Diagonalization

Suppose that $$A \in \mathbb{R}^{n\times n}$$ has a linearly independent set of eigenvectors $$v_1, \dots, v_n$$:

$$Av_i = \lambda_i v_i$$ for $$i=1,\dots,n$$.

We can define $$T:=[v_1 \dots v_n]$$ and $$\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n)$$. Using the eigenvalue identities listed above, we can write:

$$AT = T\Lambda$$

and finally (since $$T$$ has a set of linearly independent eigenvectors):

$$T^{-1}A T = \Lambda$$.

In other words, similarity transformation by $$T$$ diagonalizes $$A$$. The converse holds too: If there is $$T=[v_1 \dots v_n]$$ such that $$T^{-1}A T = \Lambda = \text{diag}(\lambda_1, \dots, \lambda_n)$$, then $$AT = T\Lambda$$, which implies that $$v_1,\dots, v_n$$ form a set of linearly independent eigenvectors.

When is $$A$$ diagonalizable?
$$A$$ is diagonalizable if
 * there exists $$T$$ s.t. $$T^{-1}A T = \Lambda$$ is diagonal
 * $$A$$ has a set of linearly independent eigenvectors

Alternative, $$A$$ is diagonizable also if it has distinct eigenvalues. (The converse does not necessarily hold.)