What new geometric information about a matrix does eigendecomposition of the matrix tells us? Since we break the matrix down into eigenvectors and eigenvalues, I would presume that it has some sort of geometric information. 
This video seems to suggest one https://youtu.be/9YtmGy-wfE4?t=793, however, I can't seem to relate the meaning explained with the components of the eigendecomposition.
Can someone kindly explain to me the geometric meaning of eigendecomposition?
 A: Let $\varphi$ be an endomorphism of a vector space $V$. 
If $v$ is an eigenvector of $\varphi$ corresponding to the eigenvalue $\lambda$, this simply means that $\varphi$ induces an homothety of ratio $\lambda$ on the line generated by $v$, i.e. every vector on this line is stretched out by $\lambda$. 
Note that this can be extended to the case of an eigenspace $W$. 
For instance, if $\varphi : \Bbb R^2\to \Bbb R^2$ is the linear map whose matrix in the canonical basis is 
$$\begin{bmatrix} 2 & 0\\ 0& 3 \end{bmatrix}$$
then it corresponds to stretching out all the x-axis by a factor 2 and simultaneously all the y-axis by a factor 3. 
Edit: (answer to your comment below)
Performing the eigendecomposition of a diagonalizable matrix $A$ gives you $A=PDP^{-1}$ where:


*

*$D$ is the diagonal matrix containing the eigenvalues of $A$

*$P$ is an invertible matrix whose columns form the coordinate vectors of a basis of $V$ which contains eigenvectors of $A$. 
Geometrically, multiplying by $P$ on the left and by $P^{-1}$ on the right simply corresponds to a change of basis (cf. the change of basis formula): it gives you a set of new coordinate axes where the linear map acts as the diagonal matrix $D$ (stretchings as before). 
