I will define a nice eigenvalue decomposition (eigendecomposition) as one where both the normalized eigenvector matrix $U$ and its inverse both have finite decimal representations with at most a few digits to the right of the decimal point. This makes it much easier to demonstrate this in class, but such examples are very rare for smaller matrices, and it is not necessarily obvious how to find such examples. In some cases, this author believes the descriptions below give are exhaustive.
Important: If $U$ is a matrix of linearly independent unit eigenvectors where each column is normalized, and where $U^{-1}$ has a finite decimal representation, then if any two columns of $U$ are swapped then the resulting matrix will still have an inverse that has a finite decimal representation. If you swap rows, the inverse may, however, no longer have a finite decimal representation.
None of the examples have a zero eigenvalue, meaning that all of the examples are invertible. The main purpose for this is to show the more difficult case, where the matrix $A = UDU^{-1}$ is as dense as is likely. You are of course welcome to choose appropraite values of the $n_k$ so that one of the eigenvalues becomes zero.
Almost nice matrices: If you ensure that the eigenvalues are either multiples of $10$ or $100$ in any of the below examples, and then divide the resulting $UDU^{-1}$ matrix by $10$ or $100$, respectively, the resulting matrix, while not necessarily being an integer matrix, will still have integer eigenvalues.
Warning: if you select eigenvalues that are all different, then using the standard techniques of finding the eigenvectors (solving $\lambda I_n - A = \textbf{0}_n$) will have you find these nice eigenvectors. However, if you repeat an eigenvalue, then the basis you find for the higher-dimensional eigenspace may not be these nice eigenvectors.
The diagonal matrix of eigenvectors will be $D = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$.
Two such eigenvector matrices $U$ are upper triangular, and thus $U^{-1}$ is also upper triangular, and thus the matrix $A = UDU^{-1}$ is also upper triangular, which is nice, as a student can read the eigenvalues off of the matrix $A$ itself.
$U = \begin{pmatrix} 1 & 0.6 \\ 0 & 0.8 \end{pmatrix}$ and $U^{-1} = \begin{pmatrix} 1 & -0.75 \\ 0 & 1.25 \end{pmatrix}$
U = [1, 0.6
0, 0.8]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{3}{4}(b - a) \\ 0 & b \end{pmatrix}$
If $n_1$ and $n_2$ are any two integers, and $a = n_1$ and $b = n_2 + 4n_1$, then $UDU^{-1}$ will be an integer matrix.
If $n_1$ and $n_2$ are any two integers, and $a = 10n_1$ and $b = 20n_2 + 10n_1$, then $UDU^{-1}$ will be a matrix with integer eigenvalues with entries having at most one digit beyond the decimal point.
If $n_1$ and $n_2$ are any two integers, and $a = 100n_1$ and $b = 100n_2$, then $UDU^{-1}$ will be a matrix with integer eigenvalues with entries having at most two digits beyond the decimal point.
You can see some examples here.
$U = \begin{pmatrix} 1 & -0.6 \\ 0 & 0.8 \end{pmatrix}$ and $U^{-1} = \begin{pmatrix} 1 & 0.75 \\ 0 & 1.25 \end{pmatrix}$
U = [1, -0.6
0, 0.8]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{3}{4}(a - b) \\ 0 & b \end{pmatrix}$
If $n_1$ and $n_2$ are any two integers, and $a = n_1$ and $b = n_2 + 4n_1$, then $UDU^{-1}$ will be an integer matrix.
You can see some examples here.
This author has only been able to find one full eigenvalue matrix that produces nice matrices:
$U = \begin{pmatrix} 0.6 & 0.28 \\ 0.8 & -0.96 \end{pmatrix}$ and $U^{-1} = \begin{pmatrix} 1.2 & 0.35 \\ 1 & -0.75 \end{pmatrix}$
U = [0.6, 0.28
0.8, -0.96]
$A = UDU^{-1} = \begin{pmatrix} \frac{1}{25}(18a + 7b) & \frac{21}{100}(a - b) \\ \frac{21}{100}(a - b) & \frac{1}{25}(7a + 18b) \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2$ must be integers and let the eigenvalues be $n_1+28n_2, n_1-72n_2$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $100n_1+10n_2, 10n_2$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
For symmetric matrices, this is much easier, as we need only need observe that the eigenvalue matrix $U$ can be isometric (orthogonal), and therefore if the first column is $\begin{pmatrix} \alpha \\ \beta \end{pmatrix}$, the second column must be $\begin{pmatrix} \beta \\ -\alpha \end{pmatrix}$.
Therefore, all we need are 2-dimensional unit vectors that have terminating decimal representations, and these are sufficiently rare:
$\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$, $\begin{pmatrix} 0.28 \\ 0.96 \end{pmatrix}$ and $\begin{pmatrix} 0.352 \\ 0.936 \end{pmatrix}$
Other are here, but these have more digits beyond the decimal point.
$U = \begin{pmatrix} 0.6 & 0.8 \\ 0.8 & -0.6 \end{pmatrix}$
U = [0.6, 0.8
0.8, -0.6]
$A = UDU^\textrm{T} = \begin{pmatrix} \frac{1}{25}(9a + 16b) & \frac{12}{25}(a - b) \\ \frac{12}{25}(a - b) & \frac{1}{25}(16a + 9b) \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2$ must be integers and let the eigenvalues be $n_1+25n_2, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2+50n_1, 10n_2$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.28 & 0.96 \\ 0.96 & -0.28 \end{pmatrix}$
U = [0.28, 0.96
0.96, -0.28]
$A = UDU^\textrm{T} = \begin{pmatrix} \frac{1}{625}(49a + 576b) & \frac{168}{625}(a - b) \\ \frac{168}{625}(a - b) & \frac{1}{625}(576a + 49b) \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2$ must be integers and let the eigenvalues be $n_1+625n_2, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2+1250n_1, 10n_2$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2+2500n_1, 100n_2$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2$ must be integers and let the eigenvalues be $n_1+625n_2, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2+1250n_1, 10n_2$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2+2500n_1, 100n_2$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.352 & 0.936 \\ 0.936 & -0.352 \end{pmatrix}$
U = [0.352, 0.936
0.936, -0.352]
$A = UDU^\textrm{T} = \begin{pmatrix} \frac{1}{15625}(1936a + 13689b) & \frac{5148}{15625}(a - b) \\ \frac{5148}{15625}(a - b) & \frac{1}{15625}(13689a + 1936b) \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2$ must be integers and let the eigenvalues be $n_1+15625n_2, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2+31250n_1, 10n_2$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2+62500n_1, 100n_2$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
Three-dimensional unit vectors with at most two digits beyond the decimal point include:
$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0.6 \\ 0.8 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0.28 \\ 0.96 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0.36 \\ 0.48 \\ 0.8 \end{pmatrix}$ and $\begin{pmatrix} 0.48 \\ 0.6 \\ 0.64 \end{pmatrix}$.
There are many matrices composed of columns of these vectors that have inverses that too have only at most two digits beyond the decimal point, and therefore we will only give a small sampling.
For clarity, we will use $a$, $b$ and $c$ for the eigenvalues, and not $\lambda_1$, $\lambda_2$ and $\lambda_3$.
There is one reasonably nice example of a block-diagonal matrix using one of the 2 × 2 matrices from above as the second block.
$U = \begin{pmatrix}1& 0& 0\\ 0& 0.6& 0.28\\ 0& 0.8&-0.96 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& 0& 0\\ 0& 1.2& 0.35\\ 0& 1& -0.75 \end{pmatrix}$
U = [1, 0, 0
0, 0.6, 0.28
0, 0.8, -0.96]
$A = UDU^{-1} = \begin{pmatrix} a & 0 & 0 \\ 0 & \frac{1}{25}(18b - 7c) & \frac{21}{100}(b - c) \\ 0 & \frac{24}{25}(b - c) & \frac{1}{25}(7b + 18c) \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1, 72n_2+n_3, -28n_2+n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2, 100n_1+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
The following one eigenvector matrix is upper triangular.
$U = \begin{pmatrix}1& 0.6& 0.36\\ 0& 0.8& 0.48\\ 0& 0& 0.8 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& -0.75&0\\ 0& 1.25&-0.75\\ 0& 0& 1.25 \end{pmatrix}$
U = [1, 0.6, 0.36
0, 0.8, 0.48
0, 0, 0.8]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{3}{4}(b - a) & \frac{9}{20}(c - b) \\ 0 & b & \frac{3}{5}(c - b) \\ 0 & 0 & c \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $-4n_1-20n_2+n_3, -20n_2+n_3, n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2, -20n_1+10n_2+20n_3, 10n_2+20n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
The following four eigenvector matrices are block upper triangular.
$U = \begin{pmatrix}1& 0& 0.6\\ 0& 0.6& 0.64\\ 0& 0.8&-0.48 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& -0.6& 0.45\\ 0& 0.6& 0.8\\ 0& 1& -0.75 \end{pmatrix}$
U = [1, 0, 0.6
0, 0.6, 0.64
0, 0.8, -0.48]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{3}{5}(c - a) & \frac{9}{20}(a - c) \\ 0 & \frac{1}{25}(9b + 16c) & \frac{12}{25}(b - c) \\ 0 & \frac{12}{25}(b - c) & \frac{1}{25}(16b - 9c) \\ \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1+20n_2-9n_3, n_1+16n_3, n_1-9n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_3+20n_1, 50n_2+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix}1& 0.36&0.48\\ 0& 0.48&0.64\\ 0& 0.8&-0.6 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& -0.75&0\\ 0& 0.75&0.8\\ 0& 1& -0.6 \end{pmatrix}$
U = [1, 0.36, 0.48
0, 0.48, 0.64
0, 0.8, -0.6]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{1}{100}(27b + 48c) - \frac{3}{4}a & \frac{36}{125}(b - c) \\ 0 & \frac{1}{25}(9b + 16c) & \frac{48}{125}(b - c) \\ 0 & \frac{3}{5}(b - c) & \frac{1}{25}(16b + 9c) \\ \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1, n_1+4n_2+80n_3, n_1+4n_2-45n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2, 10n_2+260n_1+20n_3, 10n_2+10n_1+20n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2, 500n_1+100n_3, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix}1& 0.6& 0.6\\ 0& 0.48&0.64\\ 0& 0.64&-0.48 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& -1.05&-0.15\\ 0& 0.75&1\\ 0& 1& -0.75 \end{pmatrix}$
U = [1, 0.6, 0.6
0, 0.48, 0.64
0, 0.64, -0.48]
$A = UDU^{-1} = \begin{pmatrix} a & \frac{1}{20}(9b + 12c) - \frac{21}{20}a & \frac{1}{20}(12b - 9c) - \frac{3}{20}a \\ 0 & \frac{1}{25}(9b + 16c) & \frac{12}{25}(b - c) \\ 0 & \frac{12}{25}(b - c) & \frac{1}{25}(16b + 9c) \\ \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1, n_1-340n_2-400n_3, n_1+260n_2+300n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $-260n_1-300n_2+10n_3, -600n_1-700n_2+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $-1300n_1-300n_2+100n_3, -3000n_1-700n_2+100n_3, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix}0.6& 0.28&0.36\\ 0.8&-0.96&0.48\\ 0& 0& 0.8 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1.2& 0.35&-0.75\\ 1& -0.75&0\\ 0& 0& 1.25 \end{pmatrix}$
U = [0.6, 0.28, 0.36
0.8, -0.96, 0.48
0, 0, 0.8]
$A = UDU^{-1} = \begin{pmatrix} \frac{1}{25}(18b + 7c) & \frac{21}{100}(a - b) & \frac{9}{20}(c - a) \\ \frac{24}{25}(a - b) & \frac{1}{25}(7b + 18b) & \frac{3}{5}(c - a)\\ 0 & 0 & c \\ \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1+100n_2, n_1, n_1+100n_2+20n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $-20n_2+10n_3, -100n_1-20n_2+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
The following eigenvector matrix is zero below the sub-diagonal.
$U = \begin{pmatrix}0.28&0.36&0.48\\ 0.96&-0.48&-0.64\\ 0& -0.8& 0.6 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}1& 0.75&0\\ 0.72&-0.21&-0.8\\ 0.96&-0.28&0.6 \end{pmatrix}$
U = [0.28, 0.36, 0.48
0.96, -0.48, -0.64
0, -0.8, 0.6]
$A = UDU^\textrm{T} = \begin{pmatrix} \frac{7}{25}a + \frac{1}{625}(162b + 288c) & \frac{21}{100}a - \frac{1}{2500}(189b + 256c) & \frac{36}{125}(c - b) \\ \frac{24}{25}a - \frac{1}{625}(216b + 384c) & \frac{18}{25}a + \frac{1}{625}(63b + 112c) & \frac{48}{125}(b - c) \\ \frac{72}{125}(c - b) & \frac{21}{125}(b - c) & \frac{1}{25}(16b + 9c) \end{pmatrix}$ and
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $100n_1-35n_2+n_3, 45n_2+n_3, -80n_2+n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $100n_1+90n_2+10n_3, 250n_2+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2, -1700n_1-1600n_3+100n_2, 800n_1+900n_3+100n_2$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
This is only full eigenvector matrix, and unfortunately, the eigenvalues must be quite large for the resulting matrix $UDU^{-1}$ to have few digits beyond the decimal point, let alone an integer matrix.
$U = \begin{pmatrix}0.36&0.48&0.64\\ 0.48&0.64&-0.48\\ 0.8&-0.6&-0.6 \end{pmatrix}$ $U^{-1} = \begin{pmatrix}0.84&0.12&0.8\\ 0.12&0.91&-0.6\\ 1& -0.75&0 \end{pmatrix}$
U = [0.36, 0.48, 0.64
0.48, 0.64, -0.48
0.8, -0.6, -0.6]
$A = UDU^\textrm{T} = \frac{1}{2500}\begin{pmatrix} 756a + 144b + 1600c & 108a + 1092b - 1200c & 720(a - b) \\ 1008a + 192b - 1200c & 144a + 1456b + 900c & 960(a - b) \\ 1680a - 180b - 1500c & 240a - 1365b + 1125c & 1600a + 900b \end{pmatrix}$ and
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $1413n_1+2988n_2+4400n_3, 288n_1+613n_2+900n_3, -692n_1-1467n_2-2160n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $26730n_1+29880n_2+4400n_3, 5480n_1+6130n_2+900n_3, -13120n_1-14670n_2-2160n_3$, then define $A = \frac{1}{10}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $292500n_1+298800n_2+22000n_3, 60000n_1+61300n_2+4500n_3, -143600n_1-146700n_2-10800n_3$, then define $A = \frac{1}{10^2}UDU^{-1}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
Unfortunately, there is only one eigendecomposition of three-by-three matrices where the eigenvectors have a finite representation with only at most two digits after the decimal point. The eigenvalues must, however, be somewhat large. With the one zero entry, it is a little easier to calculate the characteristic polynomial.
$U = \begin{pmatrix}0.6& 0.48&0.64\\ 0.8&-0.36&-0.48\\ 0& -0.8& 0.6 \end{pmatrix}$
U = [0.6, 0.48, 0.64
0.8, -0.36, -0.48
0, -0.8, 0.6]
$A = UDU^\textrm{T} = \begin{pmatrix} \frac{9}{25}a + \frac{1}{625}(144b + 256c) & \frac{12}{25}a - \frac{1}{625}(108b + 192c) & \frac{48}{125}(c - b) \\ \frac{12}{25}a - \frac{1}{625}(108b + 192c) & \frac{16}{25}a + \frac{1}{625}(81b + 144c) & \frac{36}{125}(b - c) \\ \frac{48}{125}(c - b) & \frac{36}{125}(b - c) & \frac{1}{25}(16b + 9c) \\ \end{pmatrix}$ and
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $n_1+16n_2, n_1+71n_2+80n_3, n_1-54n_2-45n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $140n_1+90n_2+10n_3, 250n_1+250n_2+10n_3, 10n_3$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_2+2500n_3$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
There is one reasonably nice example, where the eigenvalues are 1, -15 and 10, producing a reasonable nice characteristic polynomial and two zeros on the super-anti-diagonal, making the calculation of the characteristic polynomial easier:
$A = \begin{pmatrix} 1 & 0 & 9.6 \\ 0 & 1 & -7.2 \\ 9.6 & -7.2 & -6 \end{pmatrix}$
$(\lambda - 1)((\lambda - 1)(\lambda + 6) - 7.2^2) - 9.6^2(\lambda - 1) = (\lambda - 1)(\lambda^2 + 5\lambda - 6 - 51.84) - 92.16(\lambda - 1)$ which expands to $\lambda^3 + 4\lambda^2 - 155\lambda + 150$. One may observe that $1 + 4 - 155 + 150 = 0$ and $1000 + 400 - 1550 + 150 = 0$, so $1$ and $10$ are eigenvalues, and as the product of the eigenvalues is $-150$, it follows that the last eigenvalue is $-15$.
However, if you are looking at giving students a sort-of nice example, you must resort to using fractional arithmetic. There are two orthogonal matrices you can consider using that may be reasonable in class.
$U = \begin{pmatrix} \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & -\frac{1}{3} & -\frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \end{pmatrix}$
U = [2, 2, 1
2, -1, -2
1, -2, 2]/3
$A = UDU^\textrm{T} = \frac{1}{9} \begin{pmatrix} 4a + 4b + c & 4a - 2b - 2c & 2a - 4b + 2c \\ 4a - 2b - 2c & 4a + b + 4c & 2a + 2b - 4c \\ 2a - 4b + 2c & 2a + 2b - 4c & a + 4b + 4c \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $4n_1-4n_2-3n_3, n_1-4n_2, -2n_1+5n_2+3n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_1, 10n_1+30n_2, 10n_1+60n_2+90n_3$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_1+300n_2, 100n_1+600n_2+900n_3$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} \frac{3}{7} & \frac{6}{7} & \frac{2}{7} \\ \frac{6}{7} & -\frac{2}{7} & -\frac{3}{7} \\ \frac{2}{7} & -\frac{3}{7} & \frac{6}{7} \end{pmatrix}$.
U = [3, 6, 2
6, -2, -3
2, -3, 6]/7
$A = UDU^\textrm{T} = \frac{1}{49} \begin{pmatrix} 9a + 36b + 4c & 18a - 12b - 6c & 6a - 18b + 12c \\ 18a - 12b - 6c & 36a + 4b + 9c & 12a + 6b - 18c \\ 6a - 18b + 12c & 12a + 6b - 18c & 4a + 9b + 36c \end{pmatrix}$
respectively.
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3$ must be integers and let the eigenvalues be $-27n_1-63n_2+28n_3, 8n_1+28n_2-7n_3, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2+350n_3+490n_1, 10n_2, 10n_2+70n_3$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_1+700n_2, 100n_1+3500n_2+4900n_3$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
There are many $4 \times 4$ matrices composed of independent but not mutually orthogonal unit vectors where the inverse also has a nice representation. Rather that try to enumerate these, instead, the example file below simply lists many of such matrices giving the matrix $U$, its inverse $U^{-1}$ and one example where the eigenvalues are $1$, $2$, $3$ and $4$.
You can see some examples here.
There are two categories. The first is based on a pattern, and then the second is an example that does not follow this pattern.
First, given any normalized vector $\begin{pmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{pmatrix}$, you can trivially create an orthogonal matrix, the columns of which, are orthonormal, by creating the matrix
$\begin{pmatrix} \alpha & \beta & \gamma & \delta \\ \beta & -\alpha & -\delta & \gamma \\ \gamma & \delta & -\alpha & -\beta \\ \delta & -\gamma & \beta & -\alpha \end{pmatrix}$
We provide five such examples.
$U = \begin{pmatrix} 0.5 & 0.5 & 0.5 & 0.5 \\ 0.5 & -0.5 & -0.5 & 0.5 \\ 0.5 & 0.5 & -0.5 & -0.5 \\ 0.5 & -0.5 & 0.5 & -0.5 \end{pmatrix}$
[0.5, 0.5, 0.5, 0.5 0.5, -0.5, -0.5, 0.5 0.5, 0.5, -0.5, -0.5 0.5, -0.5, 0.5, -0.5]
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $-n_1+2n_2+2n_4, n_1+2n_3-2n_4, n_1, -n_1+2n_2-2n_3$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_1, 10n_2, 10n_3, 10n_3+10n_1+10n_2+20n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.2 & 0.4 & 0.4 & 0.8 \\ 0.4 & -0.2 & -0.8 & 0.4 \\ 0.4 & 0.8 & -0.2 & -0.4 \\ 0.8 & -0.4 & 0.4 & -0.2 \end{pmatrix}$
[0.2, 0.4, 0.4, 0.8 0.4, -0.2, -0.8, 0.4 0.4, 0.8, -0.2, -0.4 0.8, -0.4, 0.4, -0.2]
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $-79n_1-235n_2+20n_3-315n_4, 21n_1+65n_2-5n_3+85n_4, -24n_1-75n_2-100n_4, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $50n_4+50n_1+50n_2-40n_3, 10n_3, -240n_4-150n_1-200n_2, 10n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.1 & 0.1 & 0.7 & 0.7 \\ 0.1 & -0.1 & -0.7 & 0.7 \\ 0.7 & 0.7 & -0.1 & -0.1 \\ 0.7 & -0.7 & 0.1 & -0.1 \end{pmatrix}$
[0.1, 0.1, 0.7, 0.7 0.1, -0.1, -0.7, 0.7 0.7, 0.7, -0.1, -0.1 0.7, -0.7, 0.1, -0.1]
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $-49n_1-98n_2+50n_3+50n_4, -49n_1-50n_3+50n_4, n_1, n_1+2n_2$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $-490n_4+50n_1+50n_2, -490n_3-50n_1+50n_2, 10n_3, 10n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.1 & 0.5 & 0.5 & 0.7 \\ 0.5 & -0.1 & -0.7 & 0.5 \\ 0.5 & 0.7 & -0.1 & -0.5 \\ 0.7 & -0.5 & 0.5 & -0.1 \end{pmatrix}$
[0.1, 0.5, 0.5, 0.7 0.5, -0.1, -0.7, 0.5 0.5, 0.7, -0.1, -0.5 0.7, -0.5, 0.5, -0.1]
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $-2599n_1-1150n_2-500n_3-3750n_4, 551n_1+250n_2+110n_3+800n_4, -449n_1-200n_2-90n_3-650n_4, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $-25990n_4-1150n_1-500n_2-3750n_3, 5510n_4+250n_1+110n_2+800n_3, -4490n_4-200n_1-90n_2-650n_3, 10n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $-8900n_4-100n_1-700n_2-300n_3, 100n_2, 100n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
$U = \begin{pmatrix} 0.1 & 0.3 & 0.3 & 0.9 \\ 0.3 & -0.1 & -0.9 & 0.3 \\ 0.3 & 0.9 & -0.1 & -0.3 \\ 0.9 & -0.3 & 0.3 & -0.1 \end{pmatrix}$
[0.1, 0.3, 0.3, 0.9 0.3, -0.1, -0.9, 0.3 0.3, 0.9, -0.1, -0.3 0.9, -0.3, 0.3, -0.1]
$UDU^\textrm{T} = \frac{1}{100} \begin{pmatrix} a+ 9b+ 9c+81d & 3a-3b-27c+27d & 3a+27b-3c-27d & 9a- 9b+ 9c-9d \\ 3a- 3b-27c+27d & 9a+ b+81c+ 9d & 9a- 9b+9c- 9d & 27a+ 3b-27c-3d \\ 3a+27b- 3c-27d & 9a-9b+ 9c- 9d & 9a+81b+ c+ 9d & 27a-27b- 3c+3d \\ 9a- 9b+ 9c- 9d & 27a+3b-27c- 3d & 27a-27b-3c+ 3d & 81a+ 9b+ 9c+ d \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $-809n_1-2830n_2+90n_3-3640n_4, 91n_1+320n_2-10n_3+410n_4, -99n_1-350n_2-450n_4, n_1$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $100n_4+50n_1+50n_2-90n_3, 10n_3, -990n_4-350n_1-450n_2, 10n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_2, 100n_3, -2900n_4-100n_1-700n_2-6300n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
There are many nice examples, including one where the eigenvalues are $\pm 2$ and $\pm 3$, producing a reasonable nice characteristic polynomial $(\lambda^2 - 4)(\lambda^2 - 9) = \lambda^4 - 13\lambda^2 + 36$ and with one zero in each row and column, making the calculation of the characteristic polynomial easier:
$A = \begin{pmatrix} 1.2 & 1.5 & -0.9 & 0 \\ 1.5 & -2 & 0 & 1.5 \\ -0.9 & 0 & -1.2 & 1.5 \\ 0 & 1.5 & 1.5 & 2 \end{pmatrix}$
There is one other matrix not of this format that has at most one digit beyond the decimal point:
$U = \begin{pmatrix} 0.5 & 0.5 & 0.1 & 0.7 \\ 0.5 & 0.5 & -0.1 & -0.7 \\ 0.5 & -0.5 & -0.7 & 0.1 \\ 0.5 & -0.5 & 0.7 & -0.1 \end{pmatrix}$
[0.5, 0.7, 0.1, 0.5 0.5, -0.7, -0.1, 0.5 0.5, 0.1, -0.7, -0.5 0.5, -0.1, 0.7, -0.5]
$UDU^\textrm{T} = \frac{1}{100} \begin{pmatrix} 25a+25b+ c+49d & 25a+25b- c-49d & 25a-25b- 7c+7d & 25a-25b+ 7c-7d \\ 25a+25b- c-49d & 25a+25b+ c+49d & 25a-25b+ 7c-7d & 25a-25b- 7c+7d \\ 25a-25b-7c+ 7d & 25a-25b+7c- 7d & 25a+25b+49c+ d & 25a+25b-49c- d \\ 25a-25b+7c- 7d & 25a-25b-7c+ 7d & 25a+25b-49c- d & 25a+25b+49c+ d \end{pmatrix}$
For $A$ to be an integer matrix with integer eigenvalues, $n_1, n_2, n_3, n_4$ must be integers and let the eigenvalues be $n_1, n_1+2n_2, -49n_1-98n_2+50n_3-50n_4, -n_1+2n_3+2n_4$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 1 digit beyond the decimal point, if the diagonal entries of $D$ are $10n_2, 10n_3, 250n_2-490n_3-100n_1+250n_4, 10n_4$, then define $A = \frac{1}{10}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10$.
Otherwise, for $A$ to be an matrix with integer eigenvalues with at most 2 digits beyond the decimal point, if the diagonal entries of $D$ are $100n_1, 100n_2, 100n_3, 100n_4$, then define $A = \frac{1}{10^2}UDU^\textrm{T}$ and the eigenvalues are the diagonal entries divided by $10^2$.
You can see some examples here.
nice singular-value decompositions integer matrices with integer eigenvalues integer matrices with integer singular values integer-normed vectors integer-normed complex vectors geometric sequences with nice norms