The following matrices have previously appeared in Exercises 24.7 under the heading Calculating eigenvalues and eigenspaces, where you were previously asked to compute the characteristic polynomial (in factored form), eigenvalues, and eigenspaces for each matrix. Following up on that information, for each matrix:
Confirm Corollary 25.6.6 by demonstrating that the full collection of calculated basis vectors from all the eigenspaces, taken all together, form a linearly independent set.
State whether the matrix is diagonalizable and, if so, produce a transition matrix \(P \) and diagonal matrix \(D \) so that \(\inv{P} A P = D \) is true, where \(A \) is the given matrix. (Note: You should not need to calculate \(\inv{P} \) in order to produce \(D \text{.}\))
The geometric multiplicity is \(\dim E_{-3}(A) = 1 \) and the algebraic multiplicity is \(2 \text{.}\) This is consistent with Theorem 25.6.7 as the geometric multiplicity is not greater than the algebraic multiplicity.
For each of the two eigenvalues both the geometric and algebraic multiplicities are equal to \(1 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
For each of the three eigenvalues both the geometric and algebraic multiplicities are equal to \(1 \text{.}\) This is consistent with Theorem 25.6.7 as in no case is the geometric multiplicity is greater than the algebraic multiplicity.
For eigenvalue \(\lambda = 0 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) and for eigenvalue \(\lambda = -7 \) both the geometric and algebraic multiplicities are \(2 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
For eigenvalue \(\lambda = -3 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) but for eigenvalue \(\lambda = -1 \) the algebraic multiplicity is \(2 \) while the geometric multiplicity is only \(1 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
Since there is only a single eigenvalue, we know that the basis vectors for that single eigenspace that we have already computed (see the provided answers to Exercise 24.7.10) form an independent set.
For the single eigenvalue the algebraic multiplicity is \(3 \) while the geometric multiplicity is only \(2 \text{.}\) This is consistent with Theorem 25.6.7 as the geometric multiplicity is not greater than the algebraic multiplicity.
Since there is only a single eigenvalue, we know that the single basis vector for that single eigenspace that we have already computed (see the provided answers to Exercise 24.7.11) forms an independent set.
For the single eigenvalue the algebraic multiplicity is \(3 \) while the geometric multiplicity is only \(1 \text{.}\) This is consistent with Theorem 25.6.7 as the geometric multiplicity is not greater than the algebraic multiplicity.
For eigenvalue \(\lambda = -1 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) and for eigenvalue \(\lambda = -4 \) both the geometric and algebraic multiplicities are \(3 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
For each of the eigenvalues \(\lambda = 0 \) and \(\lambda = 4 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) but for eigenvalue \(\lambda = -3 \) the algebraic multiplicity is \(2 \) while the geometric multiplicity is only \(1 \text{.}\) This is consistent with Theorem 25.6.7 as in no case is the geometric multiplicity is greater than the algebraic multiplicity.
For eigenvalue \(\lambda = 2 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) but for eigenvalue \(\lambda = -2 \) the algebraic multiplicity is \(3 \) while the geometric multiplicity is only \(2 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
For eigenvalue \(\lambda = 2-6 \) both the geometric and algebraic multiplicities are \(1 \text{,}\) but for eigenvalue \(\lambda = 4 \) the algebraic multiplicity is \(3 \) while the geometric multiplicity is only \(1 \text{.}\) This is consistent with Theorem 25.6.7 as in neither case is the geometric multiplicity is greater than the algebraic multiplicity.
Since there is only a single eigenvalue, we know that the single basis vector for that single eigenspace that we have already computed (see the provided answers to Exercise 24.7.16) forms an independent set.
For the single eigenvalue the algebraic multiplicity is \(4 \) while the geometric multiplicity is only \(1 \text{.}\) This is consistent with Theorem 25.6.7 as the geometric multiplicity is not greater than the algebraic multiplicity.
In each case, assume that \(A \) is similar to the given diagonal matrix. State the eigenvalues of \(A \text{,}\) the algebraic multiplicity of each eigenvalue, and the characteristic polynomial of \(A \text{.}\)
must be diagonalizable, by Statement 2 of Corollary 25.6.9. In particular, if \(P \) is a \(4 \times 4 \) matrix constructed by an eigenvector for each of the eigenvalues \(\lambda = 1, 2, 3, 4 \text{,}\) in order, for its columns, then it will be that
The letter \(E \) has been chosen deliberately to represent the matrix to be used to “transform” the initial transition matrix \(P \) into a new transition matrix \(Q \text{.}\) Recall that the result of a matrix product \(E P \text{,}\) with an elementary matrix\(E \) on the left, is the same as if an elementary row operation had been performed on \(P \text{.}\) It turns out that the result of a matrix product \(P E \) with elementary \(E \) on the right is the same as if an “elementary column operation” had been performed on \(P \text{.}\) Apply this new pattern for elementary matrices, combined with the summary information in the concluding sentences of Procedure 25.4.1
In each case, determine whether the pair of matrices are similar. If they are, exhibiting a specific transition matrix that achieves a similarity relationship between the two. If they are not, provide convincing reasoning to justify this answer based on facts from Section 25.6.
Based on the exercises under the heading Similarity of diagonal matrices above, make a conjecture about the precise condition(s) under which two diagonal matrices of the same size are similar.
Does the pattern of similarity between diagonal matrices, expressed using your reworded formulation from Task b, also accurately characterize when two diagonalizable (but not necessarily diagonal) matrices are similar?
More generally, does the pattern of similarity from Task b also accurately characterize when any two (not necessarily diagonal nor diagonalizable) matrices are similar?
In each case, determine whether the pair of matrices are similar. If they are, exhibiting a specific transition matrix that achieves a similarity relationship between the two. If they are not, provide convincing reasoning to justify this answer based on facts from Section 25.6.
Let \(B \) represent the second matrix provided in this exercise. It can be determined that \(B \) is also diagonalizable, and following Procedure 25.4.1 leads to a transition matrix and corresponding diagonal form:
Reconsider the information in the hint to Exercise 25 in the following way: if \(E \) is an elementary matrix, then the result of \(\inv{E} A E \) can be re-interpreted as applying both an elementary column operation and the corresponding reverse elementary row operation.
