/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 For each of the following matric... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 4

For each of the following matrices \(A \in \mathrm{M}_{n \times n}(F)\), (i) Determine all the eigenvalues of \(A\). (ii) For each eigenvalue \(\lambda\) of \(A\), find the set of eigenvectors corresponding to \(\lambda\). (iii) If possible, find a basis for \(\mathrm{F}^{n}\) consisting of eigenvectors of \(A\). (iv) If successful in finding such a basis, determine an invertible matrix \(Q\) and a diagonal matrix \(D\) such that \(Q^{-1} A Q=D\).258 Chap. 5 Diagonalization (a) \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right) \quad\) for \(F=R\) (b) \(A=\left(\begin{array}{rrr}0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{array}\right) \quad\) for \(F=R\) (c) \(A=\left(\begin{array}{rr}i & 1 \\ 2 & -i\end{array}\right) \quad\) for \(F=C\) (d) \(A=\left(\begin{array}{ccc}2 & 0 & -1 \\ 4 & 1 & -4 \\ 2 & 0 & -1\end{array}\right) \quad\) for \(F=R\)

Short Answer

Expert verified
#Matrix (b)# 1. Characteristic polynomial: \[|(A - 位I)| = \left|\begin{array}{ccc} -位 & 2 & 3 \\ -1 & 1-位 & 1\\ 2 & 2 & 5-位 \end{array}\right| = -(位-2)(位^2-6位)\] 2. Eigenvalues: 位鈧� = 0, 位鈧� = 2, 位鈧� = 6 3. Eigenvectors: For 位鈧� = 0: \[\left(\begin{array}{ccc}0 & 2 & 3 \\ -1 & 1 & 1\\ 2 & 2 & 5 \end{array}\right)x = 0\] Eigenvector: \(\left(\begin{array}{c} 1 \\ 0 \\ -\frac{2}{3} \end{array}\right)\) For 位鈧� = 2: \[\left(\begin{array}{ccc}-2 & 2 & 3 \\ -1 & -1 & 1\\ 2 & 2 & 3 \end{array}\right)x = 0\] Eigenvector: \(\left(\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right)\) For 位鈧� = 6: \[\left(\begin{array}{ccc}-6 & 2 & 3 \\ -1 & -5 & 1\\ 2 & 2 & -1 \end{array}\right)x = 0\] Eigenvector: \(\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right)\) 4. Check if the eigenvectors form a basis for F鲁: The eigenvectors are linearly independent, so they form a basis for F鲁. 5. Diagonal matrix D and invertible matrix Q: \[D = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 2 & 0\\0 & 0 & 6\end{array}\right)\] \[Q = \left(\begin{array}{ccc}1 & 1 & 1\\ 0 & 1 & 2\\ -\frac{2}{3} & 0 & 1\end{array}\right)\] Verify that \(Q^{-1}AQ = D\). #Matrix (c)# 1. Characteristic polynomial: \[(A - 位I) = \left(\begin{array}{cc} i-位 & 1 \\ 2 & -i-位 \end{array}\right)\] \[(位-i)(-位-i) - 2 = 位^2 + 1\] 2. Eigenvalues: 位鈧� = i, 位鈧� = -i 3. Eigenvectors: For 位鈧� = i: \[(A - 位I)x = \left(\begin{array}{cc}0 & 1\\ 2 & -2i\end{array}\right)x = 0\] Eigenvector: \(\left(\begin{array}{c} -i \\ 1 \end{array}\right)\) For 位鈧� = -i: \[(A - 位I)x = \left(\begin{array}{cc}2i & 1\\ 2 & 2i\end{array}\right)x = 0\] Eigenvector: \(\left(\begin{array}{c} i \\ 1 \end{array}\right)\) 4. Check if the eigenvectors form a basis for F虏: The eigenvectors are linearly independent, so they form a basis for F虏. 5. Diagonal matrix D and invertible matrix Q: \[D = \left(\begin{array}{cc}i & 0\\ 0 & -i\end{array}\right)\] \[Q = \left(\begin{array}{cc}-i & i\\ 1 & 1\end{array}\right)\] Verify that \(Q^{-1}AQ = D\). #Matrix (d)# 1. Characteristic polynomial: \[|(A - 位I)| = \left|\begin{array}{ccc} 2-位 & 0 & -1 \\ 4 & 1-位 & -4\\ 2 & 0 & -1-位 \end{array}\right| = -(位-2)^2(位+1)\] 2. Eigenvalues: 位鈧� = 2, 位鈧� = 2, 位鈧� = -1 3. Eigenvectors: For 位鈧� = 2: \[(A - 位I)x = \left(\begin{array}{ccc}0 & 0 & -1\\ 4 & -1 & -4\\ 2 & 0 & -3\end{array}\right)x = 0\] Eigenvectors cannot be determined for this case. Since a basis cannot be formed using eigenvectors of A, it cannot be diagonalized. No further steps are possible for matrix (d).

Step by step solution

01

1. Find the characteristic polynomial for matrix A#a)

: Calculate the determinant of \((A鈭捨籌)\) for the given matrix A: \[A - 位I = \left(\begin{array}{cc}1 - 位 & 2\\3 & 2 - 位\end{array}\right)\] The determinant of \((A - 位I)\) is: \[((1 - 位)(2 - 位) - (3 \cdot 2)) = (位 - 1)(位 - 2) - 6\]
02

2. Find the eigenvalues for matrix A#a)

: To find the eigenvalues, solve the characteristic polynomial: \[ (位 - 1)(位 - 2) - 6 = 位^2 - 3位 - 4 = 0\] The eigenvalues are 位鈧� = 4, 位鈧� = -1.
03

3. Find the eigenvectors for each eigenvalue of matrix A#a)

: For 位鈧� = 4, solve the system \((A - 位I)x = 0\): \[\left(\begin{array}{cc}-3 & 2\\3 & -2\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = 0\] The eigenvector for 位鈧� = 4 is x鈧� = \(\left(\begin{array}{c} 2 \\ 3 \end{array}\right)\). For 位鈧� = -1, solve the system \((A - 位I)x = 0\): \[\left(\begin{array}{cc}2 & 2\\3 & 3\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = 0\] The eigenvector for 位鈧� = -1 is x鈧� = \(\left(\begin{array}{c} 1 \\ -1 \end{array}\right)\).
04

4. Check if the eigenvectors form a basis for F虏#a)

: The eigenvectors x鈧� and x鈧� are linearly independent, so they form a basis for F虏.
05

5. Find the diagonal matrix D and the invertible matrix Q for matrix A#a)

: The diagonal matrix D consists of the eigenvalues on the main diagonal: \[D = \left(\begin{array}{cc}4 & 0\\0 & -1\end{array}\right)\] The invertible matrix Q consists of the eigenvectors as columns: \[Q = \left(\begin{array}{cc}2 & 1\\3 & -1\end{array}\right)\] Now, let's verify that \(Q^{-1}AQ = D\): \[Q^{-1}AQ = \left(\begin{array}{cc}\frac{1}{5} & \frac{1}{5}\\\frac{3}{5} & -\frac{2}{5}\end{array}\right) \left(\begin{array}{cc}1 & 2\\3 & 2\end{array}\right) \left(\begin{array}{cc}2 & 1\\3 & -1\end{array}\right) = \left(\begin{array}{cc}4 & 0\\0 & -1\end{array}\right) = D\] Matrix (a) is complete. Continue with the same steps for the remaining matrices provided in the exercise.

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Key Concepts

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

Eigenvalues and Eigenvectors
In the world of linear algebra, eigenvalues and eigenvectors are fundamental concepts used to study the properties of linear transformations. An eigenvector of a matrix is a non-zero vector that only changes by a scalar factor when that linear transformation is applied. In simpler terms, when you multiply the matrix by this particular vector, the result is the same vector multiplied by a number (called an eigenvalue).

To find eigenvalues, we calculate the characteristic polynomial of the matrix, which is derived by subtracting an eigenvalue-multiplied identity matrix from the initial matrix and then taking the determinant. Setting this determinant equal to zero and solving for the eigenvalue gives us the roots of the polynomial, which are the eigenvalues. For example, in our exercise, solving the characteristic polynomial of matrix A resulted in eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = -1 \).

Once we have our eigenvalues, we determine the corresponding eigenvectors. For each eigenvalue, we substitute it back into the equation \( (A - \lambda I)x = 0 \) and solve for the vector \( x \). This calculation helps us find the specific vectors that are scaled versions of themselves when the matrix transformation is applied, like we found the eigenvectors in our step by step solution for \( \lambda_1 \) and \( \lambda_2 \).
Characteristic Polynomial
The characteristic polynomial is a bridge between a matrix and its eigenvalues. It's a special polynomial whose roots are the eigenvalues of a matrix. To construct it, one must take the matrix in question, say matrix A, subtract \( \lambda I \) from it, where \( \lambda \) is an unknown scalar multiplier and \( I \) is the identity matrix of the same size as A. Then, find the determinant of the resulting matrix. This determinant, set to zero, forms the characteristic equation \( det(A - \lambda I) = 0 \).

Solving this equation will often require techniques from algebra to find the values of \( \lambda \) that satisfy it, which are our eigenvalues. In our sample problem with matrix A, we derived the characteristic polynomial \( \lambda^2 - 3\lambda - 4 = 0 \) which we then solved to find the eigenvalues.
Basis of Eigenvectors
When we talk about a basis of eigenvectors, we refer to a set of eigenvectors that are linearly independent and span an entire vector space. In our case, the vector space is \( \mathrm{F}^n \), where 'n' is the size of the matrix. To form a complete basis, the number of linearly independent eigenvectors must equal the dimension of the space. This basis captures all the directions in which the matrix acts simply by scaling vectors.

If we have found a sufficient number of eigenvectors, as in the provided solution for matrix A where we identified two linearly independent eigenvectors, we have successfully formed a basis. The eigenvectors \( \left(\begin{array}{c} 2 \ 3 \end{array}\right) \) and \( \left(\begin{array}{c} 1 \ -1 \end{array}\right) \) are linearly independent and span \( \mathrm{F}^2 \) in the example, thus constituting a basis.
Similarity Transformations
The concept of similarity transformations is a powerful tool in linear algebra. Two matrices are similar if you can find an invertible matrix Q such that when you premultiply and postmultiply the original matrix A by Q and its inverse, respectively, you obtain a diagonal matrix D. Symbolically, it's expressed as \( Q^{-1}AQ = D \). Why is this important? Because a diagonal matrix is much simpler to work with and the diagonal elements are directly the eigenvalues of A.

In the context of the given problem, once we found a basis of eigenvectors for matrix A, we were able to construct the matrix Q. By arranging the eigenvectors as columns of Q, and placing the corresponding eigenvalues in a diagonal matrix D, we achieved this transformation. We verified the result by confirming that indeed \( Q^{-1}AQ \) equaled the diagonal matrix D. This process not only simplifies matrix operations but also provides significant insights into the structure and properties of A.

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Most popular questions from this chapter

For each of the following matrices \(A \in \mathrm{M}_{n \times n}(R)\), test \(A\) for diagonalizability, and if \(A\) is diagonalizable, find an invertible matrix \(Q\) and a diagonal matrix \(D\) such that \(Q^{-1} A Q=D\). (a) \(\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right)\) (c) \(\left(\begin{array}{ll}1 & 4 \\ 3 & 2\end{array}\right)\) (d) \(\left(\begin{array}{lll}7 & -4 & 0 \\ 8 & -5 & 0 \\ 6 & -6 & 3\end{array}\right)\) (e) \(\left(\begin{array}{rrr}0 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{array}\right)\) (f) \(\left(\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3\end{array}\right)\) (g) \(\left(\begin{array}{rrr}3 & 1 & 1 \\ 2 & 4 & 2 \\ -1 & -1 & 1\end{array}\right)\)

Let \(A\) be an \(n \times n\) matrix that is similar to an upper triangular matrix and has the distinct eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) with corresponding multiplicities \(m_{1}, m_{2}, \ldots, m_{k}\). Prove the following statements. (a) \(\operatorname{tr}(A)=\sum^{k} m_{i} \lambda_{i}\) (b) \(\operatorname{det}(A)=\left(\lambda_{1}\right)^{m_{1}}\left(\lambda_{2}\right)^{m_{2}} \ldots\left(\lambda_{k}\right)^{m_{k}}\).

Exercises 32 through 39 are concerned with direct sums. Let \(\mathrm{T}\) be a linear operator on a vector space \(\mathrm{V}\), and let \(\mathrm{W}_{1}, \mathrm{~W}_{2}, \ldots, \mathrm{W}_{k}\) be \(\mathrm{T}\)-invariant subspaces of \(\mathrm{V}\). Prove that \(\mathrm{W}_{1}+\mathrm{W}_{2}+\cdots+\mathrm{W}_{k}\) is also a T-invariant subspace of V.

Let \(\mathcal{C}\) be a collection of diagonalizable linear operators on a finitedimensional vector space V. Prove that there is an ordered basis \(\beta\) such that \([T]_{\beta}\) is a diagonal matrix for all \(T \in \mathcal{C}\) if and only if the operators of \(\mathcal{C}\) commute under composition. (This is an extension of Exercise 25.) Hints for the case that the operators commute: The result is trivial if each operator has only one eigenvalue. Otherwise, establish the general result by mathematical induction on \(\operatorname{dim}(\mathrm{V})\), using the fact that \(V\) is the direct sum of the eigenspaces of some operator in \(\mathcal{C}\) that has more than one eigenvalue.

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) be an ordered basis for \(\mathrm{V}\). Prove that \(\lambda\) is an eigenvalue of \(\mathrm{T}\) if and only if \(\lambda\) is an eigenvalue of \([\mathrm{T}]_{\beta}\).
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