Chapter 6: Problem 14
Let \(A\) be a \(2 \times 2\) matrix. If \(\operatorname{tr}(A)=8\) and \(\operatorname{det}(A)=\) \(12,\) what are the eigenvalues of \(A ?\)
Short Answer
Expert verified
The eigenvalues of the matrix A are 2 and 6.
Step by step solution
01
Set Up The Equations
To find the eigenvalues of the matrix A, we can utilize the trace and determinant of A. Let 位鈧 and 位鈧 be the eigenvalues of A. Then the trace of A, tr(A), is equal to 位鈧 + 位鈧, and the determinant of A, det(A), is equal to 位鈧佄烩倐.
From the given information, we have:
\[tr(A) = \lambda_1 + \lambda_2 = 8\]
\[det(A) = \lambda_1 \cdot \lambda_2 = 12\]
02
Solve for One of the Eigenvalues
We now have a system of two equations with two unknowns (位鈧 and 位鈧). Let's solve for one of the eigenvalues, say 位鈧, from the trace equation.
\[\lambda_1 = 8 - \lambda_2\]
03
Substitute to Find the Other Eigenvalue
Substitute the expression for 位鈧 into the determinant equation:
\[(8 - \lambda_2) \cdot \lambda_2 = 12\]
Now, solve for 位鈧:
\(8\lambda_2 - \lambda_2^2 = 12\)
\(\lambda_2^2 - 8\lambda_2 + 12 = 0\)
Using the quadratic formula to find the roots, we get:
\[\lambda_2 = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2}\]
\[\lambda_2 = \frac{8 \pm \sqrt{64 - 48}}{2}\]
\[\lambda_2 = \frac{8 \pm \sqrt{16}}{2}\]
\[\lambda_2 = 2, 6\]
04
Find the First Eigenvalue
Now that we have found 位鈧, we can substitute it back into the expression for 位鈧 to find the two possible values for 位鈧:
For 位鈧 = 2:
\[\lambda_1 = 8 - 2 = 6\]
For 位鈧 = 6:
\[\lambda_1 = 8 - 6 = 2\]
Since both values of 位鈧 yield the same result for 位鈧, both cases are valid, and the eigenvalues are:
\[\lambda_1 = 2, \lambda_2 = 6\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It is foundational for understanding various aspects of mathematics and practical applications in science, engineering, computer science, economics, and more. At the core of linear algebra is the concept of a matrix, which can represent such a system of linear equations. Matrices have certain characteristics that can be used to derive more properties of the system they represent. One such characteristic is calculating eigenvalues, which give us insight into the transform properties that the matrix offers.
Trace of a Matrix
The trace of a matrix is the sum of all the diagonal elements in a square matrix. It is denoted by \(tr(A)\) for a matrix A. This mathematical property holds significance because it is invariant under a change of basis in linear algebra, meaning that the trace of a matrix is the same no matter how you rotate or scale the coordinate system in which the matrix is described. Interestingly, the trace can inform us about the eigenvalues of a matrix, as their sum is equal to the trace of the matrix. This relationship helps in solving for eigenvalues, as demonstrated in our example exercise.
Determinant of a Matrix
The determinant is another important concept in linear algebra, represented by \(det(A)\) for a matrix A. It's a scalar value that can be computed from the elements of a square matrix. The determinant gives us information about the geometric properties of the matrix such as scaling factor and whether the matrix is invertible. A matrix with a determinant of zero is said to be singular, which means it does not have an inverse. In the context of eigenvalue calculation, the product of the eigenvalues is equal to the determinant of the matrix. This piece of information is vital when deducing the eigenvalues as shown in our exercise.
Quadratic Formula
When faced with a quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula gives a straightforward method for finding the roots of the equation. This formula is expressed as \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). In our exercise, this formula was used to calculate the eigenvalues from the quadratic equation obtained by substituting the value we had from the trace into the determinant equation. The discriminant \(\sqrt{{b^2 - 4ac}}\) within the formula is key to determining the nature of the roots鈥攚hether they're real or complex.
Eigenvalue Calculation
Eigenvalue calculation is fundamental in linear algebra and has practical implementations in various fields such as physics, statistics, and machine learning. An eigenvalue of a matrix A is a number \(\lambda\) such that when A acts on a vector \(v\), named an eigenvector, it produces \(v\) scaled by \(\lambda\), or \(Av = \lambda v\). To calculate the eigenvalues of a matrix, one often uses its trace and determinant because they are equal to the sum and product of the eigenvalues, respectively. Solving these two equations together, potentially using the quadratic formula, allows you to find the eigenvalues for a square matrix. In the exercise we looked at, this process revealed the eigenvalues of a 2x2 matrix by utilizing its trace and determinant.