Eigenvalues and Diagnolisation

Eigenvalues and Eigenvectors

We note that for some matrix \(\mathcal{A}\), the eigenvalue \(\lambda\) and eigenvector \(\vec v\) are defined such that

\[\mathcal{A}\vec v = \lambda \vec v\]

From here, we can multiply each side by the Identity Matrix, \(I\), to get the following results:

\[\begin{align*}\mathcal{A} \vec v &= \lambda I \vec v \\ \left(\mathcal{A} - \lambda I\right)\vec v &= \vec 0\end{align*}\]

We note that the only solutions here are that \(\vec v = \vec 0\) or \(det(\mathcal{A} - \lambda I) = 0\), which can be taken into account this determinant. Hence, we solve for \(\lambda\) such that \(det(\mathcal{A} - \lambda I) = 0\).

We note that we expect a characteristic polynomial from this determinant, which can be represented as \(\chi(\lambda)\). This gives a set of \(\lambda\) values such that \(\chi(\lambda) = 0\). These are the eigenvalues.

To determine the eigenvectors, we evaluate a vector space \(X\) such that \((\mathcal{A} - \lambda I)\vec X_i = \vec 0\). This is simply the nullspace of \(\mathcal{A} - \lambda I\). There you have the eigenvalues, and combining these vector spaces gives you the eigenspace, \(E_\lambda(\mathcal{A})\).

Diagnolisation

In diagnolisation, we derive a diagonal matrix \(\mathcal{D}\) and invertible matrix \(\mathcal{P}\) such that

\[\begin{align*}\mathcal{A} &= \mathcal{P}\mathcal{D}\mathcal{P}^{-1} \\ \mathcal{A}^2 &= \mathcal{P}\mathcal{D}\mathcal{P}^{-1}\mathcal{P}\mathcal{D}\mathcal{P}^{-1} = \mathcal{P}\mathcal{D}^2\mathcal{P}^{-1}\\ \mathcal{A}^n &= \mathcal{P}\mathcal{D}\mathcal{P}^{-1}\times \cdots \times \mathcal{P}\mathcal{D}\mathcal{P}^{-1} = \mathcal{P}\mathcal{D}^n\mathcal{P}^{-1}\end{align*}\]

Here, we consider a matrix \(\mathcal{A}\) that has a eigenspace of dimension \(n\), i.e. it has \(n\) linearly independent basis vectors, \(v_1\) to \(v_n\). If a matrix does not satisfy this requirement, then it CANNOT be diagnolised.

From here we are able to say this:

\[\begin{align*}\mathcal{P} &=\begin{bmatrix} \vert & \vert & &\vert \\ v_1 & v_2 & \cdots & v_n \\ \vert & \vert & &\vert \end{bmatrix} \\\\ \mathcal{D} &=\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \end{align*}\]

From here, if you're asked to determine \(\mathcal{A}^n\), this should be an ideal way to make your life easier. And that's it.