# Why does Eigendecomposition of a matrix change the matrix?

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: $$A = Q\Lambda š¯‘„^{-1}$$ 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?

• This site uses MathJax and Markdown to format its content. (There's a help button labeled "?" that elaborates on how to use markdown in the editor; it has a link to MathJax help too I believe.) Please use them in future posts. Jun 14, 2021 at 4:45

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

• but why, im thinking right shear then scale then left shear = scaled origin | right shear then left shear then scaled = scaled origin. I'm still confused Jun 14, 2021 at 5:08
• do you know any website where i could visually see the 3d matrix transformation, it's kinda weird that i couldn't find any Jun 14, 2021 at 5:16
• @iZner You would be right if it happened that all the diagonal entries of $\Lambda$ were the same. They are not though, so $\Lambda$ is not a uniform scaling; it acts differently on each component. Jun 14, 2021 at 6:38