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Find the characteristic equation of A, the eigenvalues of A, and a basis for the eigenspace corresponding to each eigenvalue.
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$(a). The characteristic equation is $\;|A-\lambda I|=0\;$\\ That is $\;(-5-\lambda)(1-\lambda)(3-\lambda)=0\;$\\ Therefore, the characteristic equation is $\;|A-\lambda I|=0\;$\\ That is $\;-\lambda^{3}-\lambda^{2} +17\lambda-15=0\;$\\ \\(b). Therefore,the eigen values are $\;\;-5\;,\;1\;,\;3\;.\;$\\ \\(c). If $\;\lambda_{j}\;$ is a given eigen value, then an eigen vector corresponding to $\;\lambda_{j}\;$ is a non-zero solution vector of the linear system$\;(A-\lambda_{j}I)X=\bar{0}\;$\\ Therefore, we get $\;u=(1,0,0)^{T}\;,\;v=(9,8,16)^{T}\;,\;w=(1,6,0)^{T}\;$ as eigen vectors corresponding to the eigen values $\;-5,\;1,\;3\:$ respectively.\\ Therefore, \\A basis for the eigen space of $\:\lambda_{1}=-5\:$ is $\;B_{1}=\{(1,0,0)^{T}\}\;$\\ A basis for the eigen space of $\:\lambda_{2}=1\;$ is $\;B_{2}=\{(9,8,16)^{T}\}\;$\\ A basis for the eigen space of $\:\lambda_{3}=3\;$ is $\;B_{3}=\{(1,6,0)^{T}\}\;$\\

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