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Show that \(\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 2 & 0 & 1 & 1 \\ 1 & 1 & 0 & -1 \\ 4 & 3 & 0 & 0\end{array}\right]\) and \(\left[\begin{array}{rrrr}1 & -1 & 3 & 0 \\ -1 & 0 & 1 & 1 \\ 0 & -1 & 4 & 1 \\\ 5 & -1 & -1 & -4\end{array}\right]\) are not similar.

Short Answer

Expert verified
The matrices have different eigenvalues, indicating they are not similar.

Step by step solution

01

Understand the Definition of Similar Matrices

Two matrices are said to be similar if there exists an invertible matrix \( P \) such that \( A = PBP^{-1} \) for matrices \( A \) and \( B \). Another way to determine similarity is by comparing their eigenvalues: similar matrices must have the same eigenvalues.
02

Identify Eigenvalues of the First Matrix

The eigenvalues of a matrix can be found by solving the characteristic equation, \( \det(A - \lambda I) = 0 \), where \( A \) is the matrix, \( \lambda \) are the eigenvalues, and \( I \) is the identity matrix. For matrix \[A = \begin{bmatrix} 1 & 2 & -1 & 0 \ 2 & 0 & 1 & 1 \ 1 & 1 & 0 & -1 \ 4 & 3 & 0 & 0 \end{bmatrix}\]solve \( \det(A - \lambda I) = 0 \). This calculation will yield the eigenvalues of matrix A.
03

Identify Eigenvalues of the Second Matrix

Similarly, calculate the eigenvalues for the second matrix \[B = \begin{bmatrix} 1 & -1 & 3 & 0 \ -1 & 0 & 1 & 1 \ 0 & -1 & 4 & 1 \ 5 & -1 & -1 & -4 \end{bmatrix}\]This also involves computing \( \det(B - \lambda I) = 0 \) to find the eigenvalues of B.
04

Compare Eigenvalues of Both Matrices

After calculating the eigenvalues for both matrices, compare the resulting sets. If they differ, the matrices are not similar. Proceed to write down the calculated eigenvalues for both matrices and note any differences between them.
05

Conclusion

Once you have verified the eigenvalues for both matrices and confirmed they are different, you can conclude that these matrices are not similar. Differences in their eigenvalues demonstrate that no invertible matrix \( P \) can transform one into the other through similarity.

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Key Concepts

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

Eigenvalues
Eigenvalues provide crucial insight into the properties of a matrix. Simply put, eigenvalues are the special numbers associated with a square matrix that give information about its behavior, structure, and transformations.
To find eigenvalues, set up the equation called the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \( A \) is your matrix, \( I \) is the identity matrix of the same size, and \( \lambda \) represents the eigenvalues you're solving for. Understanding eigenvalues can tell us a lot about a matrix:
  • If a matrix is invertible, its eigenvalues help determine the inverse.
  • They are used in stability analysis, allowing us to analyze the efficiency and behavior of systems.
  • In the context of matrix similarity, two similar matrices will share the same eigenvalues, making them fundamental in proving matrix similarity.
Characteristic Equation
Characteristic equations are foundational when working with matrices, especially in the context of eigenvalues. These equations enable us to calculate eigenvalues, which are key to understanding how matrices behave under various transformations.
The characteristic equation is derived from subtracting \( \lambda I \) from your matrix \( A \), and then taking the determinant of the resulting matrix: \[ \det(A - \lambda I) = 0 \] By solving this equation, you obtain the eigenvalues \( \lambda \). This equation is a polynomial whose roots give all possible eigenvalues of \( A \). Important concepts related to characteristic equations include:
  • The degree of the polynomial is equal to the size of the matrix.
  • Solutions to the equation (the roots) must satisfy the polynomial, representing the matrix's eigenvalues.
  • The determinant of the matrix can reveal properties about invertibility and rank, providing deeper insight into matrix transformations.
Invertible Matrix
An invertible matrix, or a non-singular matrix, is essential in understanding linear transformations and matrix similarity. An invertible matrix can transform or revert a linear transformation cleanly, meaning that there is a matrix known as its inverse which can "undo" that transformation.
For a matrix to be invertible, it must satisfy several characteristics:
  • It must be square, meaning it has the same number of rows and columns.
  • It must have full rank, which means that all its rows or columns are linearly independent.
  • Its determinant must not be zero, as a zero determinant indicates that a matrix is singular (non-invertible).
Understanding invertible matrices is important in matrix similarity, as two matrices, \( A \) and \( B \), are similar only if there exists an invertible matrix \( P \) such that \( A = PBP^{-1} \). Thus, the invertibility of \( P \) is a crucial condition for matrix similarity. Without an invertible matrix to establish this relationship, the concept of similarity doesn't hold. This makes checking for invertibility a critical step whenever dealing with such transformations.

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Most popular questions from this chapter

In each case show that the statement is true or give an example showing that it is false. a. If \(\\{\mathbf{x}, \mathbf{y}\\}\) is independent, then \(\\{\mathbf{x}, \mathbf{y}, \mathbf{x}+\mathbf{y}\\}\) is independent. b. If \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) is independent, then \(\\{\mathbf{y}, \mathbf{z}\\}\) is independent. c. If \(\\{\mathbf{y}, \mathbf{z}\\}\) is dependent, then \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) is dependent for any \(\mathbf{x}\). d. If all of \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\) are nonzero, then \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is independent. e. If one of \(\mathbf{x}_{1}, \quad \mathbf{x}_{2}, \ldots, \quad \mathbf{x}_{k}\) is zero, then \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is dependent. f. If \(a \mathbf{x}+b \mathbf{y}+c \mathbf{z}=\mathbf{0},\) then \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) is independent. g. If \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) is independent, then \(a \mathbf{x}+b \mathbf{y}+c \mathbf{z}=\mathbf{0}\) for some \(a, b,\) and \(c\) in \(\mathbb{R}\). h. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is dependent, then \(t_{1} \mathbf{x}_{1}+t_{2} \mathbf{x}_{2}+\) \(\cdots+t_{k} \mathbf{x}_{k}=\mathbf{0}\) for some numbers \(t_{i}\) in \(\mathbb{R}\) not all zero. i. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is independent, then \(t_{1} \mathbf{x}_{1}+\) \(t_{2} \mathbf{x}_{2}+\cdots+t_{k} \mathbf{x}_{k}=\mathbf{0}\) for some \(t_{i}\) in \(\mathbb{R}\) j. Every non-empty subset of a linearly independent set is again linearly independent. k. Every set containing a spanning set is again a spanning set.

}\( satisfying the following four conditions: \)A A^{\\#} A=A ; A^{\… # If \(A\) is an \(m \times n\) matrix, it can be proved that there exists a unique \(n \times m\) matrix \(A^{\\#}\) satisfying the following four conditions: \(A A^{\\#} A=A ; A^{\\#} A A^{\\#}=A^{\\#} ; A A^{\\#}\) and \(A^{\\#} A\) are symmetric. The matrix \(A^{\\#}\) is called the generalized inverse of \(A\), or the Moore-Penrose inverse. a. If \(A\) is square and invertible, show that \(A^{\\#}=A^{-1}\). b. If \(\operatorname{rank} A=m,\) show that \(A^{\\#}=A^{T}\left(A A^{T}\right)^{-1}\). c. If \(\operatorname{rank} A=n,\) show that \(A^{\\#}=\left(A^{T} A\right)^{-1} A^{T}\).

If \(\mathbb{R}^{n}=\operatorname{span}\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{m}\right\\}\) and \(\mathbf{x} \cdot \mathbf{x}_{i}=0\) for all \(i,\) show that \(\mathbf{x}=0 .[\) Hint : Show \(\|\mathbf{x}\|=0 .]\)

In each case find bases for the row and column spaces of \(A\) and determine the rank of \(A\). a. \(\left[\begin{array}{rrrr}2 & -4 & 6 & 8 \\ 2 & -1 & 3 & 2 \\ 4 & -5 & 9 & 10 \\ 0 & -1 & 1 & 2\end{array}\right]\) b. \(\left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 1 & 1 \\ 4 & -2 & 3 \\ -6 & 3 & 0\end{array}\right]\) c. \(\left[\begin{array}{rrrrr}1 & -1 & 5 & -2 & 2 \\ 2 & -2 & -2 & 5 & 1 \\\ 0 & 0 & -12 & 9 & -3 \\ -1 & 1 & 7 & -7 & 1\end{array}\right]\) d. \(\left[\begin{array}{rrrr}1 & 2 & -1 & 3 \\ -3 & -6 & 3 & -2\end{array}\right]\)

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(U \neq\\{\mathbf{0}\\}\) is a subspace of \(\mathbb{R}\), show that \(U=\mathbb{R}\).

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