Given an n x n matrix [math]A[/math] and let [math]T\left(v\right)=Av[/math] for any column vector [math]v[/math] in [math]\mathbb{R}^n[/math] (in standard coordinates). Sometimes we can simplify the matrix representation of [math]T[/math] drastically if we choose a "nice" coordinate system. For example, if there are n linearly independent eigenvectors [math]v_1,v_2,\ldots,v_n[/math] of [math]A[/math], then they can form a basis for [math]\mathbb{R}^n[/math], which is called the [b]eigenbasis[/b] of [math]\mathbb{R}^n[/math]. Using it to define a new coordinate system, the new matrix representation of T will be [math]B=\begin{pmatrix}\lambda_1& 0 & \cdots & 0\\0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0&0&\cdots&\lambda_n\end{pmatrix}[/math], where [math]Av_i=\lambda_iv_i[/math] i.e. [math]v_i[/math] is an eigenvector corresponding to the eigenvalue [math]\lambda_i[/math] for [math]i=1,\ldots,n[/math]. In other words, [math]B=P^{-1}AP[/math], where [math]P=\left(v_1 \ v_2 \ \cdots \ v_n\right)[/math] i.e. the matrix formed by the column vectors of the eigenvectors (in standard coordinates). Hence, [math]A[/math] is similar to the diagonal matrix [math]B[/math] and this process is called [b]diagonalization[/b]. A matrix is said to be [b]diagonalizable[/b] if it is similar to a diagonal matrix. The following theorem gives a characterization of diagonalizable matrices:[br][br][u]Theorem[/u]: An n x n matrix [math]A[/math] is diagonalizable if and only if [math]A[/math] has n linearly independent eigenvectors.[br][br][br][u]Example 1[/u]: Recall the previous [url=https://www.geogebra.org/m/rsamte2c#material/m5ww8s3k]example[/url] [math]A=\begin{pmatrix}2&3\\3&-6\end{pmatrix}[/math]. We have already known that [math]\begin{pmatrix}3\\1\end{pmatrix}[/math] is an eigenvector of [math]A[/math] corresponding to [math]\lambda=3[/math] and [math]\begin{pmatrix}-\frac13\\1\end{pmatrix}[/math] is an eigenvector of [math]A[/math] corresponding to [math]\lambda=-7[/math]. Therefore, we define [math]P=\begin{pmatrix}3&-\frac13\\1&1\end{pmatrix}[/math] and [math]D=P^{-1}AP[/math], where [math]D=\begin{pmatrix}3&0\\0&-7\end{pmatrix}[/math]. In other words,[br][br][math]\begin{pmatrix}2&3\\3&-6\end{pmatrix}=\begin{pmatrix}3&-\frac13\\1&1\end{pmatrix}\begin{pmatrix}3&0\\0&-7\end{pmatrix}\begin{pmatrix}3&-\frac13\\1&1\end{pmatrix}^{-1}[/math][br][br](Note: the matrix [math]P[/math] in the diagonalization is not unique as there are many choices of eigenvectors.)[br][br][br][u]Example 2[/u]: The matrx [math]A=\begin{pmatrix}1&1\\0&1\end{pmatrix}[/math] is not diagonalizable because its only eigenvalue is [math]1[/math] and the eigenspace is [math]\text{Span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}[/math], which is not enough to form a basis for [math]\mathbb{R}^2[/math].[br][br][br][u]Example 3[/u]: Diagonalize [math]A=\begin{pmatrix}1&3&3\\-3&-5&-3\\3&3&1\end{pmatrix}[/math], if possible.[br][br]Solution: Compute the eigenvalues by solving the characteristic equation:[br][br][math]\begin{vmatrix}1-\lambda&3&3\\-3&-5-\lambda&-3\\3&3&1-\lambda\end{vmatrix}=0[/math][br][math]\Rightarrow -\lambda^3-3\lambda^2+4=0\Rightarrow -(\lambda-1)(\lambda+2)^2=0[/math][br][br]Hence, the eigenvalues of [math]A[/math] are [math]\lambda=1, \lambda=-2[/math].[br][br]When [math]\lambda=1[/math], solve [math]\begin{pmatrix}0&3&3\\-3&-6&-3\\3&3&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}[/math]. The matrix is row reduced to the matrix in echelon form :[br][br][math]\begin{pmatrix}3&3&0\\0&-3&-3\\0&0&0\end{pmatrix}[/math][br][br]The solutions in the parametric vector form are [math]\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}s\\-s\\s\end{pmatrix}=s\begin{pmatrix}1\\-1\\1\end{pmatrix}[/math].[br][br]When [math]\lambda=-2[/math], solve [math]\begin{pmatrix}3&3&3\\-3&-3&-3\\3&3&3\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}[/math]. The matrix is row reduced to the matrix in echelon form :[br][br][math]\begin{pmatrix}3&3&3\\0&0&0\\0&0&0\end{pmatrix}[/math][br][br]The solutions in the parametric vector form are [math]\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}-s-t\\s\\t\end{pmatrix}=s\begin{pmatrix}-1\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\0\\1\end{pmatrix}[/math].[br][br]We have three eigenvectors [math]\left\{\begin{pmatrix}1\\-1\\1\end{pmatrix},\begin{pmatrix}-1\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix}\right\}[/math] that are linearly independent. Therefore, they form an eignebasis for [math]\mathbb{R}^3[/math] and [math]A[/math] is diagonalizable:[br][br][math]\begin{pmatrix}1&3&3\\-3&-5&-3\\3&3&1\end{pmatrix}=PDP^{-1}=\begin{pmatrix}1&-1&-1\\-1&1&0\\1&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&-2&0\\0&0&-2\end{pmatrix}\begin{pmatrix}1&-1&-1\\-1&1&0\\1&0&1\end{pmatrix}^{-1}[/math][br][br](Note: The eigenvalues in the diagonal matrix [math]D[/math] must be in the same order as the eigenvectors in [math]P[/math].)[br][br]