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Chapter 9: Problem 17

Let \(T: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) be defined by \(T(x, y, z)=(2 x+y-2 z, \quad 2 x+3 y-4 z, \quad x+y-z) .\) Find all eigenvalues of \(T,\) and find a basis of each eigenspace. Is \(T\) diagonalizable? If so, find the basis \(S\) of \(\mathbf{R}^{3}\) that diagonalizes \(T,\) and find its diagonal representation \(D\)

Short Answer

Expert verified
The eigenvalues of the linear transformation $T$ are 1 and 2. A basis for each eigenspace is: for $\lambda=1$, it is spanned by the vector (1, 0, 1); for $\lambda=2$, it is spanned by the vector (2, 2, 1). The transformation $T$ is diagonalizable with the basis $S = \begin{bmatrix} 1 & 2 \\ 0 & 2 \\ 1 & 1 \end{bmatrix}$ and the diagonal representation $D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$.

Step by step solution

01

Calculate matrix A

First, create the matrix A associated with the given transformation. The transformation has the following form: $$ T(x,y,z) = (2x+y-2z,\; 2x+3y-4z,\; x+y-z) $$ Let's write down the coefficients of x, y, and z as rows of the matrix A: $$ A = \begin{bmatrix} 2 & 1 & -2 \\ 2 & 3 & -4 \\ 1 & 1 & -1 \\ \end{bmatrix} $$
02

Compute the characteristic polynomial p(t)

Next, compute the characteristic polynomial p(t) of matrix A. This is found as the determinant of the difference of a scalar and the matrix A (tI - A): $$ \Delta(t) = \det(tI - A) = \det \begin{bmatrix} t-2 & -1 & 2 \\ -2 & t-3 & 4 \\ -1 & -1 & t+1 \\ \end{bmatrix} $$ After calculating the determinant, we find that $$ \Delta(t) = t^3 - 4t^2 + 5t - 2 = (t-1)^2 (t-2) $$
03

Find eigenvalues

Find the eigenvalues by solving the equation \(\Delta(t) = 0\): $$ (t-1)^2 (t-2) = 0 $$ The eigenvalues are \(\lambda = 1\) and \(\lambda = 2\).
04

Find a basis for each eigenspace

To find the eigenspace for each eigenvalue, we'll solve the equation (A - 位I)v = 0 for each value of 位. 1. For 位 = 1: $$ (A - I)v = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 2 & -4 \\ 1 & 1 & -2 \\ \end{bmatrix}v = 0 $$ The eigenspace for 位=1 is spanned by the vector v = (1, 0, 1). 2. For 位 = 2: $$ (A - 2I)v = \begin{bmatrix} 0 & 1 & -2 \\ 2 & 1 & -4 \\ 1 & 1 & -3 \\ \end{bmatrix}v = 0 $$ The eigenspace for 位=2 is spanned by the vector v = (2, 2, 1).
05

Check if T is diagonalizable and find the basis

Since we have found 3 linearly independent eigenvectors (1 for 位=1 and 2 for 位=2), we can conclude that T is diagonalizable. The basis S that diagonalizes T is formed by the eigenvectors found in the previous step: $$ S = \begin{bmatrix} 1 & 2 \\ 0 & 2 \\ 1 & 1 \\ \end{bmatrix} $$
06

Find the diagonal representation D

The diagonal matrix D has the eigenvalues on its main diagonal: $$ D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ \end{bmatrix} $$ In conclusion, T is diagonalizable with the basis S and diagonal representation D given above.

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Key Concepts

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

Diagonalization
Diagonalization is akin to translating a language we find complex into one we understand effortlessly. When dealing with linear transformations or matrices, it simplifies complex problems by converting a matrix into a diagonal matrix, which has its values only on the main diagonal and zeros elsewhere.

For a square matrix representing a linear transformation to be diagonalizable, we must be able to find a set of eigenvectors that forms a basis for the underlying vector space. In the exercise, the transformation matrix A is diagonalizable because we have found three linearly independent eigenvectors associated with its eigenvalues \(\lambda = 1\) and \(\lambda = 2\). The transformation matrix can be expressed in terms of these eigenvectors and their corresponding eigenvalues in a simpler, diagonal form.
Characteristic Polynomial
Picture the characteristic polynomial as the DNA of a matrix鈥攅ncapsulating its fundamental properties in a single, albeit abstract, mathematical expression. It is created from the determinant of an expression of the form \(tI - A\), where \(t\) is a scalar, \(I\) is the identity matrix, and \(A\) is our matrix of interest.

In this exercise, the characteristic polynomial \(\Delta(t)\) unveils the eigenvalues of matrix A, representing the linear transformation T. The roots of this polynomial equation are the eigenvalues, which are the scalars that describe the behavior of the corresponding eigenvectors under transformation.
Linear Transformation
A linear transformation is a rule that takes a vector in one space and maps it to another, while preserving the operations of vector addition and scalar multiplication. One way to represent such transformations is through matrices. In the context of our exercise, the matrix A represents the linear transformation T from \(\mathbf{R}^{3}\) to itself.

Diagonalization leverages the simplicity of linear transformations by finding a basis in which the transformation is represented by a diagonal matrix. This greatly simplifies operations such as computing powers of the matrix, determining the transformation's behavior, and analyzing the system's dynamics.

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Most popular questions from this chapter

Let \(A=\left[\begin{array}{ll}3 & -4 \\ 2 & -6\end{array}\right]\) (a) Find all eigenvalues and corresponding eigenvectors. (b) Find matrices \(P\) and \(D\) such that \(P\) is nonsingular and \(D=P^{-1} A P\) is diagonal. (a) First find the characteristic polynomial \(\Delta(t)\) of \(A\) : $$ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}+3 t-10=(t-2)(t+5) $$ The roots \(\lambda=2\) and \(\lambda=-5\) of \(\Delta(t)\) are the eigenvalues of \(A\). We find corresponding eigenvectors. (i) Subtract \(\lambda=2\) down the diagonal of \(A\) to obtain the matrix \(M=A-2 I\), where the corresponding homogeneous system \(M X=0\) yields the eigenvectors corresponding to \(\lambda=2 .\) We have $$ M=\left[\begin{array}{ll} 1 & -4 \\ 2 & -8 \end{array}\right], \quad \text { corresponding to } \quad \begin{array}{r} x-4 y=0 \\ 2 x-8 y=0 \end{array} \text { or } x-4 y=0 $$ The system has only one free variable, and \(v_{1}=(4,1)\) is a nonzero solution. Thus, \(v_{1}=(4,1)\) is an eigenvector belonging to (and spanning the eigenspace of) \(\lambda=2\). (ii) Subtract \(\lambda=-5\) (or, equivalently, add 5) down the diagonal of \(A\) to obtain $$ M=\left[\begin{array}{ll} 8 & -4 \\ 2 & -1 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} &8 x-4 y=0 \\ &2 x-y=0 \end{aligned} \text { or } 2 x-y=0 $$ The system has only one free variable, and \(v_{2}=(1,2)\) is a nonzero solution. Thus, \(v_{2}=(1,2)\) is an eigenvector belonging to \(\lambda=5\) (b) Let \(P\) be the matrix whose columns are \(v_{1}\) and \(v_{2}\). Then $$ P=\left[\begin{array}{ll} 4 & 1 \\ 1 & 2 \end{array}\right] \quad \text { and } \quad D=P^{-1} A P=\left[\begin{array}{rr} 2 & 0 \\ 0 & -5 \end{array}\right] $$ Note that \(D\) is the diagonal matrix whose diagonal entries are the eigenvalues of \(A\) corresponding to the eigenvectors appearing in \(P\).

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 \(A\).

Suppose \(f(t)\) is an irreducible monic polynomial for which \(f(A)=0\) for a matrix \(A\). Show that \(f(t)\) is the minimal polynomial of \(A\).

Find the characteristic and minimal polynomials of each of the following matrices: (a) \(A=\left[\begin{array}{rrr}3 & 1 & -1 \\ 2 & 4 & -2 \\ -1 & -1 & 3\end{array}\right]\), (b) \(\quad B=\left[\begin{array}{lll}3 & 2 & -1 \\ 3 & 8 & -3 \\ 3 & 6 & -1\end{array}\right]\)

Let \(B\) be the matrix in Example 9.12(a) that has 1 's on the diagonal, \(a\) 's on the superdiagonal, where \(a \neq 0\), and 0 's elsewhere. Show that \(f(t)=(t-\lambda)^{n}\) is both the characteristic polynomial \(\Delta(t)\) and the minimum polynomial \(m(t)\) of \(A\). Because \(A\) is triangular with \(\lambda\) 's on the diagonal, \(\Delta(t)=f(t)=(t-\lambda)^{n}\) is its characteristic polynomial. Thus, \(m(t)\) is a power of \(t-\lambda\). By Problem 9.29, \((A-\lambda I)^{r-1} \neq 0\). Hence, \(m(t)=\Delta(t)=(t-\lambda)^{n^{\prime}}\).
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