Are Eigen Vectors the Same for Eigen Values of Transpose A and A?

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SUMMARY

The eigenvalues of a square matrix A are identical to those of its transpose A^t, as established by the property that det(A) = det(A^t). For any eigenvalue k of matrix A, the corresponding eigenvectors remain unchanged when considering A^t. This is proven through the characteristic equation det(A - kI_n) = 0, which holds true for both A and A^t, confirming that the solution space P(A) is equivalent to P(A^t).

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  • Understanding of eigenvalues and eigenvectors in linear algebra
  • Familiarity with the characteristic equation and determinants
  • Knowledge of square matrices and their properties
  • Basic concepts of linear transformations and vector spaces
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nhrock3
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A is a square matrix over F field

if k is the eigen value of A



prove that k is eigen value of A^t too

and has the same eigen vectors

??



eigen vectors are the solution space P(A)

is found by solving (A-kI)x=0

dim P(A)=dim n -dim (ro(a))

rho(a)=rho(a^t)

|A|=|A^t|



these are the laws i maneged to come up with to solve it



??
 
Last edited:
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We know that the eigenvalues of a matrix A are all the scalars k that satisfy the characteristic equation det( A - k In ) = 0.

We also know that det( A ) = det( A^t ) for any square matrix A.

Can we use these facts to complete a proof?
 
ok i prooved that the eigen values are the same
hw to prove that the eigen vectors are the same to for every eigen value
 

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