An Example of Manually Calculating Eigenvalues and Eigenvectors of the Square Matrix
Apr. 22, 2025 • Updated Apr. 26, 2025
Consider a $3\times 3$ symmetric matrix:
\[\boldsymbol{\mathrm{A}}=\begin{bmatrix} -2 & 2 & 2\\ 2 & 1 & 4\\ 2 & 4 & 1\\ \end{bmatrix}\notag\]Firstly, we can calculate its eigenvalues:
\[\begin{split} \vert\boldsymbol{\mathrm{A}}-\lambda\boldsymbol{\mathrm{I}}\vert =&\begin{vmatrix} -2-\lambda & 2 & 2\\ 2 & 1-\lambda & 4\\ 2 & 4 & 1-\lambda\\ \end{vmatrix}\\ =&(-2-\lambda)(1-\lambda)^2+16+16\\ &-4(1-\lambda)-16(-2-\lambda)-4(1-\lambda)\\ =&-\lambda^3+27\lambda+54\\ =&(\lambda-6)(\lambda+3)^2 \end{split} \notag\]Let $\vert\boldsymbol{\mathrm{A}}-\lambda\boldsymbol{\mathrm{I}}\vert$ equals to zero, we’ll have:
\[\lambda_1=6,\ \lambda_2=\lambda_3=-3\label{eq1}\](1) When $\lambda_1=6$:
\[(\boldsymbol{\mathrm{A}}-\lambda_1\boldsymbol{\mathrm{I}})\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}} \Rightarrow\begin{bmatrix} -8 & 2 & 2\\ 2 & -5 & 4\\ 2 & 4 & -5\\ \end{bmatrix}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}} \Rightarrow\begin{bmatrix} 2 & 0 & -1\\ 0 & -1 & 1\\ 0 & 0 & 0\\ \end{bmatrix}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}}\\\notag\]then we can get a solution:
\[\boldsymbol{\mathrm{x}}= k_1\begin{bmatrix} 1 \\ 2 \\ 2\\ \end{bmatrix}\notag\notag\]where an eigenvector is:
\[\xi_1 =\begin{bmatrix} 1 \\ 2 \\ 2\\ \end{bmatrix}\notag\]Next, unitize it as:
\[\xi_1 = \begin{bmatrix} 1/\sqrt{1+4+4} \\ 2/\sqrt{1+4+4} \\ 2/\sqrt{1+4+4}\\ \end{bmatrix}= \begin{bmatrix} 1/3 \\ 2/3 \\ 2/3\\ \end{bmatrix}\label{eq2}\](2) When $\lambda_2=\lambda_3=-3$, we have:
\[(\boldsymbol{\mathrm{A}}-\lambda_1\boldsymbol{\mathrm{I}})\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}} \Rightarrow\begin{bmatrix} 1 & 2 & 2\\ 2 & 4 & 4\\ 2 & 4 & 4\\ \end{bmatrix}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}} \Rightarrow\begin{bmatrix} 1 & 2 & 2\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{0}}\\\notag\]whose solution is:
\[\boldsymbol{\mathrm{x}}=k_2\xi_2+k_3\xi_3=k_2\begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix}+k_3 \begin{bmatrix} -2 \\ 0 \\ 1\\ \end{bmatrix} \notag\]At this point, we can get two eigenvectors:
\[\xi_2 = \begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix},\ \xi_3 = \begin{bmatrix} -2 \\ 0 \\ 1\\ \end{bmatrix}\notag\]Then, use the Gram-Schmidt algorithm1 to make these two eigenvectors orthonormal:
\[\begin{split} \xi_2^{\prime}=&\xi_2=\begin{bmatrix} -2 \\ 1 \\ 0 \\ \end{bmatrix}\\ \xi_3^{\prime}=& \xi_3-\dfrac{\langle\xi_3,\xi_2\rangle}{\langle\xi_2,\xi_2\rangle}\xi_2 =\begin{bmatrix} -2 \\ 0 \\ 1\\ \end{bmatrix}-\dfrac{\langle \begin{bmatrix} -2 \\ 0 \\ 1\\ \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix}\rangle}{\langle \begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix}\rangle} \cdot\begin{bmatrix} -2 \\ 1 \\ 0\\ \end{bmatrix} =\begin{bmatrix} -2/5 \\ -4/5 \\ 1\\ \end{bmatrix} \end{split}\notag\] \[\begin{split} \xi_2^{\prime}&=\begin{bmatrix} -2/\sqrt{4+1}\\ 1/\sqrt{4+1}\\ 0\\ \end{bmatrix}=\begin{bmatrix} -2/\sqrt5\\ 1/\sqrt5\\ 0\\ \end{bmatrix}\\ \xi_3^{\prime}&= \begin{bmatrix} (-2/5)/(\sqrt{4/25+16/25+1})\\ (-4/5)/(\sqrt{4/25+16/25+1})\\ 1/(\sqrt{4/25+16/25+1})\\ \end{bmatrix}=\begin{bmatrix} -2/3\sqrt5\\ -4/3\sqrt5\\ 5/3\sqrt5\\ \end{bmatrix} \end{split}\notag\]Finally, we can get two eigenvectors:
\[\xi_2 = \begin{bmatrix} -2/\sqrt5 \\ 1/\sqrt5 \\ 0\\ \end{bmatrix},\ \xi_3 = \begin{bmatrix} -2/3\sqrt5 \\ -4/3\sqrt5 \\ 5/3\sqrt5\\ \end{bmatrix}\label{eq3}\]After obtaining all eigenvalues and eigenvectors, we can define one matrix:
\[\boldsymbol{\mathrm{U}} = \begin{bmatrix} \xi_1 & \xi_2 & \xi_3\\ \end{bmatrix} =\begin{bmatrix} 1/3 & -2/\sqrt5 & -2/3\sqrt5\\ 2/3 & 1/\sqrt5 & -4/3\sqrt5\\ 2/3 & 0 & 5/3\sqrt5\\ \end{bmatrix}\label{eq4}\]and a diagonal matrix:
\[\boldsymbol{\Lambda}= \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3\\ \end{bmatrix}= \begin{bmatrix} 6 & 0 & 0\\ 0 & -3 & 0\\ 0 & 0 & -3\\ \end{bmatrix}\notag\]then we can get:
\[\boldsymbol{\mathrm{U}}\boldsymbol{\Lambda}\boldsymbol{\mathrm{U}}^T= \begin{bmatrix} 1/3 & -2/\sqrt5 & -2/3\sqrt5\\ 2/3 & 1/\sqrt5 & -4/3\sqrt5\\ 2/3 & 0 & 5/3\sqrt5\\ \end{bmatrix} \begin{bmatrix} 6 & 0 & 0\\ 0 & -3 & 0\\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} 1/3 & 2/3 & 2/3\\ -2/\sqrt5 & 1/\sqrt5 & 0\\ -2/3\sqrt5 & -4/3\sqrt5 & 5/3\sqrt5 \end{bmatrix}=\boldsymbol{\mathrm{A}}\label{eq5}\]For the sake of simplicity, we can compute it in MATLAB:
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clc, clear, close all
U = [1/3, -2/sqrt(5), -2/(3*sqrt(5));
2/3, 1/sqrt(5), -4/(3*sqrt(5));
2/3, 0, 5/(3*sqrt(5))];
Lambda = diag([6, -3, -3]);
disp(U*Lambda*U')
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-2.0000 2.0000 2.0000
2.0000 1.0000 4.0000
2.0000 4.0000 1.0000
On the basis of this example, we can also verify some properties about eigenvalues and eigenvectors of a symmetric matrix:
- According to $\eqref{eq1}$, we can verify that, $n\times n$ symmetric matrix has $n$ real eigenvalues (counted by their multiplicities), and for each eigenvalue, a real eigenvector is associated with it (so, when $\lambda=-3$, we can find two eigenvectors, see $\eqref{eq3}$).2
- From $\eqref{eq2}$ and $\eqref{eq3}$, we have $\langle\xi_1,\xi_2\rangle=0$ and $\langle\xi_1,\xi_3\rangle=0$, which is because “eigenvectors corresponding to different eigenvalues of a real symmetric matrix are orthogonal.”2
- Besides $\langle\xi_1,\xi_2\rangle=0$ and $\langle\xi_1,\xi_3\rangle=0$, due to that we made a Gram-Schmidt process, we can see $\langle\xi_2,\xi_3\rangle=0$ from $\eqref{eq3}$. As a result, matrix $\boldsymbol{\mathrm{U}}$ in $\eqref{eq4}$ is an orthonormal matrix3, i.e. $\boldsymbol{\mathrm{U}}^T\boldsymbol{\mathrm{U}}=\boldsymbol{\mathrm{U}}\boldsymbol{\mathrm{U}}^T=\boldsymbol{\mathrm{I}}$, or rather, $\boldsymbol{\mathrm{U}}^T=\boldsymbol{\mathrm{U}}^{-1}$. Furthermore, based on $\eqref{eq5}$ we can verify that, “real symmetric matrices are diagonalizable by orthogonal matrices”.4
Reference