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Let us consider a square matrix A of order n with Eigen values of a, b, c then the Eigen values of the matrix AT could be.
Correct answer is 'a,b,c'. Can you explain this answer?
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Let us consider a square matrix A of order n with Eigen values of a, b...
According to the property of the Eigen values, any square matrix A and its transpose AT have the same Eigen values.
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Let us consider a square matrix A of order n with Eigen values of a, b...
Understanding Eigenvalues of a Matrix and its Transpose
When we have a square matrix A of order n, its eigenvalues represent the scalar values for which there exists a non-zero vector (the eigenvector) such that the equation A*v = λ*v holds, where λ is an eigenvalue and v is an eigenvector.
Eigenvalues and Transpose
- The transpose of a matrix, denoted as AT, is obtained by swapping its rows and columns.
- An essential property of matrices is that the eigenvalues of a matrix and its transpose are the same.
Proof Concept
- If λ is an eigenvalue of A, then there exists a non-zero vector v such that A*v = λ*v.
- Taking the transpose of this equation, we have (A*v)T = (λ*v)T.
- Using the property of transposes: (AB)T = BTAT, we can rewrite this as vT * AT = λ*vT.
- This shows that if v is an eigenvector of A corresponding to eigenvalue λ, then vT is an eigenvector of AT corresponding to the same eigenvalue λ.
Conclusion
- Therefore, for a square matrix A with eigenvalues a, b, and c, the eigenvalues of the transpose matrix AT are also a, b, and c.
- This result holds true for any square matrix, confirming the equality of eigenvalues for a matrix and its transpose.
In summary, the eigenvalues of the transpose matrix AT are the same as those of the original matrix A, which are a, b, and c.
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