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Chapter 11: Problem 278

Given: $$ \begin{array}{rl} \mathrm{A}=\mid 1 & 1 \mid \\ \mid 0 & 1 \mid \end{array} \text { and } \mathrm{P}=\mid \begin{array}{cc} 1 & 1 \mid \\ 1 & -1 \mid \end{array} $$ (a) Find \(\mathrm{P}^{-1}\). (b) Find \(\mathrm{P}^{-1 \mathrm{AP}}\). (c) Verify that, if \(\mathrm{B}\) is similar to \(\mathrm{A}\), then \(\mathrm{A}\) is similar to \(\mathrm{B}\). (d) Show that \(\mathrm{B}^{\mathrm{k}}=\mathrm{P}^{-1} \mathrm{~A}^{\mathrm{k}} \mathrm{P}\) if \(\mathrm{B}=\mathrm{P}^{-1} \mathrm{AP}\) where \(\mathrm{k}\) is any positive integer.

Short Answer

Expert verified
In summary, we find that the inverse of matrix P is \(P^{-1} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\). Then, by calculating \(P^{-1}AP\), we find matrix B to be \(B = \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}\). It is shown that A is similar to B if B is similar to A. Finally, for any positive integer k, \(B^k = P^{-1}A^kP\).

Step by step solution

01

Find P鈦宦�

To find the inverse of matrix P, we can use the formula for 2x2 matrices: \(P^{-1} = \dfrac{1}{\text{det}(P)} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}\) Where \(a, b, c,\) and \(d\) are the elements of P: \(P = \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}\) Calculate the determinant of P: \(\text{det}(P) = ad - bc = (1)(-1)-(1)(1) = -2\) Now, find P鈦宦�: \(P^{-1} = \dfrac{1}{-2} \begin{bmatrix} -1 & -1\\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\)
02

Find P鈦宦笰P

Now that we have P鈦宦�, we can find P鈦宦笰P: \( P^{-1}AP = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \) Perform the matrix multiplications: \(P^{-1}A = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{3}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\) \(P^{-1}AP = \begin{bmatrix} \dfrac{1}{2} & \dfrac{3}{2}\\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}\) So, \(P^{-1}AP = B = \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}\)
03

Verify that A is similar to B if B is similar to A

By definition of similar matrices, if B is similar to A, there exists an invertible matrix P such that \(B = P^{-1}AP\). We want to show that A is also similar to B by finding an invertible matrix Q such that \(A = Q^{-1}BQ\). We know that B = P鈦宦笰P, so let's multiply both sides by P to get A: \( PBP^{-1} = P(P^{-1}AP)P^{-1} = (PP^{-1})A(PP^{-1}) = AAP^{-1} \) Since we started with A, we can now let Q = P: \( A = Q^{-1}BQ \text{ with } Q = P \) Thus, A is similar to B if B is similar to A.
04

Show that B岬�=P鈦宦笰岬廝 for any positive integer k

Let k be any positive integer. We want to show that \(B^k = P^{-1}A^kP\). If we start with B = P鈦宦笰P, we can raise both sides to the power of k: \((P^{-1}AP)^k = B^k\) Using the properties of matrix multiplication, the left side of the equation can be expanded as follows: \[ (P^{-1}AP)(P^{-1}AP)\cdots(P^{-1}AP) = P^{-1}A^kP \] Therefore, \(B^k = P^{-1}A^kP\) for any positive integer k.

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Key Concepts

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

Matrix Inversion
Understanding the concept of matrix inversion is important when dealing with linear algebra problems, especially those involving transformations or solving linear systems. Inverting a matrix essentially finds another matrix that, when multiplied with the original, results in the identity matrix. For a 2x2 matrix \(P\):
  • The formula for finding the inverse is \(P^{-1} = \dfrac{1}{\text{det}(P)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\), where \(det(P)\) is the determinant of matrix \(P\).
  • Determinants help determine if an inverse exists since a matrix must have a non-zero determinant to be invertible.
For matrix \(P = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}\), we calculated:
  • Determinant \(\text{det}(P) = (1)(-1) - (1)(1) = -2\).
  • Inverse \(P^{-1} = \begin{bmatrix} \dfrac{-1}{-2} & \dfrac{-1}{-2} \ \dfrac{-1}{-2} & \dfrac{1}{-2} \end{bmatrix} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2} \ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\).
Matrix inversion is crucial for solving problems where you need to "reverse" the effects of a transformation represented by a matrix.
Matrix Multiplication
Matrix multiplication is a key operation that forms the backbone of many matrix manipulations. This operation is not as straightforward as multiplying individual numbers. Here's what you need to consider:
  • The number of columns in the first matrix must match the number of rows in the second matrix to perform the multiplication.
  • The resulting matrix will have dimensions that are the rows of the first matrix by the columns of the second matrix.
To calculate the product, multiply the rows of the first matrix by the columns of the second matrix, summing the products for each entry. In our example, we first found \(P^{-1}A\):
  • \(P^{-1}A = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2} \ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{3}{2} \ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\).
Then, by multiplying the result by matrix \(P\):
  • \(P^{-1}AP = \begin{bmatrix} \dfrac{1}{2} & \dfrac{3}{2} \ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}\begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}\).
This operation allowed us to find a matrix similar to the original, showing how transformations relate through multiplication.
Determinant Calculation
Determinants are scalar values that can be computed from a square matrix, providing key insights about the matrix. They are particularly valuable in determining invertibility and understanding transformation properties. Here鈥檚 how you calculate the determinant for a 2x2 matrix \(P = \begin{bmatrix} a & b \ c & d \end{bmatrix}\):
  • The determinant formula is \(\text{det}(P) = ad - bc\).
  • If the determinant is non-zero, the matrix is invertible; otherwise, it is not.
In our specific example:
  • \(P = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}\), we calculated \(\text{det}(P) = 1(-1) - 1(1) = -2\).
  • The negative determinant confirms the matrix is invertible, allowing us to proceed with inverse calculation and subsequent transformations.
Determinants help link matrix algebra to geometrical interpretations, such as scaling properties and volume transformations.

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Most popular questions from this chapter

If \(\mathrm{e}_{1}=(1,0), \mathrm{e}_{2}=(0,1), \mathrm{f}_{1}=(1,3), \mathrm{f}_{2}=(2,5)\), then \(\left\\{\mathrm{e}_{1}, \mathrm{e}_{2}\right\\}\) and \(\left\\{\mathrm{f}_{1}, \mathrm{f}_{2}\right\\}\) are based on \(\mathrm{R}^{2}\) (a) Find the transaction matrix \(\mathrm{P}\) from \(\left\\{\mathrm{e}_{1}\right\\}\) to \(\left\\{\mathrm{f}_{\mathrm{i}}\right\\}\). (b) Find the transaction matrix \(\mathrm{Q}\) from \(\left\\{\mathrm{f}_{\mathrm{i}}\right\\}\) to \(\left\\{\mathrm{e}_{\mathrm{i}}\right\\}\) and verify that \(\mathrm{Q}=\mathrm{P}^{-1}\) (c) Show that \(\left[\mathrm{v}_{\mathrm{f}}\right]=\mathrm{P}^{-1}[\mathrm{v}]_{\mathrm{e}}\) for any vector \(\mathrm{v} \in \mathrm{R}^{2}\). (d) Show that \([\mathrm{T}]_{\mathrm{f}}=\mathrm{P}^{-1}[\mathrm{~T}]_{\mathrm{e}} \mathrm{P}\) for the operator \(\mathrm{T}\) defined on \(\mathrm{R}^{2}\) by \(\mathrm{T}(\mathrm{x}, \mathrm{y})=(2 \mathrm{y}, 3 \mathrm{x}-\mathrm{y})\)

Find a nonsingular matrix \(\mathrm{P}\) such that \(\mathrm{P}^{-1} \mathrm{AP}\) is diagonal, given that $$ \mathrm{A}=\begin{array}{rr} 1 & 1 \\ & \mid 3 & -1 \mid \end{array} $$

Find an orthogonal matrix P that diagonalizes $$ \mathrm{A}=\begin{array}{rrr} 14 & 2 & 2 \mid \\ 2 & 4 & 2 \mid \\ \mid 2 & 2 & 4 \end{array} $$

Prove that \(\mathrm{n} \times \mathrm{n}\) matrix \(\mathrm{A}\) is diagonalizable if and only if it has n linearly independent eigenvectors. In this case \(\mathrm{A}\) is similar to a matrix D whose diagonal elements are the eigenvalues of \(\mathrm{A}\).

Define a symmetric matrix. Is every symmetric matrix similar to a diagonal matrix?
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