A square matrix A is diagonalizable if A is similar to a diagonal matrix

\begin{equation} A = PDP^{-1} \end{equation}

where D is a diagonal matrix

This is especially useful when one is confronted with large matrix operations such as $A^{42}$

\begin{equation} A^{42} = (PDP^{-1})^{42} = (PDP^{-1})(PDP^{-1})(PDP^{-1})...(PDP^{-1}) \end{equation}

It may be noted that there are very many $P^{-1}P$ terms which reduces to the identity matrix and reduces the whole equation down to

\begin{equation} PD^{42}P^{-1} \end{equation}

And D is much easier to raise to a power than multiplying A by itself 42 times. This is great for computers.

So, if D is something along the lines of $\begin{bmatrix} 2&0\\0&1 \end{bmatrix}$, this $A^{42}$ becomes much easier. In fact, it just becomes

\begin{align} A^{42} = PD^{42}P^{-1} = P\begin{bmatrix} 2^{42}&0\\0&1^{42} \end{bmatrix}P^{-1} \end{align}

Also, consequently, the Diagonal Matrix is comprised of the eigenvalues, and the P matrix is comprised of the eigenvectors IN THE SAME ORDER OF THE EIGENVALUES. Thus, in our previous example, the first eigenvalue in our D matrix was 2. We can therefore conclude that the first vector element of P is the eigenvector corresponding to that eigenvalue of 2.

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