To find the diagonal matrix D, we need to compute the determinant of the matrix A-tI.
![{\displaystyle {\begin{aligned}\det(A-tI)&=\det \left(\left[{\begin{array}{cccc}2-t&0&3\\0&3-t&0\\4&0&1-t\\\end{array}}\right]\right)\\&=(3-t)\det \left(\left[{\begin{array}{cc}2-t&3\\4&1-t\end{array}}\right]\right)\\&=(3-t)\left((2-t)(1-t)-12\right)\\&=(3-t)(t-5)(t+2)\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5d1663501cb938f827959182984899ab2e9e5094)
Which means that eigenvalues are 3, 5 and -2. To find the eigenvectors (which will be the columns of the matrix P), we pick vectors of the kernels of 
where t = 3, 5, -2.
Eigenvalue t = 3.
![{\displaystyle \left[{\begin{array}{cccc}2-3&0&3\\0&3-3&0\\4&0&1-3\\\end{array}}\right]=\left[{\begin{array}{cccc}-1&0&3\\0&0&0\\4&0&-2\\\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f06cb1345059859ccaaba2a60bb7fcbdc5ed2166)
Thus an eigenvector corresponding to the eigenvalue t = 3 is given by
![{\displaystyle \left[{\begin{array}{c}0\\1\\0\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b52ab4cb272d5f8f89e5350782d5a7698294e245)
Eigenvalue t = 5.
![{\displaystyle \left[{\begin{array}{cccc}2-5&0&3\\0&3-5&0\\4&0&1-5\\\end{array}}\right]=\left[{\begin{array}{cccc}-3&0&3\\0&-2&0\\4&0&-4\\\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0058848b13408ed347d76d9bd04284edc207e74c)
Thus an eigenvector corresponding to the eigenvalue t = 5 is given by
![{\displaystyle \left[{\begin{array}{c}1\\0\\1\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e390a3208b8d5f68c206edc5f94cdfbdcff48f48)
Eigenvalue t = -2.
![{\displaystyle \left[{\begin{array}{cccc}2-(-2)&0&3\\0&3-(-2)&0\\4&0&1-(-2)\\\end{array}}\right]=\left[{\begin{array}{cccc}4&0&3\\0&5&0\\4&0&3\\\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/529e5f99057c1547c23ffe96fb11b7af48751b9f)
Thus an eigenvector corresponding to the eigenvalue t = -2 is given by
![{\displaystyle \left[{\begin{array}{c}3\\0\\-4\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d24c51135b0e1bc3aa93154b1cac15c97370c105)
can be chosen as an eigenvector.
So the matrices D and P are given by
![{\displaystyle D=\left[{\begin{array}{ccc}3&0&0\\0&5&0\\0&0&-2\end{array}}\right]\quad P=\left[{\begin{array}{ccc}0&1&3\\1&0&0\\0&1&-4\\\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/713895b4945251b1ea8725cd0b12988b0df4821c)
Now, as 
, then
![{\displaystyle {\begin{aligned}A^{-k}&=\left(PDP^{-1}\right)^{-k}\\&=PD^{-k}P^{-1}\\&=P\left[{\begin{array}{ccc}3&0&0\\0&5&0\\0&0&-2\\\end{array}}\right]^{-k}P^{-1}\\&=P\left[{\begin{array}{ccc}3^{-k}&0&0\\0&5^{-k}&0\\0&0&(-2)^{-k}\\\end{array}}\right]P^{-1}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8146098b8d99170df7464ea02b8da2ddae5816ce)
So we can see that as k gets large, 
get closers to the zero matrix. Hence the right hand side, and with it the matrix A-k, also approaches the zeros matrix.
