We can use the Jordan normal form of a linear operator defined over \(\CC\) to establish what might be called a ‘real Jordan normal form’ for a linear operator, \(T\), on a vector space, \(V\), over \(\RR\). The idea is to use the basis corresponding to the Jordan normal form of \(T\)’s complexification, \(T_\CC:V_\CC\mapto V_\CC\), to identify a distinguished basis for \(V\) in which the matrix representation of \(T\) has a ‘nice’ form.

Recall that the complexification, \(V_\CC\), of \(V\) consists of pairs \((x,y)\) such that \(x,y\in V\) and \((a+ib)(x,y)=(ax-by,ay+bx)\) for \(a,b\in\RR\) and that \(T_\CC\) acts on \(V_\CC\) according to \(T_\CC(x,y)=(Tx,Ty)\). This operator may have both real and complex eigenvalues, with its real eigenvalues corresponding to eigenvalues of \(T\) and complex eigenvalues, the ‘missing’ eigenvalues of \(T\), coming in conjugate pairs. Clearly Jordan bases for the generalised eigenspaces of \emph{real} eigenvalues of \(T_\CC\) can be identified as images, through the natural inclusion, \(v\mapsto(v,0)\) of \(V\) in \(V_\CC\), of the Jordan bases of the generalised eigenspaces in \(V\) of those eigenvalues of \(T\).

So we focus on the conjugate pairs, \((\lambda_i,{\lambda_i}^*)\), of complex eigenvalues of \(T_\CC\). As we’ve seen, the number and size of the Jordan blocks corresponding to an eigenvalue \(\lambda_i\) is determined by the dimensions of the spaces \(\ker(T_\CC-\lambda_i\id_{V_{\CC}})^k\) for \(k\leq\dim V\). But taking complex conjugates, we see that \((T_\CC-\lambda_i\id_{V_{\CC}})^kv=0\) is equivalent to \((T_\CC-{\lambda_i}^*\id_{V_{\CC}})^kv^*=0\). That is, the map \(v\mapsto v^*\) determines a one-to-one correspondence between \(\ker(T_\CC-\lambda_i\id_{V_{\CC}})^k\) and \(\ker(T_\CC-{\lambda_i}^*\id_{V_{\CC}})^k\). In other words, there is a one-to-one correspondence between the Jordan blocks corresponding to conjugate eigenvalues \(\lambda_i\) and \({\lambda_i}^*\) of \(T_\CC\). So if \((x_i,y_i)\), \(i=1\dots n_i\), is a basis for the generalised eigenspace of \(\lambda_i\) in \(V_\CC\) then \((x_i,-y_i)\), \(i=1\dots n_i\), is the basis for the generalised eigenspace of \({\lambda_i}^*\) in \(V_\CC\). But this means that the \(x_i,y_i\in V\) are linearly independent in \(V\). So we may consider the matrix representation of the restriction of \(T\) to \(\Span(x_1,y_1,\dots,x_{n_i},y_{n_i})\) with respect to this basis. It is not difficult to see that starting with the \(n_i\times n_i\) Jordan matrix of the restriction of \(T_\CC\) to the generalised eigenspace of \(\lambda_i\), this is just the \(2n_i\times2n_i\) matrix obtained by replacing
\begin{equation*}
\lambda_i\mapsto\begin{pmatrix}
a_i&b_i\\
-b_i&a_i
\end{pmatrix}
\end{equation*}
where \(\lambda_i=a_i+ib_i\), on the diagonal and
\begin{equation*}
1\mapsto\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}
\end{equation*}
on the superdiagonal.

In summary, we have obtained the following

Theorem (Real Jordan normal form) For any linear operator \(T\in\mathcal{L}(V)\) on a finite dimensional real vector space \(V\), there is a basis for \(V\) such that the matrix representation of \(T\), \(\mathbf{T}\), has the form
\begin{equation*}
\mathbf{T}=\begin{pmatrix}
\mathbf{T}_1& &\mathbf{0}\\
&\ddots& \\
\mathbf{0}& &\mathbf{T}_m
\end{pmatrix}
\end{equation*}
where each \(\mathbf{T}_i\) has the form
\begin{equation*}
\mathbf{T}_i=\begin{pmatrix}
\boldsymbol{\lambda}_i&\mathbf{1}& &\mathbf{0}\\
&\ddots&\ddots& \\
& &\ddots&\mathbf{1}\\
\mathbf{0}& & &\boldsymbol{\lambda}_i
\end{pmatrix},
\end{equation*}
with \(\boldsymbol{\lambda_i}\) and \(\mathbf{1}\) being simply the eigenvalue \(\lambda_i\) and the number \(1\) respectively in the case that \(\lambda_i\) is a (real) eigenvalue of \(T\) and
\begin{equation*}
\boldsymbol{\lambda}_i=\begin{pmatrix}
a_i&b_i\\
-b_i&a_i
\end{pmatrix}
\quad\text{and}\quad\mathbf{1}=\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}
\end{equation*}
for each complex conjugate pair, \((\lambda_i,{\lambda_i}^*)\), \(\lambda_i=a_i+ib_i\), of (complex) eigenvalues of the complexified operator \(T_\CC\).