Opening the book…
- Find eigenvalues and eigenvectors for similar matrices, elementary matrices, and triangular matrices.
There are three special kinds of matrices which we can use to simplify the process of finding eigenvalues and eigenvectors. Throughout this section, we will discuss similar matrices, elementary matrices, as well as triangular matrices.
Similar Matrices
We begin with a definition.
Let \(A\) and \(B\) be \(n \times n\) matrices. Suppose there exists an invertible matrix \(P\) such that \[A = P^{-1}BP\nonumber \] Then \(A\) and \(B\) are called similar matrices.
It turns out that we can use the concept of similar matrices to help us find the eigenvalues of matrices. Consider the following lemma.
Let \(A\) and \(B\) be similar matrices, so that \(A=P^{-1}BP\) where \(A,B\) are \(n\times n\) matrices and \(P\) is invertible. Then \(A,B\) have the same eigenvalues.
- Proof
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Suppose \(A = P^{-1}BP\) and \(\lambda\) is an eigenvalue of \(A\), that is \(A\vec{x}=\lambda \vec{x}\) for some \(\vec{x}\neq \vec{0}.\) Then \[P^{-1}BP\vec{x}=\lambda \vec{x}\nonumber \] and so \[BP\vec{x}=\lambda P\vec{x}\nonumber \]
Since \(P\) is an invertible matrix (hence corresponds to a one-to-one transformation), and since \(\vec{x} \neq 0\), it follows that \(P\vec{x} \neq 0\). Here, \(P\vec{x}\) plays the role of the eigenvector in this equation. Thus \(\lambda\) is also an eigenvalue of \(B\). One can similarly verify that any eigenvalue of \(B\) is also an eigenvalue of \(A\), and thus both matrices have the same eigenvalues as desired.
Note that this proof also demonstrates that the eigenvectors of \(A\) and \(B\) will (generally) be different. We see in the proof that \(A\vec{x} = \lambda \vec{x}\), while \(B \left(P\vec{x}\right)=\lambda \left(P\vec{x}\right)\). Therefore, for an eigenvalue \(\lambda\), \(A\) will have the eigenvector \(\vec{x}\) while \(B\) will have the eigenvector \(P\vec{x}\).
Elementary Matrices
The second special type of matrices we discuss in this section is elementary matrices. Recall that an elementary matrix \(E\) is obtained by applying one row operation to the identity matrix.
It is possible to use elementary matrices to find a simpler similar matrix before searching for its eigenvalues and eigenvectors. This is illustrated in the following example.
Find the eigenvalues for the matrix \[A = \left[ \begin{array}{rrr} 33 & 105 & 105 \\ 10 & 28 & 30 \\ -20 & -60 & -62 \end{array} \right]\nonumber \]
Solution
This matrix has big numbers and therefore we would like to simplify as much as possible before computing the eigenvalues.
We will do so using row operations. First, add \(2\) times the second row to the third row. To do so, left multiply \(A\) by \(E \left(2\times2 + 3\right)\). Then right multiply \(A\) by the inverse of \(E \left(2\times2 + 3\right)\) as illustrated. \[\left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right] \left[ \begin{array}{rrr} 33 & 105 & 105 \\ 10 & 28 & 30 \\ -20 & -60 & -62 \end{array} \right] \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array} \right] =\left[ \begin{array}{rrr} 33 & -105 & 105 \\ 10 & -32 & 30 \\ 0 & 0 & -2 \end{array} \right]\nonumber \] By Lemma \(\PageIndex{1}\), the resulting matrix has the same eigenvalues as \(A\) where here, the matrix \(E \left(2\times2 + 3\right)\) plays the role of \(P\).
We do this step again, as follows. In this step, we use the elementary matrix obtained by adding \(-3\) times the second row to the first row, or \(E \left(-3\times2 + 1\right)\). \[\left[ \begin{array}{rrr} 1 & -3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{rrr} 33 & -105 & 105 \\ 10 & -32 & 30 \\ 0 & 0 & -2 \end{array} \right] \left[ \begin{array}{rrr} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] =\left[ \begin{array}{rrr} 3 & 0 & 15 \\ 10 & -2 & 30 \\ 0 & 0 & -2 \end{array} \right] \label{elemeigenvalue}\] Again by Lemma \(\PageIndex{1}\), this resulting matrix has the same eigenvalues as \(A\). At this point, we can easily find the eigenvalues. Let \[B = \left[ \begin{array}{rrr} 3 & 0 & 15 \\ 10 & -2 & 30 \\ 0 & 0 & -2 \end{array} \right]\nonumber \] Then, we find the eigenvalues of \(B\) (and therefore of \(A\)) by solving the equation \(\det \left( \lambda I - B \right) = 0\). You should verify that this equation becomes \[\left(\lambda +2 \right) \left( \lambda +2 \right) \left( \lambda - 3 \right) =0\nonumber \] Solving this equation results in eigenvalues of \(\lambda_1 = -2, \lambda_2 = -2\), and \(\lambda_3 = 3\). Therefore, these are also the eigenvalues of \(A\).
Through using elementary matrices, we were able to create a matrix for which finding the eigenvalues was easier than for \(A\). At this point, you could go back to the original matrix \(A\) and solve \(\left( \lambda I - A \right) \vec{x} = 0\) to obtain the eigenvectors of \(A\).
Notice that when you multiply on the right by an elementary matrix, you are doing the column operation defined by the elementary matrix. In \(\eqref{elemeigenvalue}\) multiplication by the elementary matrix on the right merely involves taking three times the first column and adding to the second. Thus, without referring to the elementary matrices, the transition to the new matrix in \(\eqref{elemeigenvalue}\) can be illustrated by \[\left[ \begin{array}{rrr} 33 & -105 & 105 \\ 10 & -32 & 30 \\ 0 & 0 & -2 \end{array} \right] \rightarrow \left[ \begin{array}{rrr} 3 & -9 & 15 \\ 10 & -32 & 30 \\ 0 & 0 & -2 \end{array} \right] \rightarrow \left[ \begin{array}{rrr} 3 & 0 & 15 \\ 10 & -2 & 30 \\ 0 & 0 & -2 \end{array} \right]\nonumber \]
Triangular Matrices
The third special type of matrix we will consider in this section is the triangular matrix. Recall that an upper (lower) triangular matrix contains all zeros below (above) the main diagonal. Remember that finding the determinant of a triangular matrix is a simple procedure of taking the product of the entries on the main diagonal. It turns out that this leads to a simple way to find the eigenvalues of a triangular matrix.
In the next example we will demonstrate that the eigenvalues of a triangular matrix are the entries on the main diagonal.
Let \(A=\left[ \begin{array}{rrr} 1 & 2 & 4 \\ 0 & 4 & 7 \\ 0 & 0 & 6 \end{array} \right] .\) Find the eigenvalues of \(A\).
Solution
We need to solve the equation \(\det \left( \lambda I - A \right) = 0\) as follows \[\begin{aligned} \det \left( \lambda I - A \right) = \det \left[ \begin{array}{ccc} \lambda -1 & -2 & -4 \\ 0 & \lambda -4 & -7 \\ 0 & 0 & \lambda -6 \end{array} \right] =\left( \lambda -1 \right) \left( \lambda -4 \right) \left( \lambda -6 \right) =0\end{aligned}\]
Solving the equation \(\left( \lambda -1 \right) \left( \lambda -4 \right) \left( \lambda -6 \right) = 0\) for \(\lambda \) results in the eigenvalues \(\lambda_1 = 1, \lambda_2 = 4\) and \(\lambda_3 = 6\). Thus the eigenvalues are the entries on the main diagonal of the original matrix.
The same result is true for lower triangular matrices. For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. To find the eigenvectors of a triangular matrix, we use the usual procedure.
In the next section, we explore an important process involving the eigenvalues and eigenvectors of a matrix.
This page titled 6.2: Eigenvalues and Eigenvectors for Special Matrices was last modified on Wed, 27 Aug 2025 04:30:10 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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