6  Eigenvalues and Eigenvectors

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Let \(\mathbb{F}\) be either the real numbers or the complex numbers. A nonzero vector \(\vec v\) in \(\mathbb{F}^n\) is called an of an \(n\times n\) matrix \(A\) if \(A \vec v\) is a scalar multiple of \(\vec v\), that is \(A \vec v= \lambda \vec v\) for some scalar \(\lambda\). Note that this scalar \(\lambda\) may be zero. The scalar \(\lambda\) is called the eigenvalue associated with the eigenvector \(\vec v\). Even though, \(A\vec 0=\lambda \vec 0\) we do not call \(\vec 0\) an eigenvector. Of course a matrix need not have any eigenvalues or eigenvectors, but notice if \(\vec v\) is an eigenvector of matrix \(A\), then \(\vec v\) is an eigenvector of matrices \(A^2\), \(A^3\), as well, with \(A^t\vec v=\lambda^t \vec v,\) for all positive integers \(t\). If \(\mathbb{F}=\mathbb{C}\), then counting multiplicities, every \(n\times n\) matrix has exactly \(n\) eigenvalues.

If \(\vec v\) is an eigenvector of the \(n\times n\) matrix \(A\) with associated eigenvalue \(\lambda\), what can you say about \(\ker(A-\lambda I_n)\)? Is the matrix \(A-\lambda I_n\) invertible? We know \(A \vec v=\lambda \vec v\) so \((A-\lambda I_n)\vec v=A \vec v-\lambda I_n\vec v=\lambda\vec v-\lambda \vec v=0\). Thus a nonzero vector \(\vec v\) is in the kernel of \((A-\lambda I_n)\). Therefore, \(\ker(A-\lambda I_n)\neq \{\vec 0\}\) and so \(A-\lambda I_n\) is not invertible.

Lemma 6.1 Let \(A\) be an \(n\times n\) matrix \(A\) and \(\lambda\) a scalar. Then \(\lambda\) is an eigenvalue of \(A\) if and only if \(\det(A-\lambda I_n)=0\).

Proof. The proof follows from the chain of equivalent statements:

  • \(\lambda\) is an eigenvalue of \(A\),
  • there exists a nonzero vector \(\vec v\) such that \((A -\lambda I_n ) \vec v=0,\)
  • \(\ker(A -\lambda I_n )\neq \{\vec 0\}\),
  • matrix \(A-\lambda I_n\) fails to be invertible, and
  • \(\det(A -\lambda I_n )=0\).

Example 6.1 Find all eigenvectors and eigenvalues of the identity matrix \(I_n\). Since \(I_n \vec v = \lambda \vec v= 1 \vec v\) for all \(\vec v\in \mathbb{R}^n\), all nonzero vectors in \(\mathbb{R}^n\) are eigenvectors of \(I_n\), with eigenvalues \(\lambda=1\).

Lemma 6.2 The eigenvalues of a triangular matrix are its diagonal entries.

Proof. Let \(A\) be a triangular matrix. Then \(A-\lambda I_n\) is also a triangular matrix, and so \(\det(A-\lambda I_n)\) is the product of its diagonal entries. Let \(a_{ii}\) be any diagonal entry of \(A\). Then \(a_{ii}-\lambda\) is the corresponding diagonal entry of \(A-\lambda I_n\). Thus \(\lambda\) is an eigenvalue of \(A\) if and only if \(a_{ii}-\lambda=0\) by \(\ref{eigenprop}\).

Example 6.2 Find a basis of the linear space \(V\) of all \(2\times 2\) matrices for which \(\vec e_1\) is an eigenvector. For an arbitrary \(2\times 2\) matrix we want \[ \begin{bmatrix} a & b \\c & d\end{bmatrix} \vectortwo{1}{0}=\vectortwo{a}{c}=\vectortwo{\lambda}{0}=\lambda \vectortwo{1}{0} \] for any \(\lambda\). Hence \(a, b, d\) are free and \(c=0\); thus a desired basis of \(V\) is \[ \left( \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix} \right). \]

Example 6.3 Find a basis of the linear space \(V\) of all \(4\times 4\) matrices for which \(\vec e_2\) is an eigenvector. We want to find all \(4 \times 4\) matrices \(A\) such that \(A \vec e_2=\lambda e_2\). Thus the second column of an arbitrary \(4 \times 4\) matrix \(A\) must be of the form \(\vectorfour{0}{\lambda}{0}{0}^T\), so \[ A=\begin{bmatrix} a & 0 & c & d \\ e & \lambda & f & g \\ h & 0 & i & j \\ k & 0 & l & m\end{bmatrix}. \] Let \(E_{ij}\) denote the \(4\times 4\) matrix with all entries zero except for a 1 in the \(i\)-th row and \(j\)-th column. Then a basis for \(V\) is \[ \left( E_{11}, E_{21}, E_{31}, E_{41}, E_{22}, E_{13}, E_{23}, E_{33}, E_{34}, E_{41}, E_{42}, E_{43}, E_{44} \right) \] and so the dimension of \(V\) is 13.

Example 6.4 Find the eigenvalues and find a basis for each eigenspace given \(A=\begin{bmatrix}1 & 0 & 0 \\ -5 & 0 & 2 \\ 0 & 0 & 1 \end{bmatrix}\). Find an eigenbasis for \(A\). The eigenvalues are \(\lambda_1=0\) and \(\lambda_2=\lambda_3=1\). A basis for \(E_{0}\) is \(\left(\vectorthree{0}{1}{0}\right)\). A basis for \(E_{1}\) is \(\left(\vectorthree{1}{-5}{0},\vectorthree{0}{2}{1}\right)\). An eigenbasis for \(A\) is \(\left(\vectorthree{0}{1}{0},\vectorthree{1}{-5}{0},\vectorthree{0}{2}{1}\right)\).

Example 6.5 Find a basis of the linear space \(V\) of all \(2\times 2\) matrices \(A\) for which \(\vectortwo{1}{-3}\) is an eigenvector. For an arbitrary \(2\times 2\) matrix we want \[ \begin{bmatrix} a & b \\c & d \end{bmatrix} \vectortwo{1}{-3}=\lambda\vectortwo{1}{-3}=\vectortwo{\lambda}{-3\lambda}. \] Thus \(a-3b=\lambda\), \(c-3d=-3\lambda\) and so \(c=-3a+9b+3d\). Thus \(A\) must be of the form \[ \begin{bmatrix} a & b \\ -3a+9b+3d & d\end{bmatrix}=a\begin{bmatrix} 1 & 0 \\-3 & 0 \end{bmatrix}+b \begin{bmatrix} 0 & 1 \\ 9 & 0 \end{bmatrix}+ d\begin{bmatrix} 0 & 0 \\ 3 & 1\end{bmatrix}. \] Thus a basis of \(V\) is \[ \left( \begin{bmatrix} 1 & 0 \\ -3 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 9 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 3 & 1\end{bmatrix} \right) \] and so the dimension of \(V\) is \(3\).

Example 6.6 Find a basis of the linear space \(V\) of all \(3\times 3\) matrices \(A\) for which both \(\vectorthree{1}{0}{0}^T\) and \(\vectorthree{0}{0}{1}^T\) are eigenvectors. Since \(A \vectorthree{1}{0}{0}^T\) is simply the first column of \(A\), the first column must be a multiple of \(\vec e_1\). Similarly, the third column must be a multiple of \(\vec e_3\). There are no other restrictions on the form of \(A\), meaning it can be any matrix of the form \[ \begin{bmatrix} a & b & 0 \\ 0 & c & 0 \\ 0 & d & e \end{bmatrix} \] Thus a basis of \(V\) is \[ \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) \] and so the dimension of \(V\) is 5.

Theorem 6.1 If \(A\) is an \(n\times n\) matrix, then \(\det(A-\lambda I_n)\) is a polynomial of degree \(n\), of the form \[\begin{equation} \label{chpo} f_A(\lambda)=(-\lambda)^n+\text{trace} (A) (-\lambda)^{n-1}+\cdots +\det(A). \end{equation}\]

Proof. This proof is left for the reader.

The equation \(\det(A-\lambda I_n)=0\) is called the characteristic equation of \(A\). The polynomial in \(\ref{chpo}\) is called the characteristic polynomial and is denoted by \(f_A(\lambda)\). We say that an eigenvalue \(\lambda_0\) of a square matrix \(A\) has algebraic multiplicity \(k\) if \(\lambda_0\) is a root of multiplicity \(k\) of the characteristic polynomial \(f_A(\lambda)\) meaning that we can write \[ f_A(\lambda)=(\lambda_0-\lambda)^k g(\lambda) \] for some polynomial \(g(\lambda)\) with \(g(\lambda_0)\neq 0\).

Example 6.7 Find the characteristic equation for a $2 $ matrix \(A\). The characteristic equation of \(A=\begin{bmatrix} a& b \\c & d\end{bmatrix}\) is \[ f_A(\lambda) =\det \begin{bmatrix} a-\lambda& b \\c & d-\lambda \end{bmatrix} %=(a-\lambda)(d-\lambda)-bc =\lambda^2-(a+d)\lambda +(ad-bc)=0. \]

Example 6.8 Use the characteristic polynomial \(f_A(\lambda)\) to determine the eigenvalues and their multiplicities of \[ A=\begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{bmatrix}. \] The characteristic equation is \(f_A(\lambda)=-\lambda^2(\lambda+3)\). So \(\lambda_1=0\) with algebraic multiplicity of 2 and \(\lambda_2=-3\) with algebraic multiplicity of 1 are the eigenvalues of \(A\).

Example 6.9 Consider the matrix \(A=\begin{bmatrix} a & b \\ b & c \end{bmatrix}\), where \(a, b, c\) are nonzero constants. For which values of $a, b, c $ does \(A\) have two distinct eigenvalues? The characteristic equation is \(f_A(\lambda)=\lambda^2-(a+c)\lambda+(a c-b^2)\). The discriminant of this quadratic equation is \[ (a+c)^2-4(ac-b^2)=a^2+2ac+c^2-4ac+4b^2=(a-c)^2+4b^2. \] The discriminant is always positive since \(b\neq 0\). Thus, the matrix \(A\) there will always have two distinct real eigenvalues.

Example 6.10 In terms of eigenvalues of \(A\), which \(2\times 2\) matrices \(A\) does there exist an invertible matrix \(S\) such that \(AS=SD\), where \(D=\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\)? If we let \(S=\begin{bmatrix} \vec v_1 & \vec v_2\end{bmatrix}\), then \(AS=\begin{bmatrix} A \vec v_1 & A \vec v_2\end{bmatrix}\) and \(SD=\begin{bmatrix} 2 \vec v_1 & 3\vec v_2\end{bmatrix}\). So that \(\vec v_1\) must be an eigenvector of \(A\) with eigenvalue 2, and \(\vec v_2\) must be an eigenvector of \(A\) with eigenvalue 3. Thus, the matrix \(S\) will exist and will have first column has an eigenvector of \(A\) with eigenvalue 2, and have second column is an eigenvector of \(A\) with eigenvalue of 3. Therefore, \(A\) can be any matrix satisfying these requirements.

Example 6.11 Let \(A\) be a matrix with eigenvalues \(\lambda_1, \ldots, \lambda_k\).

  • Show the eigenvalues of \(A^T\) are \(\lambda_1, \ldots, \lambda_k\).
  • Show the eigenvalues of \(\alpha A\) are \(\alpha\lambda_1, \ldots, \alpha \lambda_k\).

Show \(A^{-1}\) exists if and only if \(\lambda_1 \cdots \lambda_k\neq 0\). - Also, show that if \(A^{-1}\) exists then its eigenvalues are \(1/\lambda_1,\ldots,1/\lambda_k\).

Example 6.12 Let \(A\) be a matrix with eigenvalues \(\lambda_1, \ldots, \lambda_k\) and let \(m\) be a positive integer. Show that the eigenvalues of \(A^m\) are \(\lambda^m_1, \ldots, \lambda^m_k\).

Example 6.13 By using the matrix \[ \begin{bmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ \frac{-a_n}{a_0} & \frac{-a_{n-1}}{a_0}& \frac{-a_{n-2}}{a_0} & \cdots & \frac{-a_1}{a_0} \end{bmatrix} \] Show that any given polynomial \(a_o \lambda^n+a_1\lambda^{n-1}+\cdots +a_{n-1}\lambda +a_n\) where \(a_0\neq 0\), of degree \(n\) may be regarded as the characteristic polynomial of a matrix of order \(n\). This matrix is called the companion matrix of the given polynomial.

Example 6.14 Let \(A\) and \(B\) be \(n\times n\) matrices. Show that \(AB\) and \(BA\) have the same eigenvalues.

Example 6.15 Let \(A\) and \(B\) be real \(n\times n\) matrices with distinct eigenvalues. Prove that \(AB=BA\) if and only if \(A\) and \(B\) have the same eigenvectors.

Example 6.16 Prove that the characteristic polynomial of the block-triangular matrix \(A=\begin{bmatrix} B & C\\ 0 & D\end{bmatrix}\) is the product of the characteristic polynomials of \(B\) and \(D\).

Example 6.17 Suppose that \(A\) is an invertible \(n\times n\) matrix. Prove that \[\begin{equation} f_{A^{-1}}(x)=(-x)^n\det(A^{-1})f_A\left(\frac{1}{x}\right). \end{equation}\]

Example 6.18 Let \(A\) be an \(n\times n\) matrix. Prove that \(A\) and \(A^T\) have the same characteristic polynomial and hence the same eigenvalues.

If\(\lambda\) is an eigenvalue of an \(n\times n\) matrix \(A\), then the kernel of the matrix \(A-\lambda I_n\) is called the eigenspace associated with \(\lambda\) and is denoted by \(E_\lambda\). The dimension of the eigenspace is called the geometric multiplicity of eigenvalue \(\lambda\). In other words, the geometric multiplicity is the nullity of the matrix \(A-\lambda I_n\).

Theorem 6.2 Let \(A\) be an \(n\times n\) matrix. If \(\lambda_1, \ldots, \lambda_k\) are distinct eigenvalues of \(A\), and \(\vec v_1, \ldots, \vec v_k\) are any nonzero eigenvectors associated with these eigenvalues respectively, then \(\vec v_1, \ldots, \vec v_k\) are linearly independent.

Proof. Suppose there exists constants \(c_1, \ldots, c_k\) such that \[\begin{equation} \label{eigenveclin} c_1 \vec v_1+\cdots +c_k \vec v_k=0 \end{equation}\] Using the fact that \(A \vec v_i=\lambda_i \vec v_i\) we multiply \(\ref{eigenveclin}\) by \(A\) to obtain \[\begin{equation} \label{eigenveclin2} c_1 \lambda_1\vec v_1+\cdots c_k \lambda_k \vec v_k=0. \end{equation}\] Repeating this again we obtain \[\begin{equation} \label{eigenveclin3} c_1 \lambda^2_1\vec v_1+\cdots +c_k \lambda^2_k \vec v_k=0. \end{equation}\] Repeating, we are lead to the system in the vector unknowns \(\vec v_1, \ldots, \vec v_k\) \[\begin{equation} \begin{bmatrix} c_1 \vec v_1 & \cdots & c_k \vec v_k \end{bmatrix}_{n\times k} \begin{bmatrix} 1 & \lambda_1 & \lambda_1^2 & \cdots \lambda_1^{k-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots \lambda_2^{k-1} \\ 1 & \lambda_3 & \lambda_3^2 & \cdots \lambda_3^{k-1} \\ \vdots & \vdots & \vdots & \ddots \vdots \\ 1 & \lambda_k & \lambda_k^2 & \cdots \lambda_k^{k-1} \\ \end{bmatrix}_{k\times k} =0_{n\times k}. \end{equation}\] Since the eigenvalues are distinct, the coefficient matrix is an invertible Vandermonde matrix. Multiplying on the right by its inverse shows that \[ \begin{bmatrix} c_1 \vec v_1 & \cdots & c_k \vec v_k \end{bmatrix} =0_{n\times k}. \] It follows that every \(c_i\) must be zero. Hence \(\vec v_1, \ldots, \vec v_k\) are linearly independent.

A basis of \(\mathbb{F}^n\) consisting of eigenvectors of \(A\) is called an eigenbasis for \(A\).
In particular, if an \(n\times n\) matrix \(A\) has \(n\) distinct eigenvalues, then there exists an eigenbasis for \(A\), namely, construct an eigenbasis by finding an eigenvector for each eigenvalue.

Example 6.19 Find the characteristic equation, the eigenvalues, and a basis for the eigenspace. \[A=\begin{bmatrix} 3 & 2 & 4 \\ 2 & 0 & 2\\ 4 & 2 & 3 \end{bmatrix} \] The eigenvalues are \(\lambda_1=8\) (with algebraic multiplicity 1) and \(\lambda_2=-1\) (with algebraic multiplicity 2) since \[ \det(A-\lambda I) =\begin{vmatrix} 3-\lambda & 2 & 4 \\ 2 & -\lambda & 2\\ 4 & 2 & 3-\lambda \end{vmatrix} =-\lambda^3+6\lambda^2+15\lambda+8=0. \] For \(\lambda_1=8\) we obtain \[ \begin{bmatrix} -5 & 2 & 4 \\ 2 & -8 & 2\\ 4 & 2 & -5 \end{bmatrix} \vectorthree{x_1}{x_2}{x_3}=\vectorthree{0}{0}{0} \] and we obtain the eigenvector \(\vec v_1=\vectorthree{2}{1}{2}^T\) with \(E_8=\text{span}(\vec v_1)\). Therefore the geometric multiplicity of \(\lambda_1=8\) is 1. For \(\lambda_1=-1\) we obtain \[ \begin{bmatrix} 4 & 2 & 4 \\ 2 & 1 & 2\\ 4 & 2 & 4 \end{bmatrix} \vectorthree{x_1}{x_2}{x_3}=\vectorthree{0}{0}{0} \] and we obtain the eigenvectors \(\vec v_2=\vectorthree{1}{-2}{0}^T\) and \(\vec v_3=\vectorthree{0}{-2}{1}^T\) with \(E_{-1}=\text{span}(\vec v_2, \vec v_3)\). Therefore the geometric multiplicity of \(\lambda_2=-1\) is 2.

Example 6.20 Show that for each of the following matrices, \(\lambda=3\) is an eigenvalue of algebraic multiplicity 4. In each case, compute the geometric multiplicity of \(\lambda\). \[ \begin{bmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 \end{bmatrix} \qquad \begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 \end{bmatrix} \qquad \begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 \end{bmatrix} \qquad \begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 \end{bmatrix} \]

Theorem 6.3 Similar matrices \(A\) and \(B\) have the same determinant, trace, characteristic polynomial, rank, nullity, and the same eigenvalues with the same algebraic multiplicities.

Proof. The case for the determinant and trace are proven in \(\ref{propdettrace}\). Since \(A\) and \(B\) are similar, there exists an invertible matrix \(P\) such that \(B=P^{-1}AP\). Using \(\ref{propdettrace}\) we find \[\begin{align*} \det(B-\lambda I) &=\det(P^{-1}AP-\lambda I) =\det(P^{-1}AP-P^{-1}\lambda I P) =\det(P^{-1}(A-\lambda I) P) \\ & =\det(P^{-1})\det(A-\lambda I) \det(P) =\det(P^{-1})\det(P) \det(A-\lambda I) \\ & =\det(P^{-1}P)\det(A-\lambda I) =\det(A-\lambda I) \end{align*}\] Thus \(A\) and \(B\) have the same characteristic equation. Therefore also the same eigenvalues with the same algebraic multiplicities according to \(\ref{eigenprop}\) and \(\ref{charform}\).

In light of \(\ref{invsimmatrix}\), if \(T\) is a linear transformation from \(V\) to \(V\) then a scalar \(\lambda\) is called an eigenvalue of \(T\) if there exists a nonzero element \(\vec v\) in \(V\) such that \(T(\vec v)=\lambda \vec v\). Assuming \(V\) is finite-dimensional then a basis \(\mathcal{D}\) of \(V\) consisting of eigenvectors of \(T\) is called an eigenbasis for \(T\).

Theorem 6.4 Let \(T\) be a linear transformation on a finite-dimensional vector space \(V\), and let \(\lambda\) be an eigenvalue of \(T\). The geometric multiplicity of \(\lambda\) is less than or equal to the algebraic multiplicity of \(\lambda\).

Proof. Let \(k\) represent the geometric multiplicity of \(\lambda\) and assume \(\dim V=n\). First notice, by definition, the eigenspace \(E_{\lambda}\) must contain at least one nonzero vector, and thus \(k=\dim E_{\lambda} \geq 1\). Choose a basis \(\vec v_1, \ldots,\vec v_k\) for \(E_{\lambda}\) and, by \(\ref{basisspacethm}\), extend it to a basis \(\mathcal{B}=(\vec v_1, \ldots, \vec v_k, \vec v_{k+1},\ldots,\vec v_n)\) of \(V\). For \(1\leq i \leq k\), notice \[ [T(\vec v_i)]_{\mathcal{B}} =[\lambda \vec v_i]_{\mathcal{B}} =\lambda[\vec v_i]_{\mathcal{B}} =\lambda \vec e_i. \] Thus the matrix representation for \(T\) with respect to \(\mathcal{B}\) has the form \[\begin{equation} B= \begin{bmatrix} \lambda I_k & C\\ 0 & D \end{bmatrix} \end{equation}\] where \(C\) is a \(k\times (n-k)\) submatrix, \(O\) is an \((n-k)\times k\) zero submatrix, and \(D\) is an \((n-k)\times (n-k)\) submatrix.
Using \(\ref{blockdetprod}\) we determine the characteristic polynomial of \(T\) \[ f_T(x) %=f_B(x) =|x I_n-B| =\left|xI_n- \begin{bmatrix} \lambda I_k & C\\ 0 & D \end{bmatrix} \right| = \begin{vmatrix} (x-\lambda)I_k & C\\ 0 & xI_{n-k}-D \end{vmatrix} =(x-\lambda)^k f_D(x). \] It follows that \(f_T(x)=(x-\lambda)^{k+m} g(x)\) where \(g(\lambda)\neq 0\) and \(m\) is the number of factors of \(x-\lambda\) in \(f_D(x)\). Hence \(k\leq k+m\) leading to the desired conclusion.

Example 6.21 Let \(T(M)=M-M^T\) be a linear transformation from \(\mathbb{R}^{2\times2}\) to \(\mathbb{R}^{2\times2}\). For each eigenvalue find a basis for the eigenspace and state the geometric multiplicity. Since \(A=A^T\) for every symmetric matrix, we notice \(T(M)=M-M^T=M-M=0\) whenever \(M\) is a symmetric matrix. Thus the nonzero symmetric matrices are eigenmatrices with eigenvalue \(0\).
Also notice the nonzero skew-symmetric matrices have eigenvalue \(2\) since \(L(M)=M-M^T=M+M=2M\). For eigenvlaue \(\lambda=0\) we have eigenspace \(E_0\) with basis \[ \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right). \] This follow from the condition \(A=A^T\) in \(\mathbb{R}^{2\times2}\). Therefore the geometric multiplicity of \(\lambda=0\) is 3. For eigenvalue \(\lambda=2\) we have eigenspace \(E_2\) with basis \[ \left( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \right). \] which follows from the condition \(A=-A^T\) in \(\mathbb{R}^{2\times2}\). Therefore the geometric multiplicity of \(\lambda=2\) is 1. By \(\ref{eigenveceigenvallemma}\) we have an eigenbasis \[ \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \right). \] for \(T\).

6.1 Diagonalization

An \(n\times n\) matrix \(A\) is called diagonalizable if \(A\) is similar to some diagonal matrix \(D\). If the matrix of a linear transformation \(T\) with respect to some basis is diagonal then we call \(T\) diagonalizable .

Theorem 6.5 An \(n\times n\) matrix \(A\) is diagonalizable if and only if it has \(n\) linearly independent eigenvectors. In that case, the diagonal matrix \(D\) is similar to \(A\) and is given by \[\begin{equation} \label{diagmat} D= \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots& \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \end{equation}\] where \(\lambda_1, ..., \lambda_n\) are the eigenvalues of \(A\). If \(C\) is a matrix whose columns are linearly independent eigenvectors of \(A\), then \(D=C^{-1}A C\).

Proof. The proof is left for the reader.

Corollary 6.1 Let \(T\) be a linear transformation given by \(T(\vec x)=A\vec x\) where \(A\) is a square matrix. If \(\mathcal{D}=(\vec v_1, ....,\vec v_n)\) is an eigenbasis for \(T\), with \(A\vec v_i=\lambda_i \vec v_i\), then the \(\mathcal{D}\)-matrix \(D\) of \(T\) given in \(\ref{diagmat}\) is \(D=[\vec v_1, ...., \vec v_n]^{-1}A [\vec v_1, ...., \vec v_n]\).

Proof. The proof follows from \(\ref{dialineigvec}\) and \(\ref{eigenveceigenvallemma}\).

Corollary 6.2 A matrix \(A\) is diagonalizable if andy only if there exists an eigenbasis for \(A\). In particular, if an \(n\times n\) matrix \(A\) has \(n\) distinct eigenvalues, then \(A\) is diagonalizable.

Proof. The proof follows from \(\ref{dialineigvec}\) and \(\ref{eigenveceigenvallemma}\).

Example 6.22 Let \(T: \mathcal{P}_2\to \mathcal{P}_2\) be the linear transformation defined by
\[ T(a_0+a_1 x+a_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2. \] Show that \(T\) is not diagonalizable. The matrix of \(T\) with respect to the usual basis \((, x, x^2)\) for \(\mathcal{P}_2\) is easily seen to be \[ A= \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}. \] The characteristic polynomial is \(f_A(x)=-(x-1)^3\) since \(A\) is upper triangular. So \(T\) has only one (repeated) eigenvalue \(\lambda=1\). A nonzero polynomial \(g\) with \(g(x)=a_0+a_1 x+a_2 x^2\) is an eigenvector if and only if \[ \label{notdiageq} \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \vectorthree{a_0}{a_1}{a_2}=\vectorthree{0}{0}{0}. \] Thus \(a_1=0\) and \(a_2=0\), so there is only one linearly independent eigenvector for \(\lambda=1\). Thus \(T\) is not diagonalizable by \(\ref{diagonalizablechar}\).

Example 6.23 Let \(T:\mathcal{P}_2\to \mathcal{P}_2\) be the linear transformation defined by
\[\begin{equation} T(f(x))=x^2f''(x)+(3x-2)f'(x)+5 f(x). \end{equation}\] Find a basis for \(\mathcal{P}_2\) such that the matrix representation of \(T\) with respect to \(\mathcal{B}\) is diagonal. Since \(T(x^2)=13x^2-4x\), \(T(x)=8x-2\), and \(T(1)=5\) the matrix representation of \(T\) with respect to the basis \(\mathcal{B}=(x^2,x,1)\) is \[ A=\begin{bmatrix}13 & 0 & 0 \\ -4 & 8 & 0 \\ 0 & -2 & 5 \end{bmatrix}. \] Hence \[\begin{equation} f_T(x)=f_A(x)=\begin{vmatrix} x-13 & 0 & 0 \\ 4 & x-8 & 0 \\ 0 & 2 & x-5 \end{vmatrix}=(x-13)(x-8)(x-5). \end{equation}\] The eigenvalues of \(T\) are \(\lambda_1=13\), \(\lambda_2=8\) and \(\lambda_3=5\). Solving each of the homogenous systems \((A-13I_3)\vec x=\vec 0\), \((A-8I_3)\vec x=\vec 0\), and \((A-5I_3)\vec x=\vec 0\) yields the eigenvectors \(\vec v_1=5x^2-4x+1\), \(\vec v_2=3x-2\), and \(\vec v_3=1\), respectively. Notice \(\vec v_1, \vec v_2, \vec v_3\) are 3 linearly independent vectors, so by \(\ref{dialineigvec}\), \(T\) is diagonalizable. We let \(\mathcal{D}=(\vec v_1, \vec v_2, \vec v_3)\) and since \(T(\vec v_1)=13\vec v_1\),\(T(\vec v_2)=8\vec v_2\), and \(T(\vec v_3)=5\vec v_3\), the matrix representation of \(T\) with respect to \(\mathcal{D}\) is the diagonal matrix \[ \begin{bmatrix}13 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] according to \(\ref{dialineigvec}\).

Example 6.24 Let \(T:\mathcal{P}_3\to \mathcal{P}_3\) be the linear transformation defined by
\[\begin{equation} T(f(x))=xf'(x)+f(x+1). \end{equation}\] Find a basis for \(\mathcal{P}_3\) such that the matrix representation of \(T\) with respect to \(\mathcal{B}\) is diagonal. Since \(T(x^3)=4x^3+3x^2+3x+1\), \(T(x^2)=3x^2+2x+1\), \(T(x)=2x+1\), and \(T(1)=1\) the matrix representation of \(T\) with respect to the basis \(\mathcal{B}=(x^3,x^2,x,1)\) is \[ A=\begin{bmatrix} 4 & 0 & 0 & 0 \\ 3 & 3 & 0 & 0 \\ 3 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}. \] Since \(A\) is lower triangular, \(f_T(x)=f_A(x)=(x-4)(x-3)(x-2)(x-1)\); and so the eigenvalues are \(\lambda_1=4\), \(\lambda_2=3\), \(\lambda_3=2\), and \(\lambda_4=1\). Solving for a basis for each eigenspace of \(A\) yields \[ E_{\lambda_1}=\left(\vectorfour{6}{18}{27}{17}\right), \quad E_{\lambda_2}=\left(\vectorfour{0}{2}{4}{3}\right), \quad E_{\lambda_3}=\left(\vectorfour{0}{0}{1}{1}\right),\quad E_{\lambda_4}=\left(\vectorfour{0}{0}{0}{1}\right). \] By taking the polynomial corresponding to the basis vectors, we let \(\mathcal{D}=(\vec v_1, \vec v_2, \vec v_3, \vec v_4)\) where \(\vec v_1=6x^3+18x^2+27x+17\), \(\vec v_2=2x^2+4x+3\), \(\vec v_3=x+1\), and \(\vec v_4=1\). The diagonal matrix \[ \begin{bmatrix} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] is the matrix representation of \(T\) in \(\mathcal{D}\)-coordinates and has the eigenvalues of \(T\) on its main diagonal. The transition matrix \(P\) from \(\mathcal{B}\)-coordinates to \(\mathcal{D}\)-coordinates is \[ P=\begin{bmatrix} 6 & 0 & 0 & 0 \\ 18 & 2 & 0 & 0 \\ 27 & 4 & 1 & 0 \\ 17 & 3 & 1 & 1 \end{bmatrix} \]
and satisfies the required relation \(D=P^{-1}AP\) as can be verified.

Example 6.25 If \(A\) is similar to \(B\), show that \(A^n\) is similar to \(B^n\), for any positive integer \(n\).

Example 6.26 Suppose that \(C^{-1}AC=D\). Show that for any integer \(n\), \(A^n=CD^nC^{-1}\).

Example 6.27 Let \(a\) and \(b\) be real numbers. By diagonalizing \[ M= \begin{bmatrix} a & b-a \\ 0 & b \end{bmatrix}, \] prove that \[ M^n= \begin{bmatrix} a^n & b^n-a^n \\ 0 & b^n \end{bmatrix} \] for all positive integers \(n\). We need a basis of \(\mathbb{R}^2\) consisting of eigenvectors of \(M\). One such basis is \(\vec v_1=\vec e_1\) and \(\vec v_2=\vec e_1+\vec e_2\) where \(a\) and \(b\) are eigenvalues for corresponding to these eigenvectors, respectively. Let \(P=\begin{bmatrix}\vec v_1 & \vec v_2\end{bmatrix},\) then by \(\ref{eigendmatrix}\), the diagonalization is \[\begin{equation*} D=\begin{bmatrix}\vec v_1 & \vec v_2\end{bmatrix}^{-1}M\begin{bmatrix}\vec v_1 & \vec v_2\end{bmatrix}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}. \end{equation*}\] Therefore \[\begin{equation*} M^n=(PDP^{-1})^n=\underbrace{(PDP^{-1})\cdots (PDP^{-1})}_{n\text{-times}}=PD^n P^{-1} =\begin{bmatrix} a^n & b^n-a^n \\ 0 & b^n \end{bmatrix}. \end{equation*}\]