How to diagonalize a 3x3 matrix

Last updated: Jan. 1, 2019
  Diagonalize a 3x3 matrix, , and find a non-singular matrix diagonalizing $A$.

  Answer

  In general, matrix diagonalization is to find a diagonal matrix $\Lambda$ satisfying for a square matrix $A$. Here, $P$ is a non-singular matrix and is called a matrix which diagonalizes $A$.
  It is known that the diagonal elements of the 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 $$ \tag{1} $$ , which is a polynomial equation in the variable $\lambda$. Since the left-hand side is a 3x3 determinant, we have . Equation $(1)$ is expressed as By factorizing the left-hand side, we have The solution is Arranging this solution in diagonal elements, the diagonalized matrix $\Lambda$ is obtained as   In general, a non-singular matrix $P$ diagonalizing a matrix $A$ is a matrix whose columun vectors are eigenvectors of $A$. Therefore, $P$ is obtained, if eigenvector for each eigenvalue of $A$ is obtained.   In this case, the eigenvector is a vector $\mathbf{x}$ that satisfies Let $\mathbf{x}$ be . We have . This equation is a simultaneous linear equation expresseed as . Solving this gives . Therefore the eigenvector for $\lambda=1$ is , where $x_{3}$ is an arbitrary value (see Appendix). Here, we set $ x_ {3} = 1 $ for convenience, and $$ \tag{2} $$   In this case, the eigenvector is a vector $\mathbf{x}$ that satisfies Let $\mathbf{x}$ be . We have . This equation is a simultaneous linear equation expresseed as . Solving this gives . Therefore the eigenvector for $\lambda=1$ is , where $x_{3}$ is an arbitrary value (see Appendix). Here, we set $ x_ {3} = 1 $ for convenience, and $$ \tag{3} $$   In this case, the eigenvector is a vector $\mathbf{x}$ that satisfies . Let $\mathbf{x}$ be . We have . This equation is a simultaneous linear equation expresseed as . Solving this gives . Therefore the eigenvector for $\lambda=2$ is , where $x_{3}$ is an arbitrary value (see Appendix). Here, we set $ x_ {3} = 1 $ for convenience, and $$ \tag{4} $$


  As mentioned above, since the non-singular matrix $P$ diagonalizing the matrix $A$ is a matrix whose eigenvectors as column vectors, from $ (2) (3) (4) $, we obtain $$ \tag{5} $$   We will check whether the matrix $P$ obtained in equation $(5)$ actually diagonalizes the matrix $A$. that is, whether the matrix $P$ and $A$ and the diagonal matrix $\Lambda$ satisfy To do that, we need derive the inverse matrix $P^{- 1}$ of the non-singular matrix $P$.   We will derive the inverse matrix $ P^{- 1} $ by Gaussian elimination. To do this, we define a matrix in which $P$ and the identity matrix I are arranged side by side, $$ \tag{6} $$ , 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 $A ^ {- 1}$ of $A$.
  According to this method, performing the elementary row operations to the matrix $(6)$, we have Therefore, we obtain
  From this, we can calculate $P^{-1}AP$ as . We confirmed that the matrix $P$ is a matrix that diagonalizes the matrix $A$.