Eigenvalue-Eigenvector Examples
Math 5718 - Spring 2001
These notes consist of examples of the various cases that arise in working with eigenvalues and eigenvectors. The examples are intended to illustrate the theorems of Chapter 5. Recall that the author of the text approaches the eigenvalue problem through invariant subspaces, without using the traditional definition of an eigenvalue (involving determinants and characteristic polynomials). You may certainly compute eigenvalues in the usual way, if it helps you to construct examples and improve your understanding. We also know that the eigenvalues of an upper triangular matrix (which includes diagonal matrices) are the diagonal elements of the matrix. Example 1: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^3$, given by

\begin{displaymath}T = \left( \begin{array}{ccc} 1 & 2 & -2\\ 0 & 2 & -1 \\ 0 & 0 & 3 \end{array} \right). \end{displaymath}

The eigenvalues, $\lambda_1=1, \lambda_2=2$, and $\lambda_3=3$, are on the diagonal of this upper triangular matrix. To find the eigenvectors, we find a basis for the invariant subspaces $\mbox{ null } (T-\lambda_i I)$. You should verify that

\begin{displaymath}\mbox{ null }(T-\lambda_1I) = \mbox{ null }\left( \begin{arr... ...& 0 & 2 \end{array} \right) = \mbox{ span } \{(1,0,0)^T \}, \end{displaymath}


\begin{displaymath}\mbox{ null }(T-\lambda_2I) = \mbox{ null }\left( \begin{arr... ...& 0 & 1 \end{array} \right) = \mbox{ span } \{(2,1,0)^T \}, \end{displaymath}

and

\begin{displaymath}\mbox{ null }(T-\lambda_3I) = \mbox{ null }\left( \begin{arr... ... 0 & 0 \end{array} \right) = \mbox{ span } \{(2,1,-1)^T \}. \end{displaymath}

The eigenvalues are distinct and the corresponding eigenvectors are linearly independent. Each eigenvalue has an invariant subspace of dimension 1. Referring to the proof of Theorem 5.10, you can also verify that

p(T)=T3-6T2+11T-6I = (T-I)(T-2I)(T-3I)=0.

This roots of this polynomial (soon to be called the characteristic polynomial) are the eigenvalues. Example 2: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^3$, given by

\begin{displaymath}T = \left( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right). \end{displaymath}

The eigenvalues, $\lambda_1=1$ and $\lambda_2=3$, are on the diagonal. To find the eigenvectors, we find a basis for the invariant subspaces $\mbox{ null } (T-\lambda_i I)$. You should verify that

\begin{displaymath}\mbox{ null }(T-\lambda_1I) = \mbox{ null }\left( \begin{arr... ...nd{array} \right) = \mbox{ span } \{(1,0,0)^T, (0,1,0)^T \}. \end{displaymath}

and

\begin{displaymath}\mbox{ null }(T-\lambda_2I) = \mbox{ null }\left( \begin{arr... ...& 0 & 0 \end{array} \right) = \mbox{ span } \{(0,0,1)^T \}. \end{displaymath}

In this case, but not always, the double eigenvalue has an invariant subspace of dimension 2. This example illustrates the provisions of Proposition 5.21 of the text: V has a basis consisting of eigenvectors, $V=U_1 \oplus U_2 \oplus U_3$, where Ui is the span of the ith eigenvector,

\begin{displaymath}V = \mbox{ null }(T-\lambda_1I) \oplus \mbox{ null }(T-\lambda_2I), \end{displaymath}

and

\begin{displaymath}\mbox{ dim } V = \mbox{ dim null }(T-\lambda_1I) + \mbox{ dim null }(T-\lambda_2I) = 2 + 1 = 3. \end{displaymath}

Referring to the proof of Theorem 5.10, you can also verify that

p(T)=T3-3T2-T+3I = (T-I)2(T-3I)=0.

This roots of this polynomial are the eigenvalues. Example 3: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^3$, given by

\begin{displaymath}T = \left( \begin{array}{ccc} 1 & 2 & -2\\ 0 & 1 & -1 \\ 0 & 0 & 3 \end{array} \right). \end{displaymath}

The eigenvalues, $\lambda_1=1$ and $\lambda_2=3$, are on the diagonal of this upper triangular matrix. To find the eigenvectors, we find a basis for the invariant subspaces $\mbox{ null } (T-\lambda_i I)$. You should verify that

\begin{displaymath}\mbox{ null }(T-\lambda_1I) = \mbox{ null }\left( \begin{arr... ...& 0 & 2 \end{array} \right) = \mbox{ span } \{(1,0,0)^T \}. \end{displaymath}

and

\begin{displaymath}\mbox{ null }(T-\lambda_2I) = \mbox{ null }\left( \begin{arr... ... 0 & 0 \end{array} \right) = \mbox{ span } \{(3,1,-2)^T \}. \end{displaymath}

In this case, the double eigenvalue has an invariant subspace of dimension 1. Referring to the proof of Theorem 5.10, you can also verify that

p(T)=T3-3T2-T+3I = (T-I)2(T-3I)=0.

This polynomial is identical to the polynomial of Example 2, because the eigenvalues of the two matrices are the same. Example 4: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^3$, given by

\begin{displaymath}T = \left( \begin{array}{ccc} -3 & 0 & 0\\ 0 & 5 & 2 \\ 0 & 2 & 2 \end{array} \right). \end{displaymath}

Because the matrix is not upper triangular, the eigenvalues are not so evident. There are three invariant subspaces corresponding to $\lambda_1 = -3, \lambda_2=1$, and $\lambda_3=6$. To find the eigenvectors, we find a basis for the invariant subspaces $\mbox{ null } (T-\lambda_i I)$. You should verify that

\begin{displaymath}\mbox{ null }(T-\lambda_1I) = \mbox{ null }\left( \begin{arr... ...& 2 & 5 \end{array} \right) = \mbox{ span } \{(1,0,0)^T \}, \end{displaymath}


\begin{displaymath}\mbox{ null }(T-\lambda_2I) = \mbox{ null }\left( \begin{arr... ... 2 & 1 \end{array} \right) = \mbox{ span } \{(0,1,-2)^T \}, \end{displaymath}

and

\begin{displaymath}\mbox{ null }(T-\lambda_3I) = \mbox{ null }\left( \begin{arr... ... 2 & -4 \end{array} \right) = \mbox{ span } \{(0,2,1)^T \}. \end{displaymath}

The three distinct eigenvalues correspond to three linearly independent eigenvectors. Each eigenvalue has an invariant subspace of dimension 1. Referring to the proof of Theorem 5.10, you can also verify that

p(T)=T3-4T2-15T+18I = (T+3I)(T-I)(T-6I)=0.

Again, the roots of this polynomial are the eigenvalues. Example 5: Consider the matrix given by

\begin{displaymath}T = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right). \end{displaymath}

This matrix has only one eigenvalue, $\lambda=0$, and it has an invariant subspace, $\mbox{ span }\{(1,0)^T\}$, of dimension 1. The main point here is that zero can be an eigenvalue and a matrix can be deficient in eigenvalues and eigenvectors. Example 6: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^2$ or $V={\bf C}^2$, given by

\begin{displaymath}T = \left( \begin{array}{cc} 0 & -2\\ 2 & 0 \end{array} \right). \end{displaymath}

Over the complex numbers, there are two distinct eigenvalues, $\lambda_{1,2}= \pm 2i$. Associated with each eigenvalue is an invariant subspace of T of dimension 1, $\mbox{ null }(T \pm 2iI)$. The eigenvectors are multiples of $v_{1,2}= (1, \pm i)^T$. Over the real numbers, there are no (real) eigenvalues. There is an invariant subspace of T, namely $\mbox{null }T^2+4I$. In fact, you can verify that T2+4I=0, meaning that the invariant subspace is ${\bf R}^2$. Example 7: Consider the matrix $T:V \rightarrow V$, where $V= {\bf R}^2$ or $V={\bf C}^2$, given by

\begin{displaymath}T = \left( \begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{array} \right). \end{displaymath}

If you compute the eigenvalues explicitly, they turn out to be $\lambda_{1,2}=\pm i$ and $\lambda_3=2$. If we were working over the complex numbers, there would be three invariant subspaces, $\mbox{ null } (T-\lambda_i I)$, and there would be three linearly independent eigenvectors (with complex elements) associated with the eigenvalues. However, over the real numbers, there are two invariant subspaces. You can verify that

\begin{displaymath}\mbox{ null }(T^2+I) = \mbox{ null }\left( \begin{array}{ccc... ...nd{array} \right) = \mbox{ span } \{(1,0,0)^T, (0,1,0)^T \}, \end{displaymath}

and

\begin{displaymath}\mbox{ null }(T-\lambda_3I) = \mbox{ null }\left( \begin{arr... ...& 0 & 0 \end{array} \right) = \mbox{ span } \{(0,0,1)^T \}. \end{displaymath}

In this case, there is an invariant subspace of dimension 2, with a basis v1=(1,0,0)T and v2=(0,1,0)T. Notice that v1 and v2 are not themselves eigenvectors; however, Tv1=v2 and Tv2=v1. Consistent with Theorem 5.26, we are assured of one real eigenvalue. Referring to the proof of Theorem 5.24, you can also verify that

p(T)=T3-2T2+T-2I = (T2+I)(T-2I) =0.