Before we will deal with normal forms for real matrices having a characteristic polynomial with non-real roots, we will discuss the extension of a real vector space towards a complex one and the question how the original real vector space can be found in the complex one.
Let #V# be a real vector space. Take a look at the set
\[V_{\mathbb{C} }= V + \ii V\]
consisting of all vectors of the form #\vec{x}+\ii\vec{y}#, where #\vec{x}# and #\vec{y}# are vectors of #V#, and the addition is formal (hence, can also be seen as the pair #\rv{\vec{x},\vec{y}}#). This is a complex vector space if it is provided with the following operations:
\[\begin{array}{rclcl}\text{scalar multiplication}&:&(a+b\ii)(\vec{x}+\ii\vec{y}) &=&(a\cdot\vec{x}-b\cdot\vec{y})+\ii(b\cdot\vec{x}+a\cdot\vec{y})\\ \text{vector addition}&:&(\vec{x}+\ii\vec{y}) +(\vec{u}+\ii\vec{v})&=&(\vec{x}+\vec{u})+\ii(\vec{y}+\vec{v})\end{array}\] We will call this the extension of #V# to a complex vector space.
The map #\sigma : V_{\mathbb{C} }\to V_{\mathbb{C} }# defined by #\sigma(\vec{x}+\ii\vec{y}) = \vec{x}-\ii\vec{y}# for \(\vec{x},\vec{y}\) in #V# is called complex conjugate. This map is semi-linear, which means that, for all complex numbers #\lambda# and #\mu# and all vectors #\vec{u}# and #\vec{w}# of #V_{\mathbb{C} }# we have
\[\sigma(\lambda\cdot\vec{u}+\mu\cdot\vec{w}) = \overline{\lambda}\cdot\sigma(\vec{u}) +\overline{\mu}\cdot\sigma(\vec{w})\] where #\overline{\lambda}# indicates the complex conjugate of #\lambda#.
If #\alpha# is a basis for #V#, then #\alpha# is also a basis for #V_{\mathbb{C} }#. In particular #V# and #V_{\mathbb{C} }# have the same dimension.
If #A# is a real #(n\times n)#-matrix, then #L_A# can be interpreted as a linear map #V\to V#, where #V=\mathbb{R}^n# or #V = \mathbb{C}^n#. In the case #V=\mathbb{R}^n# the map #\left(L_A\right)_{\mathbb C}# is equal to the map #L_A: \mathbb{C}^n\to \mathbb{C}^n# of the second case. There is little confusion between the interpretation of #L_A# as a linear map with a real vector space as domain and as a linear map on a complex vector space.
The fact that #\sigma# is semi-linear follows from the following two calculations, where we will write #\Re(\vec{x}+\ii\vec{y})=\vec{x}# and #\Im(\vec{x}+\ii\vec{y}) =\vec{y}# for \(\vec{x},\vec{y}\) in #V#:
\[\begin{array}{rcl} \sigma(\vec{u}+\vec{w}) &=& \sigma( \Re(\vec{u})+\ii \Im(\vec{u})+\Re(\vec{w})+\ii\Im(\vec{w}) )\\&=& \sigma( \Re(\vec{u})+\Re(\vec{w})+\ii(\Im(\vec{u})+\Im(\vec{w}) ))\\&=&\Re(\vec{u})+\Re(\vec{w})-\ii(\Im(\vec{u})+\Im(\vec{w}) )\\&=&\Re(\vec{u})-\ii\Im(\vec{u})+\Re(\vec{w})-\ii\Im(\vec{w}) \\ &=& \sigma(\vec{u}) +\sigma(\vec{w})\\ \\ \sigma(\lambda\cdot\vec{u}) &=& \sigma( (\Re\lambda+\ii\Im\lambda)(\Re(\vec{u})+\ii \Im(\vec{u})) )\\&=& \sigma( \Re\lambda\cdot\Re(\vec{u}) -\Im\lambda\cdot\Im(\vec{u})+\ii (\Im\lambda\cdot\Re(\vec{u}) +\Re\lambda\cdot \Im(\vec{u})))\\ &=&\Re\lambda\cdot\Re(\vec{u}) -\Im\lambda\cdot\Im(\vec{u})-\ii (\Im\lambda\cdot\Re(\vec{u}) +\Re\lambda\cdot \Im(\vec{u}))\\&=&(\Re\lambda-\ii\Im\lambda)\cdot(\Re(\vec{u}) -\ii \Im(\vec{u}))\\ &=& \overline{\lambda}\cdot \sigma(\vec{u}) \end{array}\]
With the term complex conjugate it's possible to get confused: complex conjugation of the elements of the matrix \[A = \matrix{2+3\ii&0\\ 0&2-3\ii}\] gives \[\sigma(A) = \matrix{2-3\ii&0\\ 0&2+3\ii}\] while conjugation over #\mathbb{C}# of #A# results in a matrix of the form #T\,A\,T^{-1}# where #T# is an invertible #(2\times2)#-matrix. The second operation is explicitly not named 'complex conjugate'. In the special case where #T# is a permutation matrix corresponding to the permutation #[1,2]# the two definitions are the same.
To view a complex vector space as a real space, no expansion is needed:
A complex vector space #W# can be seen as a real vector space by limitation of scalars. This means we will only view the real numbers of #W# as scalars. We will write #\left.W\right|_{\mathbb{R}}# to indicate this real vector space.
Let #V# be a real vector space. After limitation of scalars for #W = V_{\mathbb{C} }# the semi-linear map complex conjugate #\sigma:W\to W# becomes a linear map #\left.W\right|_{\mathbb{R}}\to\left.W\right|_{\mathbb{R}}# with the property that #V# is the eigenspace of #\sigma# at eigenvalue #1# and #\ii V# the eigenspace of #\sigma# at eigenvalue #-1#.
If #\alpha# is a basis for #W#, then #\alpha\cup\ii\alpha# is a basis for #\left.W\right|_{\mathbb{R}}#. In particular, the dimension of #\left.W\right|_{\mathbb{R}}# is twice as large as the dimension of #W#.
If we start with a real vector space #V#, extend the scalars to obtain \(V_{\mathbb{C} }\) and next restrict the scalars, we get the real vector space \(U = \left.V_{\mathbb{C} }\right|_{\mathbb{R} }\) of which #V# is indeed a subspace, but of dimension twice the dimension of #V#. The question how to recover #V# from #U# can be answered in multiple ways. The statement about #\sigma# gives one method: the minimal polynomial of #\sigma # is equal to #x^2-1#, so
\[U = \ker{C-I_U}\oplus\ker{C+I_U},\phantom{xxx}\text{ where }\phantom{xxx} V =\ker{C-I_U} \]
The other term of the direct sum is #\ker{C+I_U} = \ii V#.
If #L:V\to V#, then #L# determines a unique linear map #L_{\mathbb{C}}:V_{\mathbb{C} }\to V_{\mathbb{C} }#. This map commutes with #C#, so that #V = \ker{C-I_U}# is invariant under #L_{\mathbb{C}}#. Furthermore, the limitation of #L_{\mathbb{C}}# to #V# is equal to #L#. Since the characteristic polynomial of #L_{\mathbb{C}}# is a product of linear factors, we can find a basis #\alpha# for #V_{\mathbb{C} }#, in such a way that the matrix #\left(L_{\mathbb{C}}\right)_\alpha# has a Jordan form. With help of the real and imaginary parts of this basis, we will find a basis #\beta# of #V# so that #L_\beta = \left(\left.L_{\mathbb{C}}\right|_V\right)_{\beta}# has a form closely resembling the Jordan normal form.
Take a look at the diagonal matrix \[A = \matrix{\lambda&0\\ 0&\overline{\lambda}}\] where #\lambda# is a non-real complex number and #\overline\lambda# the complex conjugate of #\lambda #. This matrix is conjugate with the real matrix \[B = \matrix{\Re\lambda & -\Im\lambda\\ \Im\lambda & \Re\lambda}\] This follow directly from the fact that the characteristic polynomials of #A# and #B# are both equal to #x^2-2\Re\lambda \cdot x+\lambda\cdot\overline\lambda#, because two #(2\times2)#-matrices with two corresponding (but distinct) eigenvalues are always conjugate.
If we start with the matrix #B#, we will find the diagonal form #A# after determining the complex eigenvalues.
Vice versa, we can, starting with #A# as above, find a real matrix conjugate to #A# (such as #B#) in the following manner: we search for a 2-dimensional real subspace #V# of #\left.{\mathbb{C}^2}\right|_{\mathbb{R}}# which is kept invariant by #A#. This boils down to determining an invertible #(2\times2)#-matrix #T=(t_{ij})# such that #V# is spanned by the columns vectors #T(\vec{e}_1)# and #T(\vec{e}_2)#. The matrix of #L_A# relative to the basis #\basis{\matrix{t_{11}\\ t_{21}},\matrix{t_{12}\\ t_{22}}}# for #V# must have real elements. This matrix is #T^{-1}AT#, hence, the complex conjugate #\sigma(T^{-1}AT)# of this matrix must be equal to #T^{-1}AT#. Furthermore, the complex conjugate #\sigma(A)# of #A# is conjugated to #A# through #P=\matrix{0&1\\ 1&0}#. We will use this to find a solution for #T#:
\[\begin{array}{rcl}\sigma(T^{-1}AT)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\text{this matrix must have real elements }}\\ \sigma(T^{-1})\sigma(A)\sigma(T)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\text{complex conjugate of sum and product is }}\\&&\phantom{xxx}\color{blue}{\text{sum and product of complex conjugate}}\\ \sigma(T^{-1})PAP^{-1}\sigma(T)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\sigma(A) = PA P^{-1}}\\ \left(T\sigma(T^{-1})P\right)A&=&A\left(T\sigma(T^{-1})P\right)\\&&\phantom{xxx}\color{blue}{\text{ on the left-hand side multiplied by }T\text{ }}\\ &&\phantom{xxx}\color{blue}{\text{ on the right-hand side multiplied by }\sigma(T)^{-1}P=\sigma(T^{-1})P}\\ T\sigma(T^{-1})P&=&I_2\\&&\phantom{xxx}\color{blue}{\text{this is sufficient for a solution (not essential)}}\\ T &=&P\sigma(T)\\&&\phantom{xxx}\color{blue}{\text{ on the right-hand side multiplied by }P^{-1}\sigma(T)\text{ ; }P^{-1} = P}\\\matrix{t_{11}&t_{12}\\ t_{21}& t_{22}} &=&\matrix{\overline{t_{21}}&\overline{t_{22}}\\ \overline{t_{11}}& \overline{t_{12}}}\\&&\phantom{xxx}\color{blue}{T=(t_{ij})\text{ substituted and worked out matrix multiplication }}\\T &=& \matrix{t_{11}&{t_{12}}\\\overline{t_{11}}& \overline{t_{12}}}\\&&\phantom{xxx}\color{blue}{\text{ solution for }t_{21},\ t_{22}\text{ substituted }}\\ T &=& \matrix{1&\ii\\ 1&-\ii}\\&&\phantom{xxx}\color{blue}{\text{ solution }t_{11}=1,\ t_{12}=\ii\text{ chosen such that }T^{-1}\text{ exists}}\\T^{-1}A\,T &=&\dfrac{1}{2\ii} \matrix{\ii&\ii\\ 1&-1}\matrix{\lambda&0\\ 0&\overline{\lambda}} \matrix{1&\ii\\ 1&-\ii}\\ &=&\dfrac{1}{2}\matrix{\lambda+\overline{\lambda}&\ii({\lambda}-\overline\lambda)\\ \ii(\overline{\lambda}-\lambda)&\lambda+\overline{\lambda}} \\&=&\matrix{\Re\lambda&-\Im\lambda\\\Im\lambda&\Re\lambda}\\ &=& B\end{array}\]With this we have found the conjugator #T# that conjugates #A# to #B#.
As we will see later, it is possible to find a unique matrix in every conjugation class of real #(n\times n)#-matrices, starting with the complex Jordan normal form. However, with the theorem below, it is easy to find out when two real #(n\times n)#-matrices are conjugate, which is the case when they are conjugate as complex matrices.
If two real #(n\times n)#-matrices are conjugate over #\mathbb{C}#, then they are also conjugate over #\mathbb{R}#.
Assume that two #(n\times n)#-matrices #A# and #B# are conjugate over #\mathbb{C}#. Then there exists a complex invertible matrix #T# so that \(B = T^{-1}\,A\,T\). After multiplying from the left by #T# on both sides we have \[A\,T =T\,B\]Write #T = K +\ii \cdot L#, where #K=\Re T# and #L = \Im T# are the real and imaginary part of #T#. We then have \[A\,K = K\,B\phantom{xxx}\text{ and }\phantom{xxx}A\,L = L\,B\]and thus
\[A\,(K+x\cdot L) =(K+x\cdot L) \,B\phantom{xxx}\text{ for each number }\phantom{xx}x\]
Hence, it is sufficient to derive the existence of a real number #x# where #K+x\cdot L# is invertible. Therefore, consider the determinant #\det(K+x\cdot L)# as a function of #x#. This is a polynomial function of degree #n# in #x# with real coefficients because #K# and #L# are real. Since \(K+\ii\cdot L=T\) is invertible, #\det(K+\ii\cdot L)\ne0# so the polynomial function is not the zero function. The polynomial function has no more than #n# zeros, so that there must exist a real value of #x# for which #\det(K+x\cdot L)\ne0#, so that #K+x\cdot L# is a real invertible matrix. This proves the fact.
In the complex case we already know that two #(n\times n)#-matrices are conjugate if and only if they have the same Jordan normal form. The theorem tells us that this is also sufficient to determine if two real #(n\times n)#-matrices are conjugate. The proof of the theorem shows us how, if this is the case, a real conjugating matrix #S# can be found, starting with a complex conjugating matrix #T#. Finding the Jordan form consists of finding a suitable basis for #V#; the complex conjugating matrix #T# consists of the columns belonging to that basis.
In that way we can determine for each pair of #(n\times n)#-matrices if they are conjugate, and even indicate a conjugating matrix. The real Jordan form of such a matrix, which we will treat later on, is (apart from a permutation of the Jordan blocks) a unique matrix from the conjugation class of #A#.
Consider the #(2\times2)#-matrix \[A = \matrix{0 & -5 \\ 1 & 2 \\ }\] The minimal polynomial of #A# is equal to the characteristic polynomial: \[m_A(x) = p_A(x) =x^2-2 x+ 5\] The roots of this polynomial are #1-2 \complexi# and #2 \complexi+1#. The corresponding eigenvectors are #\matrix{5 \\ 2 \complexi-1 \\ }# and #\matrix{5 \\ -2 \complexi-1 \\ }#, respectively. The matrix #A# is conjugate to the complex diagonal matrix \[ D = \matrix{1-2 \complexi &0\\ 0& 2 \complexi+1}\] The corresponding coordinate transformation is \[T = \matrix{5 & 5 \\ 2 \complexi-1 & -2 \complexi-1 \\ }\] so \(T^{-1}\, A\,T=D\).
The matrices #T# and #D# are not real. If we regard #L_A#, #L_D# and #L_T# as linear transformations of #\mathbb{C}^2#, then the eigenvectors (the columns of #T#) each span a complex invariant subspace of #\mathbb{C}^2# of dimension #1#. By taking the real and imaginary part of the first eigenvector #\matrix{5 \\ 2 \complexi-1 \\ }#, we find a basis for the original real vector space #\mathbb{R}^2# relative to which #L_A# has a real matrix. Determine that matrix.
#\matrix{1 & -2 \\ 2 & 1 \\ }#
By restricting scalars, we can view the linear map #L_A:\mathbb{C}^2\to \mathbb{C}^2# as a linear map #\left.\left(L_D\right)\right|_{\mathbb{R}}:U\to U#, where #U =\left.\mathbb{C}^2\right|_{\mathbb{R}}#.
As a basis for #V# we choose the real and imaginary part of the eigenvector #\matrix{5 \\ 2 \complexi-1 \\ }# with eigenvalue #1-2 \complexi#:
\[\Re \matrix{5 \\ 2 \complexi-1 \\ } = \matrix{5 \\ -1 \\ } \phantom{xxx}\text{ and }\phantom{xxx} \Im \matrix{5 \\ 2 \complexi-1 \\ } =\matrix{0 \\ 2 \\ }\] The corresponding transition matrix is #S=\matrix{5 & 0 \\ -1 & 2 \\ }#. The matrix of #A# relative to said basis is
\[\begin{array}{rcl} {S}^{-1}\,A\,S &=& {\matrix{{{1}\over{5}} & 0 \\ {{1}\over{10}} & {{1}\over{2}} \\ }}\, \matrix{0 & -5 \\ 1 & 2 \\ }\, \matrix{5 & 0 \\ -1 & 2 \\ }\\
&=& \matrix{1 & -2 \\ 2 & 1 \\ }\end{array}\]
If we work with the eigenvector of #2 \complexi+1# rather than the eigenvector of #1-2 \complexi#, we get the transition matrix #\overline{S} = \matrix{5 & 0 \\ -1 & -2 \\ }#. The matrix of #A# with respect to the corresponding basis is
\[\begin{array}{rcl} {\overline{S}}^{-1}\,A\,\overline{S} &=& {\matrix{{{1}\over{5}} & 0 \\ -{{1}\over{10}} & -{{1}\over{2}} \\ }}\, \matrix{0 & -5 \\ 1 & 2 \\ }\, \matrix{5 & 0 \\ -1 & -2 \\ }\\
&=& \matrix{1 & 2 \\ -2 & 1 \\ }\end{array}\] As the notation of the overline above #S# suggests, this basis is the complex conjugate of the columns of #S#.
The matrices #{S}^{-1}\,A\,S# and #{\overline{S}}^{-1}\,A\,\overline{S}# are conjugate by means of #\matrix{0&1\\ 1&0}#:
\[ \matrix{1 & 2 \\ -2 & 1 \\ } =\matrix{0&1\\ 1&0} \matrix{1 & -2 \\ 2 & 1 \\ } \matrix{0&1\\ 1&0} = \matrix{0&1\\ 1&0}^{-1} \matrix{1 & -2 \\ 2 & 1 \\ } \matrix{0&1\\ 1&0}\] The matrices #S^{-1} A S# and #{\overline{S}}^{-1}\,A\,\overline{S}# are the real #(2\times2)#-matrices with respect to the basis #\basis{1,\ii}# that belong to complex multiplication by #1-2 \complexi# and #2 \complexi+1#, respectively.
From real to complex vector spaces and back
