Diagonalizable.md

$A \in \mathbb{R}^{n \times n} \land A = PDP^{-1} \implies A$ is diagonalizable, where $D$ is a diagonal matrix

$A$ is diagonalizable $\iff$ $A$ has $n$ linearly independent eigenvectors. i.e. $$ \displaylines{ A = PDP^{-1}\ \iff \ A = [\vec{v_1}, ..., \vec{v_n}] \begin{bmatrix} \lambda_1 & & \ & ... & \ & & \lambda_n \end{bmatrix} [\vec{v_1}, ..., \vec{v_n}]^{-1} } $$ where $\vec{v}$ vectors are linearly independent eigenvectors, and $\lambda_1, ..., \lambda_n$ are the corresponding eigenvalues, in order.

Distinct Eigenvalues

If $A \in \mathbb{R}^{n \times n}$ and has $n$ distinct eigenvalues, then $A$ is diagonalizable

Non-distinct Eigenvalues

You check that the sum of the geometric multiplicities is equal to the size of the matrix. e.g. for $$ \begin{bmatrix} 1 & & -1 \ & 2 & \ & & 1 \end{bmatrix} $$ Find the eigenvalues: $$ \displaylines{ \begin{vmatrix} 1-\lambda & & -1 \ & 2-\lambda & \ & & 1-\lambda \end{vmatrix} = (1-\lambda)^2 (2-\lambda) = 0\ \therefore \lambda = 1,1,2 } $$ We know that geomult <= algmult. Therefore $\lambda = 2$ has 1 distinct eigenvector. This means $\lambda = 1$ has to have 2 distinct eigenvectors to form a basis, so if it doesn't then the matrix is not diagonalizable. $$ A - I = \begin{bmatrix} 0 & & -1 \ & 1 & \ & & 0 \end{bmatrix} $$ There is only one free columns here. Therefore, the dimension of the Nullspace is one, not two, which means the matrix is not diagonalizable.

Basis of Eigenvectors

$$ \displaylines{ \text{Express the vector $\vec{x}0 =$} \begin{bmatrix} 4 \ 5 \end{bmatrix} \text{ as a linear combination of the vectors }\ \vec{v}1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \text{ and } \vec{v_2} =\begin{bmatrix} 1 \ -1 \end{bmatrix} \text{ and find the coordinates of } \vec{x_0} \text{ in the basis}\ \mathcal{B} = {\vec{v_1}, \vec{v_2}}\ \ [\vec{x_0}]{\mathcal{B}} = ?\quad\quad\quad [\vec{x_0}]{\mathcal{B}} = \begin{bmatrix} 4.5 \ -0.5 \end{bmatrix}\ \ \sim \ \ \text{Let } P = [\vec{v_1}\ \vec{v_2}],\ D = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}\ \ [A^k\ \vec{x}0]{\mathcal{B}} =\ ? \quad\quad\text{where } A = PDP^{-1},\ k\in \mathbb{Z}^{+}\ \ A^k = PD^k P^{-1} = [\vec{v_1}\ \vec{v_2}] \begin{bmatrix} 1^k & \ & (-1)^k \end{bmatrix}
[\vec{v_1}\ \vec{v_2}]^{-1}\ [A^k\ \vec{x}0]{\mathcal{B}} =\ ? } $$ #todo

Misc.

$$ \displaylines{ \text{Let d(x) be "x is diagonalizable"}\\ \text{Let i(x) be "x is invertible"}\\ d(A) \land i(A) \iff d(A^{-1})\land i(A^{-1}) } $$