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Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $mathbbR$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices. Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $mathbbC$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real? Thanks in advance. matrices eigenvalues-eigenvectors spectral-theory share | cite | improve this question asked Aug 9 at 22:41 Charles 533 4 20 $lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element. – copper.hat Aug 9 at 23:00 "IfA−λI is real, then the kernel contains a purely real

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables? probability probability-theory share | cite | improve this question asked Aug 9 at 22:47 Probably 35 3 1 Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$. – Clement C. Aug 9 at 22:54 add a comment  |  up vote 0 down vote favorite I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables? probability probability-theory share | cite | improve this question asked Aug 9 at 22:47 Probably 35 3 1 Trivial counterexample: take a constant function $f$. (For instance, $f