Examples: matrix diagonalization

Answer   For a square matrix $A$, matrix diagonalization is to find a diagonal matrix $\Lambda$ satisfying , where $P$ is an invertible matrix which diagonalizes $A$.
  In the following, we find the diagonal matrix $\Lambda$ for the matrix $A$ in $(1.1)$, and the invertive matrix $P$ that diagonalizes $A$. It is known that the diagonal elements of a diagonalized matrix are the eigenvalues of the original matrix. Therefore, by obtaining eigenvalues of $A$ and arranging them in diagonal elements, diagonalized matrix $\Lambda$ is obtained.   In order to obtain the eigenvalue $\Lambda$ of $A$, we need to solve the characteristic equation \begin{eqnarray} \left| \lambda I - A \right| = 0 \end{eqnarray} $$ \tag{1.2} $$ , which is a polynomial equation in the variable (eigenvalue) $\lambda$. Since the left-hand side is a $2 \times 2$ determinant, we have Then the solutions of $(1.2)$ are By arranging these solutions in diagonal elements, we obtain the diagonalized matrix $\Lambda$ as $$ \tag{1.3} $$   The invertible matrix $P$ diagonalizing the matrix $A$ is the matrix whose columun vectors are the eigenvectors of $A$. Therefore, $P$ is obtained, if the eigenvector for each eigenvalue of $A$ is obtained. So, we will derive the eigenvectors of the eigenvalues of $A$ as follows.


Case $\lambda=5$ :
  In this case, the eigenvector $\mathbf{x}$ satisfies Let $\mathbf{x}$ be We have Rearranging this equation, we obtain Therefore, the eigenvector is expressed as , where $x_{2}$ is an arbitrary value. We set $x_2=1$ for convenience, and obtain $$ \tag{1.4} $$


Case $\lambda=-2$ :
  In this case, the eigenvector $\mathbf{x}$ satisfies Let $\mathbf{x}$ be We have Rearranging this equation, we obtain Therefore, the eigenvector is expressed as where $x_2$ is an arbitrary value. We set $x_2=1$ for convenience, and obtain $$ \tag{1.5} $$


Invertible matrix $P$
  By $(1.4)$ and $(1.5)$, we obtain the invertible matrix $P$ as $$ \tag{1.6} $$
  We will check whether the matrix $P$ in equation $(1.6)$ actually diagonalizes the matrix $A$, that is, whether $P$, $A$ and $\Lambda$ satisfy To do that, we need to derive the inverse matrix $P^{−1}$.


Derivation of $P^{-1}$
  We will derive the inverse matrix $P^{−1}$ by Gaussian elimination. We define a matrix in which $P$ and the identity matrix $I$ are arranged side by side, $$ \tag{1.7} $$ and tramsform the left half matrix to the identity matrix by the elementary row operations: As a result, the matrix appearing in the right half becomes the inverse matrix $P^{−1}$. According to this method, performing the elementary row operations to the matrix $(1.7)$, we have Therere we obtain

Check diagonalization
  Now we can check the diagonalization as follows. We see that $P$ diagonalizes $A$.

Answer   For a square matrix $A$, matrix diagonalization is to find a diagonal matrix $\Lambda$ satisfying , where $P$ is an invertible matrix which diagonalizes $A$.
  In the following, we find the diagonal matrix $\Lambda$ for the matrix $A$ in $(2.1)$, and the invertive matrix $P$ that diagonalizes $A$. It is known that the diagonal elements of a diagonalized matrix are the eigenvalues of the original matrix. Therefore, by obtaining eigenvalues of $A$ and arranging them in diagonal elements, diagonalized matrix $\Lambda$ is obtained.   In order to obtain the eigenvalue $\Lambda$ of $A$, we need to solve the characteristic equation $$ \tag{2.2} $$ , which is a polynomial equation in the variable (eigenvalue) $\lambda$. Since the left-hand side is a $3 \times 3$ determinant, we have Then the solutions of $(2.2)$ are By arranging these solutions in diagonal elements, we obtain the diagonalized matrix $\Lambda$ as $$ \tag{2.3} $$   The invertible matrix $P$ that diagonalizes the matrix $A$ is the matrix whose columun vectors are the eigenvectors of $A$. Therefore, $P$ is obtained, if the eigenvector for each eigenvalue of $A$ is obtained. So, we will derive the eigenvectors of the eigenvalues of $A$ as follows.


Case $\lambda=-1$ :
  In this case, the eigenvector $\mathbf{x}$ satisfies Let $\mathbf{x}$ be We have Rearranging this equation, we obtain Therefore the eigenvector is expressed as , where $x_{3}$ is an arbitrary value. Here, we set $ x_ {3} = 1 $ for convenience, and obtain $$ \tag{2.4} $$


Case $\lambda=1$ :
  In this case, the eigenvector $\mathbf{x}$ satisfies Let $\mathbf{x}$ be We have Rearranging this equation, we obtain Therefore the eigenvector is expressed as , where $x_{3}$ is an arbitrary value. Here, we set $ x_ {3} = 1 $ for convenience, and obtain $$ \tag{2.5} $$

Case $\lambda=2$ :
  In this case, the eigenvector $\mathbf{x}$ satisfies Let $\mathbf{x}$ be We have Rearranging this equation, we obtain Therefore the eigenvector is expressed as , where $x_{3}$ is an arbitrary value. Here, we set $ x_ {3} = 1 $ for convenience, and obtain $$ \tag{2.6} $$ Invertible matrix $P$
  By (2.4), (2.5) and (2.6), we obtain the invertible matrix $P$ as $$ \tag{2.7} $$   We will check whether the matrix $P$ in equation $(2.7)$ actually diagonalizes the matrix $A$, that is, whether $P$, $A$ and $\Lambda$ satisfy   To do that, we need to derive the inverse matrix $P^{−1}$.


Derivation of $P^{-1}$
  We will derive the inverse matrix $P^{−1}$ by Gaussian elimination. We define a matrix in which $P$ and the identity matrix $I$ are arranged side by side, $$ \tag{2.8} $$ and tramsform the left half matrix to the identity matrix by the elementary row operations: As a result, the matrix appearing in the right half becomes the inverse matrix $A^{−1}$. According to this method, performing the elementary row operations to the matrix $(2.8)$, we have Therere we obtain

Check diagonalization
  Now we can check the diagonalization as follows. We see that $P$ diagonalizes $A$.