# Exercise Solution 2.8

Consider an *n*-dimensional diagonal matrix. We identify *n* distinct eigenvectors as follows. The first has 1 for its first component and 0 for the rest of its components. The second has 1 for its second component and 0 for the rest of its components. In general, the *i ^{th}* eigenvector has 1 for its

*i*component and 0 for the rest of its components. The eigenvectors are linearly independent, so they are distinct. An

^{th}*n*-dimensional matrix cannot have more than

*n*distinct eigenvectors, so these are all the eigenvectors of our matrix. Clearly, the eigenvalue corresponding to the

*i*eigenvector is the

^{th}*i*diagonal element of the matrix. Accordingly, the diagonal elements of the matrix are its eigenvalues.

^{th}
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