Eigendecomposition

Remarks

The columns of $AQ$ are the vectors $A q_i = \lambda_i q_i$, and these are also the columns of $Q \Lambda$. Thus $AQ = Q\Lambda$, and $Q$ is invertible because its columns are independent. For example, $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ has the eigenvector $(1, 0)^T$ for $2$ and $(1, 1)^T$ for $3$. Thus $A = Q \Lambda Q^{-1}$ with $Q = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\Lambda = \operatorname{diag}(2, 3)$. The hypothesis is necessary: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has only one independent eigenvector, and it has no eigendecomposition. When $A$ is real symmetric or complex normal, $Q$ can be chosen orthogonal or unitary, by the spectral theorem.

Going deeper

For a student

The decomposition makes powers and functions of $A$ easy: $A^k = Q \Lambda^k Q^{-1}$, and the same holds for the exponential. It is the matrix form of the criterion that a matrix is diagonalisable iff it has a basis of eigenvectors. Distinct eigenvalues give enough eigenvectors, by this statement. When eigenvectors are missing, the Jordan normal form or the Schur decomposition takes its place.