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Chapter 7: Problem 8

For each of the matrices \(A\) find all (real) eigenvalues. Then find a basis of each eigenspace, and diagonalize \(A,\) if you can. Do not use technology. $$\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{array}\right]$$

Short Answer

Expert verified
The eigenvalues of matrix A are 1, 2, and 3. The eigenspaces corresponding to these eigenvalues are spanned by \(\{[1, 0, 0]^T\}\), \(\{[0, -2, 1]^T\}\), and \(\{[0, 0, 1]^T\}\), respectively. The matrix A is diagonalizable with the diagonal matrix \(D = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\).

Step by step solution

01

Find eigenvalues of the matrix

To find the eigenvalues of matrix A, we set up and solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) are the eigenvalues we are looking for. The matrix for this equation is \[\begin{pmatrix} 1-\lambda & 1 & 0 \ 0 & 2-\lambda & 2 \ 0 & 0 & 3-\lambda \end{pmatrix}\]. Calculating the determinant of this matrix yields a product of its diagonal entries, since it's an upper triangular matrix: \(\det(A - \lambda I) = (1-\lambda)(2-\lambda)(3-\lambda)\). Setting this equal to zero gives us the eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 3\).
02

Find the eigenspaces for each eigenvalue

An eigenspace for an eigenvalue is the set of all vectors \(x\) that satisfy \(Ax = \lambda x\). To find the eigenspace for each eigenvalue, solve the equation \(A - \lambda I)x = 0\) for \(x\). Starting with \(\lambda_1 = 1\), we solve \(A - I)x = 0\), which simplifies to the system of linear equations: \(\begin{align*} 0x_1 + x_2 + 0x_3 &= 0 \ 0x_1 + 1x_2 + 2x_3 &= 0 \ 0x_1 + 0x_2 + 2x_3 &= 0 \end{align*}\). From the system, we obtain the eigenvector \(x_1 = [1, 0, 0]^T\) for \(\lambda_1\). Similarly, for \(\lambda_2 = 2\), the system \(A - 2I)x = 0\) gives the eigenvector \(x_2 = [0, -2, 1]^T\). For \(\lambda_3 = 3\), the system \(A - 3I)x = 0\) yields the eigenvector \(x_3 = [0, 0, 1]^T\).
03

Diagonalize the matrix

To diagonalize the matrix A, we form a matrix P whose columns are the eigenvectors corresponding to the eigenvalues of A. From the previous step, we have the eigenvectors \(x_1 = [1, 0, 0]^T\), \(x_2 = [0, -2, 1]^T\), and \(x_3 = [0, 0, 1]^T\). Thus, the matrix P is \[P = \begin{pmatrix} 1 & 0 & 0 \ 0 & -2 & 0 \ 0 & 1 & 1 \end{pmatrix}\]. The diagonal matrix D with eigenvalues on the diagonal will be \[D = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]. Matrix A is diagonalizable since we can write \(A = PDP^{-1}\) where D is the diagonal matrix of eigenvalues and P is the matrix with columns as the eigenvectors.

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Key Concepts

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

Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is crucial when studying linear algebra, especially in the process of diagonalizing a matrix. An eigenvalue is a special scalar associated with a linear system of equations that, when multiplied by its corresponding eigenvector, does not alter the direction of this vector. Mathematically, for a square matrix A, if there exists a non-zero vector v and a scalar 位 such that Av = 位v, then 位 is the eigenvalue and v is the eigenvector.

To find eigenvalues, we typically calculate the roots of the characteristic equation. Once the eigenvalues are determined, we can find the respective eigenvectors by solving (A - 位I)v = 0, where I is the identity matrix of the same dimensions as A. These eigenvectors are vital in forming the matrix that will diagonalize the original matrix, as they will be used as the columns of the diagonalizing matrix.
Characteristic Equation
The characteristic equation is a fundamental tool in finding the eigenvalues of a matrix. It is obtained by setting up the equation det(A - 位I) = 0, where det denotes the determinant of the matrix A - 位I. This is because for an eigenvalue 位, the matrix A - 位I must be singular, meaning it has no inverse and its determinant is zero.

In our exercise example, the upper triangular form of the matrix simplifies the process, as the determinant of an upper triangular matrix is simply the product of its diagonal elements. Solving the characteristic equation provides us the specific eigenvalues for the matrix, which are crucial for further calculations, including finding eigenvectors and eventually diagonalizing the matrix.
Eigenspace
An eigenspace is a subspace of a given matrix, associated with a particular eigenvalue. It consists of all eigenvectors corresponding to that eigenvalue, along with the zero vector. Essentially, it is the solution set of the equation (A - 位I)v = 0, which we encounter after finding the eigenvalues.

Each eigenspace can be thought of as a sort of 'eigen-dimension' where each vector is a scaled version of the other, all stemming from the same eigenvalue. To find a basis for an eigenspace, we need to solve the linear system and find the vectors that span this space. In our exercise, after computing the eigenspaces for each eigenvalue, we get sets of vectors that we can use to construct the matrix P used in the diagonalization of A.
Upper Triangular Matrix
In linear algebra, an upper triangular matrix is a special type of square matrix where all the entries below the main diagonal are zero. The significance of upper triangular matrices is their simplicity in calculating the determinant, as it is the product of its diagonal elements. Furthermore, the eigenvalues of an upper triangular matrix are its diagonal entries, making the process of finding eigenvalues straightforward.

In the context of the given exercise, the originally provided matrix A is in upper triangular form. This property tremendously eases the process of solving the characteristic equation. Such matrices are often easier to work with, and in many algorithms, matrices are converted to triangular form to simplify calculations related to eigenvalues, eigenvectors, and diagonalization.

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Most popular questions from this chapter

If \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right],\) find a basis of the linear space \(V\) of all \(2 \times 2\) matrices \(S\) such that \(A S=S B,\) where \(B=\left[\begin{array}{rr}5 & 0 \\ 0 & -1\end{array}\right] .\) Find the dimension of \(V\).

Leonardo of Pisa: The rabbit problem. Leonardo of Pisa (c. \(1170-1240\) ), also known as Fibonacci, was the first outstanding European mathematician after the ancient Greeks. He traveled widely in the Islamic world and studied Arabic mathematical writing. His work is in the spirit of the Arabic mathematics of his day. Fibonacci brought the decimal-position system to Europe. In his book Liber abaci \((1202),^{11}\) Fibonacci discusses the following problem: How many pairs of rabbits can be bred from one pair in one year? A man has one pair of rabbits at a certain place entirely surrounded by a wall. We wish to know how many pairs can be bred from it in one year, if the nature of these rabbits is such that they breed every month one other pair and begin to breed in the second month after their birth. Let the first pair breed a pair in the first month, then duplicate it and there will be 2 pairs in a month. From these pairs one, namely, the first, breeds a pair in the second month, and thus there are 3 pairs in the second month. From these, in one month, two will become pregnant, so that in the third month 2 pairs of rabbits will be born. Thus, there are 5 pairs in this month. From these, in the same month, 3 will be pregnant, so that in the fourth month there will be 8 pairs. From these pairs, 5 will breed 5 other pairs, which, added to the 8 pairs, gives 13 pairs in the fifth month, from which 5 pairs (which were bred in that same month) will not conceive in that month, but the other 8 will be pregnant. Thus, there will be 21 pairs in the sixth month. When we add to these the 13 pairs that are bred in the seventh month, then there will be in that month 34 pairs [and so on, \(55,89,144,233,377, \ldots .\). Finally, there will be 377 , and this number of pairs has been born from the first-mentioned pair at the given place in one year. Let \(j(t)\) be the number of juvenile pairs and \(a(t)\) the number of adult pairs after \(t\) months. Fibonacci starts his thought experiment in rabbit breeding with one adult pair, so \(j(0)=0\) and \(a(0)=1 .\) At \(t=1,\) the adult pair will have bred a (juvenile) pair, so \(a(1)=1\) and \(j(1)=1 .\) At \(t=2,\) the initial adult pair will have bred another (juvenile) pair, and last month's juvenile pair will have grown up, so \(a(2)=2\) and \(j(2)=1\) a. Find formulas expressing \(a(t+1)\) and \(j(t+1)\) in terms of \(a(t)\) and \(j(t) .\) Find the matrix \(A\) such that \\[ \vec{x}(t+1)=A \vec{x}(t) \\] where \\[ \vec{x}(t)=\left[\begin{array}{l} a(t) \\ j(t) \end{array}\right] \\]. b. Find closed formulas for \(a(t)\) and \(j(t) .\) (Note: You will have to deal with irrational quantities here. c. Find the limit of the ratio \(a(t) / j(t)\) as \(t\) approaches infinity. The result is known as the golden section. The golden section of a line segment \(A B\) is given by the point \(P\) such that \(\frac{\overline{A B}}{\overline{A P}}=\frac{\overline{A P}}{\overline{P B}}\).

For the matrices \(A\) in Exercises 17 through \(24,\) find real closed formulas for the trajectory \(\vec{x}(t+1)=A \vec{x}(t)\) where \(\vec{x}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right] .\) Draw a rough sketch. $$A=\left[\begin{array}{rr} 2 & -3 \\ 3 & 2 \end{array}\right]$$

Find two \(2 \times 2\) matrices \(A\) and \(B\) such that \(A B\) fails to be similar to \(B A .\) Hint: It can be arranged that \(A B\) is zero, but \(B A\) isn't.

For the matrices \(A\) in Exercises 17 through \(24,\) find real closed formulas for the trajectory \(\vec{x}(t+1)=A \vec{x}(t)\) where \(\vec{x}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right] .\) Draw a rough sketch. $$A=\left[\begin{array}{rr} 0.6 & -0.8 \\ 0.8 & 0.6 \end{array}\right]$$
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