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- #1

- Feb 5, 2012

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Here's a question I got stuck. Hope you can shed some light on it.

Of course if we write the matrix of the linear transformation we get,Find all eigenvalues of a linear transformation \(f\) whose matrix in some basis is \(A^{t}.A\) where \(A=(a_1,\cdots, a_n)\).

\[A^{t}.A=\begin{pmatrix}a_1^2 & a_{1}a_2 & \cdots & a_{1}a_{n}\\a_2 a_1 & a_2^2 &\cdots & a_{2}a_{n}\\.&.&\cdots&.\\.&.&\cdots&.\\a_n a_1 & a_{n}a_2 & \cdots & a_{n}^2\end{pmatrix}\]

Now this is a symmetric matrix. So it could be written as \(A^{t}.A=QDQ^T\) where \(Q\) is a orthogonal matrix and \(D\) is a diagonal matrix. If we can do this the diagonal elements of the diagonal matrix gives all the eigenvalues we need. However I have no idea how break \(A^{t}.A\) into \(QDQ^T\). Or does any of you see a different approach to this problem which is much more easier?