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Chapter 6: Problem 34

Prove that there do not exist \(n \times n\) matrices \(A\) and \(B\) such that \\[ A B-B A=I \\] [Hint: See Exercises 10 and \(33 .\)

Short Answer

Expert verified
We are asked to prove that there do not exist \(n \times n\) matrices \(A\) and \(B\) such that \(AB-BA=I\). We find contradictions by computing the determinants and traces of both sides of the equation. For determinants, comparing both sides, we get \[-(\det(A) \det(B))^2 = 1\], which is a contradiction since the square of any real number cannot be negative. For traces, comparing both sides, we get \[\text{tr}(AB - BA) = 0 \neq n = \text{tr}(I)\], which is another contradiction. Hence, we conclude that there do not exist \(n \times n\) matrices \(A\) and \(B\) such that \(AB - BA = I\).

Step by step solution

01

Compute the determinant of both sides

First, let's compute the determinant of both sides of the equation \(AB - BA = I\). Since the determinant of a matrix product is the product of the determinants, we have: \[\det(AB) = \det(A) \det(B)\] In a similar manner, \[\det(BA) = \det(B) \det(A)\] Now, if we compute the determinant of the left side of the equation, we get: \[\det(AB - BA) = \det(AB) \det(-BA) = -\det(A) \det(B) \det(B) \det(A) = -(\det(A) \det(B))^2\] On the other hand, the determinant of the right side (identity matrix) is: \[\det(I) = 1\]
02

Compare the determinants

Now, let's compare the determinants of both sides of the equation. Remember that we have: \[-(\det(A) \det(B))^2 = 1\] For any \(n \times n\) matrices \(A\) and \(B\), their determinants are real numbers. Therefore, the square of any real number cannot be negative. This means that: \[(\det(A) \det(B))^2 \geq 0\] We have a contradiction, since \(-1\) cannot be equal to a positive number or zero.
03

Compute the trace of both sides

To strengthen our proof, let's compute the trace of both sides of the equation \(AB - BA = I\). Recall that the trace of a matrix is the sum of its diagonal elements and has the property: \(\text{tr}(AB) = \text{tr}(BA)\). Computing the trace of the left side, we get: \[\text{tr}(AB - BA) = \text{tr}(AB) - \text{tr}(BA) = 0\] And the trace of the right side (identity matrix) is: \[\text{tr}(I) = n\]
04

Compare the traces

Now, let's compare the traces of both sides of the equation. Remember that we have: \[0 = n\] This equation holds true only if \(n = 0\) which means a \(0\times 0\) matrix, but in the problem, we are given an \(n\times n\) matrix. Therefore, we have another contradiction.
05

Conclusion

Based on the contradictions obtained in step 2 and step 4, we can conclude that there do not exist \(n \times n\) matrices \(A\) and \(B\) such that \(AB - BA = I\).

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Most popular questions from this chapter

Let \(A\) be an \(n \times n\) matrix with an eigenvalue \(\lambda\) of multiplicity \(n .\) Show that \(A\) is diagonalizable if and only if \(A=\lambda I\)

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