to find T (that is, its element [1,2]) and U.

Answer 1

Looks like Schur decomposition to me. We can do it by hand at least for 2x2 matrices here is an example:

For any arbitrary matrix, A, Schur decomposition is defined as $A = UTU^{*}$ where T is upper triangular, U is a unitary matrix and $*$ is conjugate transpose.

Let,
$A = \begin{bmatrix} 7 & -2\\ 12 & -3\end{bmatrix}$

First, we find eigenvectors and eigenvalues of the matrix such that: $A = V \Lambda V^{-1}$. For our matrix $\Lambda = \begin{bmatrix} 1 & 0\\ 0 & 3\end{bmatrix}$ and $V = \begin{bmatrix} 1 & 1\\ 3 & 2\end{bmatrix}$.

Next, we can take any one of the eigenvectors and find its orthogonal vector. Lets take $v_{1} = \begin{bmatrix} 1 \\ 3\end{bmatrix}$ therefore $\{v_{1}\}^{\perp} = \begin{bmatrix} -3 \\ 1\end{bmatrix}$. Normalize them and create a unitary matrix $U = \frac{1}{\sqrt{10}}\begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix}$

Now, from Schur decomposition we know, $A = UTU^{*}$, we have found our unitary matrix $U$ we can get T as $T = U^{*}AU$. Thus, $T = \begin{bmatrix} 1 & -14 \\ 0 & 3 \end{bmatrix}$

Further reading: http://www.home.uni-osnabrueck.de/mfrankland/Math416/Math416_SchurDecomposition.pdf

Answer 2

If your matrix is not symmetric you're not guaranteed they are orthogonal. There are some matrices where if you find their eigenvectors and orthogonalize them, then they will cease to be eigenvectors.