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: $$A = Q\Lambda 𝑄^{-1}$$ 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?

  • $\begingroup$ 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. $\endgroup$ Jun 14, 2021 at 4:45

1 Answer 1


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

  • $\begingroup$ 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 $\endgroup$
    – iZner
    Jun 14, 2021 at 5:08
  • $\begingroup$ do you know any website where i could visually see the 3d matrix transformation, it's kinda weird that i couldn't find any $\endgroup$
    – iZner
    Jun 14, 2021 at 5:16
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 14, 2021 at 6:38

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