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I want to diagonalize the matrix $$A = \begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}$$ I found the characteristic polynomial $$P_A(t) = (t^2+1)(t^2-1)$$ Thus the eigenvalues of the matrix are $$\{1, -1 ,i,-i\}$$ By considering the eigenvalues $1$ and $-1$ I find the eigenvectors $$\begin{pmatrix}1\\0\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}\begin{pmatrix}0\\1\\-1\\0\end{pmatrix}$$ Symbolab tells me that the matrix $M$ formed by these 4 eigenvectors is such that $$A = M^{-1}DM$$ This is what I want, I have diagonalized the matrix $A$ but I am confused as I have completely ignored the eigenvalues of $i$ and $-i$. In this similar question posted on stackoverflow they don't ignore the complex roots of the characteristic polynomial.

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  • $\begingroup$ If I was diagonalizing over the complex numbers would I not end up with a 4x8 matrix M, which is non-invertible? $\endgroup$ – Retsek Jan 31 at 1:33
  • $\begingroup$ No. Why would you think that? $\endgroup$ – amd Jan 31 at 1:34
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    $\begingroup$ there are bigger problems. your matrix $A = \begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}$ is real symmetric so all eigenvalues are real. Also you can check $A^2=I$. This is an involution which implies all eigenvalues are -1 or +1. Nothing non-real. $\endgroup$ – user8675309 Jan 31 at 1:39
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    $\begingroup$ The characteristic polynomial is $(x^2-1)^2 $. $\endgroup$ – copper.hat Jan 31 at 3:07
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Here is a list of what has gone wrong but you recognise that something has gone wrong, which is a good start (i)You should specify the field of scalars-in this case the real numbers or the complex numbers would be the most likely choices (ii) you should specify which kind of diagonalisation is being done-in this case, you are doing diagonalisation by similarity (iii) the eigenvalues of a real symmetric matrix are real (iv)in the characteristic equation of a 4x4 matrix, the cofficient of $\lambda ^2$is the sum of the principal minors of order 2, which gives -2 for your matrix(v)in the characteristic equation of a 4x4 matrix, the constant term is the determinant of the matrix, which in this case is 1 (vi) you have found two linearly independent eigenvectors for the eigenvalue 1, so the dimension of the corresponding eigenspace is at least 2, but in your characteristic polynomial, the multiplicity of the root 1 is 1 , which would mean that the dimension of the corresponding eigenspace would be 1(vii) same problem with eigenvalue -1 (viii) the sum of the dimensions of all the eigenspaces of a 4x4 matrix is at most 4 but the sum of the dimensions of the eigenspaces for your eigenvalues would be at least 6.

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