# What is the difference between diagonalising a matrix and its eigendecomposition?

When we diagonalise a matrix, we write it in terms of a diagonal matrix $$S$$ which contains all of the eigenvalues of the matrix, a matrix $$P$$ which contains all of the corresponding eigenvectors of $$A$$, and $$P^{-1}$$.

$$A=P^{-1}SP$$

However, when we have an eigendecomposition, we also represent a matrix $$B$$ in terms a diagonal matrix $$\Lambda$$ which contains all of the eigenvalues of $$B$$, a matrix $$Q$$ (which I think is the eigenbasis) and $$Q^T$$.

$$B=Q^T \Lambda Q$$

What's the difference between the two?

• The matrix $\Lambda$ isn't always diagonal because sometime we have repeat eigenvalues. Jun 29, 2020 at 10:38

A square matrix $$A \in \mathbb{R}^{n \times n}$$ is called diagonalizable if there exists a matrix $$P$$ and a diagonal matrix $$\Lambda$$ such that $$$$A = P \Lambda P^{-1}.$$$$ Necessarily, the columns of $$P$$ are eigenvectors of $$A$$ and the diagonal elements of $$\Lambda$$ are the corresponding eigenvalues. A natural question to ask is: when is $$A$$ diagonalizable? One sufficient condition is $$A$$ is symmetric i.e. $$A^T = A$$. For symmetric $$A$$ it turns out that not only does there exist such $$P$$, we can always choose $$P$$ to be an orthogonal matrix i.e. $$P^T = P^{-1}$$, meaning that the columns of $$P$$ form an orthonormal basis for $$\mathbb{R}^n$$. Plugging into the above display you find $$$$A = P \Lambda P^T.$$$$ More generally, it turns out that a matrix $$A \in \mathbb{C}^{n \times n}$$ is unitarily diagonalizable (there exists $$P$$ with $$P^* = P^{-1}$$) if and only if $$A$$ is normal i.e. satisfies $$AA^* = A^*A$$. Here we use the notation $$A^* = \bar{A}^T$$.