[BUG] Markdown rendering problem

[atigun64]

Environment

https://v2-dev.big-agi.com/

Description

Device and browser

Device: Asus Zephyrus G14 OS: Windows 11 Browser: Firefox

Screenshots and more

No response

Willingness to Contribute

• 🙋‍♂️ Yes, I would like to contribute a fix.

[enricoros]

Thanks @atigun64 can you paste the answer also so I can see what's unexpected in the code?

[atigun64]

Thanks @atigun64 can you paste the answer also so I can see what's unexpected in the code?

# Converting Powers of a Matrix Between Bases

You're absolutely correct, and you've touched on one of the most powerful applications of change of basis in linear algebra!

## Diagonalization and Matrix Powers

When $S$ is a matrix whose columns are the eigenvectors of $A$, then $D = S^{-1}AS$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. This is called the **diagonalization** of $A$.

The key insight is that powers of a diagonal matrix are very easy to compute - you just raise each diagonal element to the power:

$$D^k = \begin{pmatrix} 
\lambda_1^k & 0 & \cdots & 0 \\
0 & \lambda_2^k & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n^k
\end{pmatrix}$$

## Converting Powers Back to Standard Basis

To find $A^k$ in the standard basis, you can indeed use:

$$A^k = S D^k S^{-1}$$

This works because:

$$A^k = (SDS^{-1})(SDS^{-1})\cdots(SDS^{-1}) = SD^kS^{-1}$$

The $S$ and $S^{-1}$ in the middle "cancel out" leaving just the outer $S$ and $S^{-1}$.

## Why This Works

This approach is powerful because:
1. Computing $D^k$ is trivial (just raise eigenvalues to power $k$)
2. You only need to compute $S$ and $S^{-1}$ once
3. It works for any power $k$, including non-integer powers in some cases

## Example

Suppose $A$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 3$ with corresponding eigenvectors forming the columns of $S$.

Then:
- $D = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$
- $D^k = \begin{pmatrix} 2^k & 0 \\ 0 & 3^k \end{pmatrix}$
- $A^k = SD^kS^{-1}$

This is much easier than computing $A·A·A·...·A$ directly $k$ times!

So yes, your formula $A^k = S(S^{-1}AS)^kS^{-1} = SD^kS^{-1}$ is exactly right!