Diagonalizable

$A\in {\mathbb{R}}^{n\times n}\wedge A=PD{P}^{-1}⟹A$ is diagonalizable, where $D$ is a diagonal matrix

$A$ is diagonalizable $⟺$ $A$ has $n$ linearly independent eigenvectors. i.e.

$$\begin{gathered} A=PD{P}^{-1} \\ ⟺ \\ A=[\overrightarrow{v_1},...,\overrightarrow{v_n}]\left[\begin{array}{ccc}{\lambda }_1& & \\ & ...& \\ & & {\lambda }_n\end{array}\right][\overrightarrow{v_1},...,\overrightarrow{v_n}{]}^{-1} \end{gathered}$$

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

$$\left[\begin{array}{ccc}1& & -1\\ & 2& \\ & & 1\end{array}\right]$$

Find the eigenvalues:

$$\left|\begin{array}{ccc}1-\lambda & & -1\\ & 2-\lambda & \\ & & 1-\lambda \end{array}\right|=(1-\lambda {)}^2(2-\lambda )=0$$ $$\therefore \lambda =1,1,2$$

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.

$$A-I=\left[\begin{array}{ccc}0& & -1\\ & 1& \\ & & 0\end{array}\right]$$

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

$$\begin{array}{c}\text{Express the vector x0=}\left[\begin{array}{c}4\\ 5\end{array}\right]\text{ as a linear combination of the vectors }\\ {\vec{v}}_1=\left[\begin{array}{c}1\\ 1\end{array}\right]\text{ and }\overrightarrow{v_2}=\left[\begin{array}{c}1\\ -1\end{array}\right]\text{ and find the coordinates of }\overrightarrow{x_0}\text{ in the basis}\\ \mathcal{B}=\{\overrightarrow{v_1},\overrightarrow{v_2}\}\\ \\ [\overrightarrow{x_0}{]}_{\mathcal{B}}=?[\overrightarrow{x_0}{]}_{\mathcal{B}}=\left[\begin{array}{c}4.5\\ -0.5\end{array}\right]\\ \\ \sim \\ \\ \text{Let }P=[\overrightarrow{v_1}\overrightarrow{v_2}],D=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\\ \\ [A^k{\vec{x}}_0{]}_{\mathcal{B}}=?\text{where }A=PD{P}^{-1},k\in {\mathbb{Z}}^{+}\\ \\ A^k=P{D}^k{P}^{-1}=[\overrightarrow{v_1}\overrightarrow{v_2}]\left[\begin{array}{cc}{1}^k& \\ & (-1{)}^k\end{array}\right][\overrightarrow{v_1}\overrightarrow{v_2}{]}^{-1}\\ [A^k{\vec{x}}_0{]}_{\mathcal{B}}=?\end{array}$$

Misc.

$$\begin{array}{c}\text{Let d(x) be "x is diagonalizable"}\\ \text{Let i(x) be "x is invertible"}\\ d(A)\wedge i(A)⟺d(A^{-1})\wedge i(A^{-1})\end{array}$$