Discussion
If characteristic equation of matrix A is same as minimum polynomial which reads as \((x - 1)^{3}(x - 4)\), then the Jordan canonical form of A is :
Answer: C
The Jordan form is \(\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&4\end{pmatrix}\).
The exponent of \((x - \lambda)\) in the minimal polynomial is the size of the largest Jordan block for \(\lambda\), and the exponent in the characteristic polynomial is the total size of all blocks for \(\lambda\). Here both are 3 for \(\lambda = 1\), so there is a single \(3 \times 3\) block for the eigenvalue 1, and a single \(1 \times 1\) block for 4.
The first option has blocks of sizes 2 and 1 for eigenvalue 1, so its minimal polynomial is \((x - 1)^{2}(x - 4)\). The second has eigenvalue 1 only twice and 4 twice, so its characteristic polynomial is wrong. The last puts a 1 above the diagonal between the eigenvalues 1 and 4, which is not a Jordan form at all.
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