choct155 / matrixcomputations Goto Github PK

Here is an example of a matrix snippet that will work in Jupyter markdown cells:

Consider the following matrices:

\begin{equation}
    A = 
    \left[
    \begin{array}{c|cc}
        1 & 3 & 2 \\
        2 & 1 & 1 \\
        \hline
        -1 & 0 & 1
    \end{array}
    \right]
\end{equation}

\begin{equation}
    X = 
    \left[
    \begin{array}{c|cc}
        1 & 0 & 1 \\
        \hline
        2 & 1 & 1 \\
        -1 & 2 & 0
    \end{array}
    \right]
\end{equation}

\begin{equation}
    B = 
    \left[
    \begin{array}{c|cc}
        5 & 7 & 4 \\
        3 & 3 & 3 \\
        \hline
        -2 & 2 & -1
    \end{array}
    \right]
\end{equation}

Show that the following holds:

$$A_{i1} X_{1j} + A_{i2} X_{2j} = B_{ij} \text{ for } j = 1, 2$$

I looked for a long time and this was the closest I found to being able to pull out the number from the node equation that does not have a variable attached. Do you have a better way? I don't know that this does what I think it should based on what I found
https://www.geeksforgeeks.org/python-sympy-atoms-method/

MWard_SupervisedPCA.pdf

For some reason this creates:
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.80000000000000]
so it is missing one of the values (x5 has a -.9)

also...do I need to actually take the negative of each of those values? (aka move to the other side of the equation)

I think that based on how we set up the CircuitNode class that the way I set up this Circuit class should actually return both my list of CircuitNodes as well as the output Tensor I've calculated, but it doesn't - it only returns the list of CircuitNodes.

Is there something special I need to do to "add a field" or something? (I can explain more what I am trying to say if you need me to).

I think the rest of the code up to here works, but now it is throwing a value error when I'm converting to a tensor. I'll try to work through this....

ValueError: Attempt to convert a value (<map object at 0x7feb5415b050>) with an unsupported type (<class 'map'>) to a Tensor.