The characteristic roots of the matrix left[b | vedclass

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(A) The characteristic roots (eigenvalues) of a square matrix $A$ are the solutions to the characteristic equation $\det(A - \lambda I) = 0$. For a tr

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$1, 3, 6$

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(A) The characteristic roots (eigenvalues) of a square matrix $A$ are the solutions to the characteristic equation $\det(A - \lambda I) = 0$. For a triangular matrix (either upper or lower),the determinant $\det(A - \lambda I)$ is simply the product of the diagonal elements minus $\lambda$. Given the matrix $A = \left[\begin{array}{ccc}1 & 0 & 0 \ 2 & 3 & 0 \ 4 & 5 & 6\end{array}ight]$,which is a lower triangular matrix,the characteristic equation is $(1 - \lambda)(3 - \lambda)(6 - \lambda) = 0$. Setting each factor to zero,we get $\lambda = 1, 3, 6$. Thus,the characteristic roots are $1, 3, 6$.

The characteristic roots of the matrix $\left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 6\end{array}\right]$ are

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