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Chapter 7: Problem 1

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. $$\left[\begin{array}{ll} 7 & 8 \\ 0 & 9 \end{array}\right]$$

Short Answer

Expert verified
The eigenvalues of the matrix A are 7 and 9, with corresponding eigenvectors \((1, 0)^T\) and \((4, -1)^T\). Thus A is diagonalizable.

Step by step solution

01

Identify the matrix A

The given matrix is: \[A = \begin{pmatrix}7 & 8 \0 & 9\end{pmatrix}\]
02

Find the eigenvalues of A

The eigenvalues of a matrix are found by solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) is an eigenvalue. Thus, we calculate the determinant of \(A - \lambda I\): \[\begin{align*}\det(A - \lambda I) &= \det\left(\begin{pmatrix}7 - \lambda & 8 \0 & 9 - \lambda\end{pmatrix}\right) \&= (7 - \lambda)(9 - \lambda) - 8\cdot 0 \&= \lambda^2 - 16\lambda + 63\end{align*}\] Setting \(\lambda^2 - 16\lambda + 63 = 0\), we solve for \(\lambda\).
03

Solve the characteristic equation

The characteristic equation \(\lambda^2 - 16\lambda + 63 = 0\) is a quadratic equation which can be factored as \((\lambda - 7)(\lambda - 9) = 0\), giving us two eigenvalues: \(\lambda_1 = 7\) and \(\lambda_2 = 9\).
04

Find the eigenspace for each eigenvalue

To find the eigenspace for \(\lambda_1 = 7\), we solve \((A - 7I)\vec{x} = \vec{0}\). Similarly, to find the eigenspace for \(\lambda_2 = 9\), we solve \((A - 9I)\vec{x} = \vec{0}\).
05

Solve for the eigenvectors

Solving the system \((A - 7I)\vec{x} = \vec{0}\), we get the eigenvector corresponding to \(\lambda_1 = 7\). For \(\lambda_1 = 7\), the system reduces to \(0\cdot x_1 + 0\cdot x_2 = 0\), which implies a free variable. Thus, any vector of the form \(\vec{v_1} = t\begin{pmatrix}1 \ 0\end{pmatrix}\) is an eigenvector corresponding to \(\lambda_1\). Similarly, \(\lambda_2 = 9\) gives us \(A - 9I = \begin{pmatrix}-2 & 8 \0 & 0\end{pmatrix}\), which leads to the vector \(\vec{v_2} = s\begin{pmatrix}4 \ -1\end{pmatrix}\).
06

Diagonalization

To diagonalize \(A\), we form a matrix \(P\) consisting of the eigenvectors and a diagonal matrix \(D\) consisting of the corresponding eigenvalues. Thus, \(P = \begin{pmatrix}1 & 4 \0 & -1\end{pmatrix}\) and \(D = \begin{pmatrix}7 & 0 \0 & 9\end{pmatrix}\). The matrix \(A\) is diagonalizable since it has two distinct eigenvalues and corresponding independent eigenvectors.

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Key Concepts

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

Eigenvalues
In the fascinating realm of linear algebra, eigenvalues are a fundamental concept that reveal much about the nature of a matrix. Simply put, an eigenvalue is a scalar that indicates how a matrix can linearly transform a vector without changing its direction. To find these powerful scalars, one needs to delve into solving the characteristic equation, which arises from setting the determinant of the matrix minus an unknown scalar times the identity matrix to zero, i.e., \(\text{det}(A - \text{lambda}\text{I}) = 0\).

In the given exercise, we discovered the eigenvalues by examining the matrix \(A = \begin{pmatrix}7 & 8 \text{0} & 9\end{pmatrix}\) and found two distinct eigenvalues, 7 and 9. These two numbers are the keys to further unlocking the secrets of matrix \(A\), as they guide us to the corresponding eigenvectors that span the eigenspaces.
Eigenvectors
Following the discovery of eigenvalues, the next step is to find eigenvectors, which are non-zero vectors that change at most by their scalar factor when the matrix is applied to them. These vectors lie in the eigenspace of their associated eigenvalue and provide the basis for the transformation property represented by that eigenvalue.

To unearth these vectors, we must solve the equation \((A - \text{lambda}\text{I})\text{vec}{x} = \text{vec}{0}\) for each eigenvalue, where \(A\) is our matrix and \(\text{lambda}\) is an eigenvalue. This task was accomplished in the exercise for eigenvalues 7 and 9, resulting in vectors \(\text{vec}{v_1}\) and \(\text{vec}{v_2}\), respectively. These vectors form the skeleton upon which we can later construct a diagonalized version of the matrix \(A\).
Characteristic Equation
The gateway to finding eigenvalues lies within the characteristic equation, a polynomial equation derived from the matrix in question. It's the equation obtained when we set the determinant of \(A - \text{lambda}\text{I}\) to zero. Mapping out the characteristic equation lays down a clear mathematical pathway to the eigenvalues, thus illuminating aspects of the matrix's structure and behavior that are otherwise not readily apparent.

In our original exercise, this equation took the form of a quadratic equation \(\text{lambda}^2 - 16\text{lambda} + 63 = 0\), which then had to be factored or solved using methods such as factoring, completing the square, or applying the quadratic formula. Upon solving this quadratic equation, we retrieved the eigenvalues, which are the roots of the characteristic equation and hold the secret to ultimately diagonalizing the matrix \(A\).

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Most popular questions from this chapter

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 \(\frac{\overline{A B}}{\overline{A P}}=\frac{\overline{A P}}{\overline{P B}}\).

Arguing geometrically, find an eigenbasis for each of the matrices \(A\) in Exercises 55 through \(63,\) and thus diagonalize A. Use Example 3 as a guide. $$A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$

Find all complex eigenvalues of the matrices in Exercises 20 through 26 (including the real ones, of course). Do not use technology. Show all your work. $$\left[\begin{array}{rr} 11 & -15 \\ 6 & -7 \end{array}\right]$$

For which \(2 \times 2\) matrices \(A\) does there exist a nonzero matrix \(M\) such that \(A M=M D,\) where \(D=\left[\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right] ?\) Give your answer in terms of the eigenvalues of \(A\).

Find all the eigenvalues and "eigenvectors" of the linear transformations. \(T(x+i y)=x-i y\) from \(\mathbb{C}\) to \(\mathbb{C} .\) Is \(T\) diagonalizable?
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