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Chapter 8: Problem 10

Prove that a square matrix is singular if and only if 0 is one of its eigenvalues.

Short Answer

Expert verified
This exercise has proven that there is indeed a connection between the singularity of a matrix and its eigenvalues. If a matrix is singular, then 0 is one of its eigenvalues and if 0 is one of the eigenvalues of a matrix, then that matrix is singular.

Step by step solution

01

Singular Matrix Implying 0 as an Eigenvalue

For a matrix \(A\), we can start by assuming that it is singular. A singular matrix does not have an inverse, so its determinant is 0, meaning \(\det(A) = 0\). The determinant of \(A\) is also the product of its eigenvalues, say \(\lambda_1, \lambda_2,...,\lambda_n\). Therefore, we have \(\lambda_1*\lambda_2*...*\lambda_n = 0\). Since this statement holds, it implies that at least one of the eigenvalues must be 0.
02

0 as an Eigenvalue Implying Matrix is Singular

In this part of the discussion, we start by assuming that 0 is an eigenvalue of the matrix \(A\). We know that the determinant of \(A\) is the product of its eigenvalues (\(\lambda_1, \lambda_2,...,\lambda_n\)). It means that since 0 is one of the eigenvalues, the determinant of matrix \(A\) is 0 (as anything multiplied by 0 is 0). And we also know that for a matrix to have an inverse, its determinant must not be equal to 0. Therefore, since determinant of \(A\) is 0, it implies that \(A\) is a singular matrix.

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Key Concepts

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

Eigenvalues
Eigenvalues are scalar values that, when multiplied by a vector, result in the same vector being obtained from a linear transformation carried out by a square matrix. More simply, if you have a matrix \(A\) and a vector \(v\), then \(\lambda\) is known as an eigenvalue if:\[ A \cdot v = \lambda \cdot v \]This equation tells us that the linear transformation of \(v\) by \(A\) results in a scaled version of \(v\), where \(\lambda\) is the scaling factor or the eigenvalue. Calculating eigenvalues involves finding solutions to the characteristic equation:\[ \text{det}(A - \lambda I) = 0 \]where \(I\) is the identity matrix of the same size as \(A\). By solving this equation, we determine the eigenvalues. Understanding eigenvalues can offer insights into the behavior of the matrix, such as stability and oscillations. Thus, eigenvalues play crucial roles in various applications like vibration analysis, stability of control systems, and quantum mechanics.
Determinant
The determinant is a scalar value derived from a square matrix. It provides significant insight into the matrix鈥檚 properties, such as whether the matrix is invertible or singular. If the determinant is zero, the matrix is singular, meaning it doesn鈥檛 have an inverse. For a matrix \(A\), the determinant is denoted as \(\text{det}(A)\). Computing the determinant involves recursive multiplication and subtraction of the elements of the matrix, a process that can be complex for large matrices. Here's why it's important:
  • The determinant of a matrix is zero if and only if the matrix is singular.
  • A non-zero determinant indicates that the matrix has full rank and is invertible.
  • The sign of the determinant indicates the orientation of the transformation described by the matrix.
Understanding the determinant enables us to answer critical questions about the matrix, especially regarding its invertibility and its behavior under transformations.
Inverse Matrix
The inverse of a matrix is another matrix that, when multiplied with the original matrix, yields the identity matrix. If you have a square matrix \(A\), the inverse is denoted as \(A^{-1}\). The formula to find the inverse involves the determinant and the adjugate of the matrix:\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]To have an inverse, the determinant of the matrix must be non-zero, which implies that singular matrices (determinant zero) do not have inverses. The identity matrix \(I\) for any matrix size, when multiplied by any matrix, results in the original matrix. Reflecting on these properties of an inverse matrix helps explain why it's vital:
  • A matrix must have full rank (means the rows or columns are linearly independent) to have an inverse.
  • In practical applications, finding an inverse is crucial in solving linear equations and optimizing problems.
  • Inverting matrices is computationally expensive, so understanding the conditions for non-existence can save resources.
This understanding is vital in many fields, including computer graphics, cryptography, and statistics.

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Most popular questions from this chapter

State and prove simplified versions of the product rule for differentiating matrix functions in the cases where the left or right factor is a constant matrix.

a. Show that \(\left[\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right]\) is similar to \(\left[\begin{array}{ll}3 & 0 \\ 0 & 2\end{array}\right]\). b. Show that \(\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 7\end{array}\right]\) is similar to \(\left[\begin{array}{lll}7 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right]\).

Let \(\bar{z}\) denote the complex conjugate of the complex number \(z\). That is, if \(z=\) \(\alpha+\beta i\), where \(\alpha, \beta \in \mathbb{R}\), then \(\bar{z}=\alpha-\beta i\). Suppose \(z_{1}\) and \(z_{2}\) are complex numbers. a. Show that \(z_{1}\) is real if and only if \(z_{1}=\bar{z}_{1}\). b. Show that \(\overline{z_{1}+z_{2}}=\bar{z}_{1}+\bar{z}_{2}\). c. Show that \(\overline{z_{1} z_{2}}=\bar{z}_{1} \bar{z}_{2}\). d. Suppose \(p\) is a polynomial with real coefficients. Show that if \(z\) is a root of \(p\), then \(\bar{z}\) is also a root of \(p\).

Use the technique developed in this section to find a basis for the solution space of the following systems of differential equations. a. \(y_{1}^{\prime}=5 y_{1}\) b. \(y_{1}^{\prime}=-2 y_{1}-2 y_{2}\) \(y_{2}^{\prime}=-4 y_{2}\) \(y_{2}^{\prime}=3 y_{1}+5 y_{2}\) c. \(\varphi^{\prime}=\left[\begin{array}{rrr}-1 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & -6 & 1\end{array}\right] \varphi\) d. \(y_{1}^{\prime}=3 y_{1}-3 y_{2}-2 y_{3}\) \(y_{2}^{\prime}=-y_{1}{ }^{\circ}+5 y_{2}+2 y_{3}\) \(y_{3}^{\prime}=y_{1}-3 y_{2}\)

Suppose \(A\) is an \(n \times n\) matrix. a. Show that \(A+A^{t}\) is symmetric. b. Show that \(A A^{t}\) is symmetric.
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