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Chapter 9: Problem 14

Let \(A=\left[\begin{array}{rrr}4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2\end{array}\right]\). (a) Find all eigenvalues of \(A\). (b) Find a maximum set \(S\) of linearly independent eigenvectors of \(A\). (c) Is A diagonalizable? If yes, find \(P\) such that \(D=P^{-1} A P\) is diagonal. (a) First find the characteristic polynomial \(\Delta(t)\) of \(A\). We have $$ \operatorname{tr}(A)=4+5+2=11 \quad \text { and } \quad|A|=40-2-2+5+8-4=45 $$ Also, find each cofactor \(A_{i i}\) of \(a_{i i}\) in \(A\) : $$ \begin{aligned} A_{11} &=\left|\begin{array}{rr} 5 & -2 \\ 1 & 2 \end{array}\right|=12, \quad A_{22}=\left|\begin{array}{rr} 4 & -1 \\ 1 & 2 \end{array}\right|=9, \quad A_{33}=\left|\begin{array}{ll} 4 & 1 \\ 2 & 5 \end{array}\right|=18 \\ \text { Hence, } \quad \Delta(t)=t^{3}-\operatorname{tr}(A) t^{2}+\left(A_{11}+A_{22}+A_{33}\right) t-|A|=t^{3}-11 t^{2}+39 t-45 \end{aligned} $$ Assuming \(\Delta t\) has a rational root, it must be among \(\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45\). Testing, by synthetic division, we get $$ 3 \mid \begin{array}{r} 1-11+39-45 \\ 3-24+45 \\ \hline 1-8+15+0 \end{array} $$ Thus, \(t=3\) is a root of \(\Delta(t)\). Also, \(t-3\) is a factor and \(t^{2}-8 t+15\) is a factor. Hence, $$ \Delta(t)=(t-3)\left(t^{2}-8 t+15\right)=(t-3)(t-5)(t-3)=(t-3)^{2}(t-5) $$ Accordingly, \(\lambda=3\) and \(\lambda=5\) are eigenvalues of \(A\) (b) Find linearly independent eigenvectors for each eigenvalue of \(A\). (i) Subtract \(\lambda=3\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{lll} 1 & 1 & -1 \\ 2 & 2 & -2 \\ 1 & 1 & -1 \end{array}\right], \quad \text { corresponding to } \quad x+y-z=0 $$ Here \(u=(1,-1,0)\) and \(v=(1,0,1)\) are linearly independent solutions. (ii) Subtract \(\lambda=5\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{rrr} -1 & 1 & -1 \\ 2 & 0 & -2 \\ 1 & 1 & -3 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} -x+y-z=0 \\ 2 x-2 z=0 \\ x+y-3 z=0 \end{aligned} \quad \text { or } \begin{aligned} x \quad-z=0 \\ y-2 z=0 \end{aligned} $$ Only \(z\) is a free variable. Here \(w=(1,2,1)\) is a solution. Thus, \(S=\\{u, v, w\\}=\\{(1,-1,0),(1,0,1),(1,2,1)\\}\) is a maximal set of linearly independent eigenvectors of \(A\).

Short Answer

Expert verified
The matrix A has eigenvalues 位鈧� = 4 and 位鈧� = 5 (with multiplicity 2). For 位鈧� = 4, we find an eigenvector \(\begin{bmatrix}1\\1\\1\end{bmatrix}\). For 位鈧� = 5, we find an eigenvector \(\begin{bmatrix}-1\\1\\1\end{bmatrix}\). However, since A has only 2 linearly independent eigenvectors, it is not diagonalizable.

Step by step solution

01

Find eigenvalues of A

First, we need to find the eigenvalues by solving the characteristic equation det(A - 位I) = 0, where det() denotes the determinant. The matrix A - 位I is \(\begin{bmatrix} 4-\lambda & 1 & -1 \\ 2 & 5-\lambda & -2 \\ 1 & 1 & 2-\lambda \end{bmatrix}\) Now, we compute its determinant: \((4-\lambda)[(5-\lambda)(2-\lambda) - (-2)(1)] - 1[2(1) - (-2)(2)] - (-1)[2 - (5-\lambda)] = 0\) After expanding and simplifying, we get: \(\lambda^3 - 11\lambda^2 + 38\lambda -40 = (\lambda-4)(\lambda-5)^2= 0\) Thus, the eigenvalues are 位鈧� = 4 and 位鈧� = 5 (with multiplicity 2).
02

Find eigenvectors for each eigenvalue

For 位鈧� = 4, we find a basis for the eigenspace by solving (A - 4I)X = 0: \(\begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & -2 \\ 1 & 1 & -2 \end{bmatrix}\) We can row reduce the matrix and obtain: \(\begin{bmatrix} 1 & 1 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\) The system of equations is: \(x_1 + x_2 - 2x_3 = 0\) \(x_2 - x_3 = 0\) So, \(x_2 = x_3\) and \(x_1 = x_3\). We can let \(x_3 = 1\), then \(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}\) is an eigenvector corresponding to 位鈧� = 4. Now, for 位鈧� = 5, we find a basis for the eigenspace by solving (A - 5I)X = 0: \(\begin{bmatrix} -1 & 1 & -1 \\ 2 & 0 & -2 \\ 1 & 1 & -3 \end{bmatrix}\) We can row reduce the matrix and obtain: \(\begin{bmatrix} 1 & 1 & -1 \\ 0 & -2 & 2 \\ 0 & 0 & 0 \end{bmatrix}\) The system of equations is: \(x_1 + x_2 - x_3 = 0\) \(-2x_2 + 2x_3 = 0\) So, \(x_2 = x_3\) and \(x_1 = -x_2\). We can let \(x_3 = 1\), then \(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}-1\\1\\1\end{bmatrix}\) is an eigenvector corresponding to 位鈧� = 5.
03

Check if A is diagonalizable

A 3x3 matrix A is diagonalizable if the sum of the dimensions of its eigenspaces is equal to 3. In our case, we have 1 eigenvector for 位鈧� and 1 eigenvector for 位鈧� (even though the multiplicity of 位鈧� is 2). Therefore, since A has only 2 linearly independent eigenvectors, A is not diagonalizable.
04

Find P and D if A is diagonalizable

As we found out in Step 3, matrix A is not diagonalizable, so there is no need to find P and D.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Polynomial
The characteristic polynomial is a pivotal concept in linear algebra, used to determine the eigenvalues of a square matrix. It is obtained by subtracting \(\lambda I\) from the matrix (where \(\lambda\) is a scalar and \(I\) is the identity matrix of the same size) and then calculating the determinant of the result. For the matrix \(A\), the characteristic polynomial is denoted as \(\Delta(t)\) and is given by the equation \(\det(A - \lambda I) = 0\).

For the given matrix \(A\), the characteristic polynomial is calculated as \(\Delta(t)=t^{3}-11 t^{2}+39 t-45\). Through testing potential rational roots, we find that \(\lambda=3\) and \(\lambda=5\) serve as eigenvalues for \(A\). Understanding the characteristic polynomial not only aids in finding eigenvalues but also informs us about the matrix's invertibility and stability properties.
Diagonalization
Diagonalization of a matrix is a way to simplify a matrix operation, whereby a matrix is expressed in the form \(D=P^{-1}AP\), where \(D\) is a diagonal matrix, and \(P\) is a matrix composed of the eigenvectors of \(A\). However, not all matrices are diagonalizable. A necessary condition for diagonalization is that the matrix must have enough linearly independent eigenvectors to form the matrix \(P\).

In the case with matrix \(A\), although we found its eigenvalues, the matrix is not diagonalizable since it does not have a full set of linearly independent eigenvectors. This means the algebraic multiplicity (from the characteristic polynomial roots) does not equal the geometric multiplicity (the actual number of independent eigenvectors that exist). Hence, there is no matrix \(P\) that can diagonalize \(A\), as the requirement for having a distinct eigenvector for each eigenvalue is not fulfilled.
Linear Independence
Linear independence is a fundamental concept which is related to vectors in a vector space. A set of vectors is considered linearly independent if no vector in the set can be written as a linear combination of the others. In other words, the only solution to the equation \(c_1v_1 + c_2v_2 + ... + c_nv_n = 0\), where \(c_i\) are coefficients and \(v_i\) are vectors, is \(c_1 = c_2 = ... = c_n = 0\).

For matrix \(A\), finding a set \(S\) of linearly independent eigenvectors corresponds to solving systems of equations after subtracting each eigenvalue down the diagonal of \(A\). The sets we get for \(\lambda=3\) and \(\lambda=5\) are \(\{u,v,w\}\), as shown. These vectors form a basis for the eigenspaces associated with their respective eigenvalues and showcase the concept of linear independence, as they cannot be represented as combinations of one another. This property is crucial for many applications, including solving systems of differential equations and transforming matrices to simpler forms.

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Most popular questions from this chapter

Prove Theorem \(9.8^{\prime}\) : Suppose \(v_{1}, v_{2}, \ldots, v_{n}\) are nonzero eigenvectors of \(T\) belonging to distinct eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\). Then \(v_{1}, v_{2}, \ldots, v_{n}\) are linearly independent.Suppose the theorem is not true. Let \(v_{1}, v_{2}, \ldots, v_{s}\) be a minimal set of vectors for which the theorem is not true. We have \(s>1\), because \(v_{1} \neq 0\). Also, by the minimality condition, \(v_{2}, \ldots, v_{s}\) are linearly independent. Thus, \(v_{1}\) is a linear combination of \(v_{2}, \ldots, v_{s}\), say, $$ v_{1}=a_{2} v_{2}+a_{3} v_{3}+\cdots+a_{s} v_{s} $$ (where some \(a_{k} \neq 0\) ). Applying \(T\) to (1) and using the linearity of \(T\) yields $$ T\left(v_{1}\right)=T\left(a_{2} v_{2}+a_{3} v_{3}+\cdots+a_{s} v_{s}\right)=a_{2} T\left(v_{2}\right)+a_{3} T\left(v_{3}\right)+\cdots+a_{s} T\left(v_{s}\right) $$ Because \(v_{j}\) is an eigenvector of \(T\) belonging to \(\lambda_{j}\), we have \(T\left(v_{j}\right)=\lambda_{j} v_{j}\). Substituting in (2) yields $$ \lambda_{1} v_{1}=a_{2} \lambda_{2} v_{2}+a_{3} \lambda_{3} v_{3}+\cdots+a_{s} \lambda_{s} v_{s} $$ Multiplying (1) by \(\lambda_{1}\) yields $$ \lambda_{1} v_{1}=a_{2} \lambda_{1} v_{2}+a_{3} \lambda_{1} v_{3}+\cdots+a_{s} \lambda_{1} v_{s} $$ Setting the right-hand sides of (3) and (4) equal to each other, or subtracting (3) from (4) yields $$ a_{2}\left(\lambda_{1}-\lambda_{2}\right) v_{2}+a_{3}\left(\lambda_{1}-\lambda_{3}\right) v_{3}+\cdots+a_{s}\left(\lambda_{1}-\lambda_{3}\right) v_{s}=0 $$ Because \(v_{2}, v_{3}, \ldots, v_{s}\) are linearly independent, the coefficients in (5) must all be zero. That is, $$ a_{2}\left(\lambda_{1}-\lambda_{2}\right)=0, \quad a_{3}\left(\lambda_{1}-\lambda_{3}\right)=0, \quad \ldots, \quad a_{s}\left(\lambda_{1}-\lambda_{s}\right)=0 $$ However, the \(\lambda_{i}\) are distinct. Hence \(\lambda_{1}-\lambda_{j} \neq 0\) for \(j>1\). Hence, \(a_{2}=0, a_{3}=0, \ldots, a_{s}=0\). This contradicts the fact that some \(a_{k} \neq 0\). The theorem is proved.

For each matrix, find a polynomial having the following matrix as a root: (a) \(\quad A=\left[\begin{array}{rr}2 & 5 \\ 1 & -3\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{ll}2 & -3 \\ 7 & -4\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 1 & 4\end{array}\right]\)

Repeat Problem \(9.14\) for the matrix \(B=\left[\begin{array}{ccc}3 & -1 & 1 \\\ 7 & -5 & 1 \\ 6 & -6 & 2\end{array}\right]\) (a) First find the characteristic polynomial \(\Delta(t)\) of \(B\). We have $$ \operatorname{tr}(B)=0, \quad|B|=-16, \quad B_{11}=-4, \quad B_{22}=0, \quad B_{33}=-8, \quad \text { so } \quad \sum_{i} B_{i i}=-12 $$ Therefore, \(\Delta(t)=t^{3}-12 t+16=(t-2)^{2}(t+4) .\) Thus, \(\lambda_{1}=2\) and \(\lambda_{2}=-4\) are the eigenvalues of \(B\) (b) Find a basis for the eigenspace of each eigenvalue of \(B\). (i) Subtract \(\lambda_{1}=2\) down the diagonal of \(B\) to obtain $$ M=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 7 & -7 & 1 \\ 6 & -6 & 0 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} x-y+z=0 \\ 7 x-7 y+z=0 \\ 6 x-6 y &=0 \end{aligned} \quad \text { or } \begin{aligned} x-y+z=0 \\ z &=0 \end{aligned} $$ The system has only one independent solution; for example, \(x=1, y=1, z=0\). Thus, \(u=(1,1,0)\) forms a basis for the eigenspace of \(\lambda_{1}=2\) (ii) Subtract \(\lambda_{2}=-4\) (or add 4) down the diagonal of \(B\) to obtain $$ M=\left[\begin{array}{rrr} 7 & -1 & 1 \\ 7 & -1 & 1 \\ 6 & -6 & 6 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} &7 x-y+z=0 \\ &7 x-y+z=0 \\ &6 x-6 y+6 z=0 \end{aligned} \quad \text { or } \quad x-\begin{array}{r} x+z=0 \\ 6 y-6 z=0 \end{array} $$ The system has only one independent solution; for example, \(x=0, y=1, z=1\). Thus, \(v=(0,1,1)\) forms a basis for the eigenspace of \(\lambda_{2}=-4\) Thus \(S=\\{u, v\\}\) is a maximal set of linearly independent eigenvectors of \(B\). (c) Because \(B\) has at most two linearly independent eigenvectors, \(B\) is not similar to a diagonal matrix; that is, \(B\) is not diagonalizable.

Let \(A=\left[\begin{array}{rrr}4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3\end{array}\right]\) and \(B=\left[\begin{array}{rrr}3 & -2 & 2 \\ 4 & -4 & 6 \\\ 2 & -3 & 5\end{array}\right] .\) The characteristic polynomial of both matrices is \(\Delta(t)=(t-2)(t-1)^{2} .\) Find the minimal polynomial \(m(t)\) of each matrix.

Prove Theorem 9.1: Let \(f\) and \(g\) be polynomials. For any square matrix \(A\) and scalar \(k\), (i) \((f+g)(A)=f(A)+g(A)\), (iii) \((k f)(A)=k f(A)\), (ii) \((f g)(A)=f(A) g(A)\), (iv) \(f(A) g(A)=g(A) f(A)\) Suppose \(f=a_{n} t^{n}+\cdots+a_{1} t+a_{0}\) and \(g=b_{m} t^{m}+\cdots+b_{1} t+b_{0} .\) Then, by definition, $$ f(A)=a_{n} A^{n}+\cdots+a_{1} A+a_{0} I \quad \text { and } \quad g(A)=b_{m} A^{m 1}+\cdots+b_{1} A+b_{0} I $$ (i) Suppose \(m \leq n\) and let \(b_{i}=0\) if \(i>m\). Then $$ f+g=\left(a_{n}+b_{n}\right) t^{n}+\cdots+\left(a_{1}+b_{1}\right) t+\left(a_{0}+b_{0}\right) $$ Hence, $$ \begin{aligned} (f+g)(A) &=\left(a_{n}+b_{n}\right) A^{n}+\cdots+\left(a_{1}+b_{1}\right) A+\left(a_{0}+b_{0}\right) I \\ &=a_{n} A^{n}+b_{n} A^{n}+\cdots+a_{1} A+b_{1} A+a_{0} I+b_{0} I=f(A)+g(A) \end{aligned} $$ (ii) By definition, \(f g=c_{n+m} t^{n+m}+\cdots+c_{1} t+c_{0}=\sum_{k=0}^{n+m} c_{k} t^{k}\), where $$ c_{k}=a_{0} b_{k}+a_{1} b_{k-1}+\cdots+a_{k} b_{0}=\sum_{i=0}^{k} a_{i} b_{k-i} $$ Hence, \((f g)(A)=\sum_{k=0}^{n+m} c_{k} A^{k}\) and $$ f(A) g(A)=\left(\sum_{i=0}^{n} a_{i} A^{i}\right)\left(\sum_{j=0}^{m} b_{j} A^{j}\right)=\sum_{i=0}^{n} \sum_{j=0}^{m} a_{i} b_{j} A^{i+j}=\sum_{k=0}^{n+m} c_{k} A^{k}=(f g)(A) $$ (iii) By definition, \(k f=k a_{n} t^{n}+\cdots+k a_{1} t+k a_{0}\), and so $$ (k f)(A)=k a_{n} A^{n}+\cdots+k a_{1} A+k a_{0} I=k\left(a_{n} A^{n}+\cdots+a_{1} A+a_{0} I\right)=k f(A) $$ (iv) By (ii), \(g(A) f(A)=(g f)(A)=\left(f_{B}\right)(A)=f(A) g(A)\)
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