Why does Eigendecomposition of a matrix change the matrix?

Question

I've been wrapping my head around eigendecomposition and i have stumbled onto something that seems to be confusing.

Given Matrix Transformation $$A = \begin{bmatrix}5&2&0\\2&5&0\\4&-1&4\end{bmatrix}$$ and Input Matrix $$Q = \begin{bmatrix}1&0&-1\\1&0&1\\1&1&5\end{bmatrix}$$

I Extracted Lambda diagonal Transformation $$\Lambda = \begin{bmatrix}7&0&0\\0&4&0\\0&0&3\end{bmatrix}$$

Formula for eigendecomposition: $$A = Q\Lambda ^{-1}$$ This means:

1. Multiply Matrix $Q$ by $\Lambda$ Transformation ($Q\Lambda$), which means this matrix would be scaled based on the given inputs.
2. Then multiply by inverse of Matrix $Q$ ($Q\Lambda Q^{-1}$)

But, everything must now be back to the origin since a transformation multiply by an inverse transformation $Q^{-1}$ cancels each other out and produces identity matrix.

Why is the resulting matrix equal to $\left[\begin{smallmatrix}5&2&0\\2&5&0\\4&-1&4\end{smallmatrix}\right]$?

\begin{align*} (QQ^{-1})\Lambda&=(\text{Identity Matrix})(\Lambda) \\ &=\Lambda \end{align*} Then lambda matrix should be the answer, right?

Answer 1

Matrix multiplication is not commutative: $Q\Lambda Q^{-1} \neq QQ^{-1}\Lambda$ in general.