(03-01-2015 04:24 AM)supernumero Wrote: [ -> ]Still, in the context of my original question, a wish for the ability to find complex eigenvalues and eigenvectors is not fulfilled.

Here's the adaption of the program from above that works with complex matrices:

Code:

`001 - 42,21,15 LBL E`

002 - 42,26,13 RESULT C

003 - 45 25 RCL I

004 - 1 1

005 - 42,23,13 DIM C

006 - 45,16,13 RCL MATRIX C

007 - 0 0

008 - 20 x

009 - 1 1

010 - 40 +

011 - 42,26,15 RESULT E

012 - 45,16,14 RCL MATRIX D

013 - 20 x

014 - 42,16, 4 MATRIX 4

015 - 45 25 RCL I

016 - 2 2

017 - 20 x

018 - 43 36 LSTx

019 - 40 +

020 - 45 25 RCL I

021 - 42,23,15 DIM E

022 - 45,16,15 RCL MATRIX E

023 - 42,16, 4 MATRIX 4

024 - 33 Rv

025 - 34 x<>y

026 - 2 2

027 - 30 -

028 - 42,23,15 DIM E

029 - 45,16,15 RCL MATRIX E

030 - 42 40 Py,x

031 - 42,16, 2 MATRIX 2

032 - 42,26,15 RESULT E

033 - 45,16,12 RCL MATRIX B

034 - 45,16,11 RCL MATRIX A

035 - 45,16,15 RCL MATRIX E

036 - 30 -

037 - 42,26,14 RESULT D

038 - 10 /

039 - 42,26,12 RESULT B

040 - 36 ENTER

041 - 42,16, 8 MATRIX 8

042 - 2 2

043 - 11 SQRT

044 - 10 /

045 - 10 /

046 - 42,26,15 RESULT E

047 - 45,16,11 RCL MATRIX A

048 - 45,16,12 RCL MATRIX B

049 - 20 x

050 - 42,26,14 RESULT D

051 - 42,16, 5 MATRIX 5

052 - 42,16, 3 MATRIX 3

053 - 43 40 Cy,x

054 - 43 32 RTN

The guess for the eigenvalue has to be stored in matrix

d 1 2. The guess for the eigenvector is expected in the matrix

b which must be in \(\tilde{Z}\) format. The same applies to matrix

A. And the dimension (i.e. 2) has to be stored in register

I.

The challenge was to figure out a means to calculate \(\lambda I\). For the transformation a mix of multiplication, transpositions and change of dimensions was used. As an example the case of 2 dimensions:

\[

\begin{bmatrix}

1 \\

1 \\

\end{bmatrix}

\begin{bmatrix}

a & b

\end{bmatrix}

= \begin{bmatrix}

a & b \\

a & b \\

\end{bmatrix}

\rightarrow \begin{bmatrix}

a & a \\

b & b \\

\end{bmatrix}

\rightarrow \begin{bmatrix}

a & a \\

b & b \\

0 & 0 \\

0 & 0 \\

0 & 0 \\

0 & 0 \\

\end{bmatrix}

\rightarrow \begin{bmatrix}

a & b & 0 & 0 & 0 & 0 \\

a & b & 0 & 0 & 0 & 0

\end{bmatrix}

\rightarrow \begin{bmatrix}

a & b & 0 & 0 \\

0 & 0 & a & b

\end{bmatrix}

\]

The other thing to consider is that the elements appear twice in the matrix. This has to be corrected when calculating the norm by dividing by \(\sqrt{2}\). Aside from this both programs are very similar.

This example was used as a test-case. After 5 iterations the result is exact to 4 places.

As a guess for the eigenvector I've used \(\begin{bmatrix}

1 + 0i \\

1 + 0i \\

\end{bmatrix}\) and \(1 + i\) as guess for the eigenvalue.

I haven't tried but I doubt that there's enough memory to handle the

3-dimensional case. The calculation should still be okay if somebody wants to test that with a

DM-15 with extended memory.

Cheers

Thomas