Upper triangular matrix/44/Jordan normal form/Example

We consider the matrix

${}M={\begin{pmatrix}3&1&0&4\\0&-1&2&1\\0&0&-1&0\\0&0&0&3\end{pmatrix}}\,$

and want to bring in in Jordan normal for. Here, we have two eigenvalues and, therefore, two two-dimensional generalized eigenspaces, which we treat separately. We have

${}M-3E_{4}={\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}\,,$

therefore, ${}{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}$ belongs to the kernel. The determinant of the upper right submatrix is not ${}0$, so the rank of the matrix is ${}3$, and its kernel is one-dimensional. The second power is

${}{\begin{aligned}{\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}^{2}&={\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}\\&={\begin{pmatrix}0&-4&2&1\\0&16&-16&-4\\0&0&16&0\\0&0&0&0\end{pmatrix}},\end{aligned}}$

a new element of the kernel is ${}{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}$. Thus, we have

${}\operatorname {GeEig} _{3}(M)=\langle {\begin{pmatrix}1\\0\\0\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}\rangle \,.$

Because of

${}{\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}={\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}\,,$

we can use the vectors ${}{\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}$ and ${}{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}$ to establish the first Jordan block.

We have

${}M+1E_{4}={\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}\,,$

therefore, ${}{\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}$ belongs to the kernel. The rank of this matrix is again ${}3$, and the kernel has dimension one. The second power is

${}{\begin{aligned}{\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}^{2}&={\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}{\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}\\&={\begin{pmatrix}16&4&2&33\\0&0&0&4\\0&0&0&0\\0&0&0&16\end{pmatrix}},\end{aligned}}$

a new element in the kernel is ${}{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}$. Thus, we have

${}\operatorname {GeEig} _{-1}(M)=\langle {\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}\rangle \,.$

Because of

${}{\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}={\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}\,,$

we can use the vectors ${}{\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}$ and ${}{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}$ to establish the second Jordan block. Altogether, the linear mapping defined by ${}M$ has, with respect the basis

${\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}\,,{\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}},$

the Jordan normal form

${\begin{pmatrix}3&1&0&0\\0&3&0&0\\0&0&-1&1\\0&0&0&-1\end{pmatrix}}.$