No eigenvectors belong to non-real zeros of the characteristic equation of a linear map #L:V\rightarrow V# of a real vector space #V# to itself. At such a zero we do however see a two-dimensional subspace. We will show that this invariant subspace can be obtained as the real part of the 2-dimensional complex subspace spanned by a complex eigenvector and the corresponding conjugate in the complex extension of #V#. After that we will discuss a real Jordan normal form where the eigenvalues on the diagonal and the ones outside the diagonal, when we are dealing with real eigenvalues, are replaced by #(2\times2)#-matrices.
Let #A# be a real #(n\times n)#-matrix. Now assume that #\alpha =\basis{\vec{a}_1,\ldots ,\vec{a}_{\ell}}# is a system of linear independent vectors in #\mathbb{C}^n# on which the matrix of #L_A# is a Jordan block with non-real eigenvalue #\mu#. Then \[\beta =\basis{\Re\vec{a}_1,\Im \vec{a}_1,\Re\vec{a}_2,\Im \vec{a}_2,\ldots,\Re\vec{a}_{\ell},\Im\vec{a}_{\ell}}\] is a basis for the span of #\alpha# and #\overline{\alpha}# and then matrix of the restriction of #L_A# to this span relative to #\beta# \[J_{\mu,\frac{n}{2}}=\matrix{B&I_2&0&\cdots &0\\ 0&B&I_2&\cdots& 0\\ \vdots&\ddots&\ddots&\ddots& \vdots\\ 0&\cdots&\ddots&B& I_2\\ 0&\cdots&\cdots&0&B} \phantom{xxx}\text{where } B = \matrix{\Re\mu&\Im\mu\\ -\Im\mu&\Re\mu}\]
To find the matrix of #L_A# relative to #\beta#, we first express the images of #\Re\vec{a}_1,\Im \vec{a}_1# in the same two vectors of #\beta#:
\[\begin{array}{rcl} L_A (\Re\vec{a}_1) &=& A (\Re\vec{a}_1)\\ &&\phantom{xxx}\color{blue}{\text{definition }L_A}\\ &=& \Re(A \,\vec{a}_1)\\ &&\phantom{xxx}\color{blue}{A \text{ is real}}\\ &=& \Re(\mu \vec{a}_1)\\ &&\phantom{xxx}\color{blue}{\vec{a}_1\text{ is an eigenvector of }A\text{ at}\mu}\\ &=&\Re( \mu)\cdot \Re(\vec{a}_1)-\Im(\mu)\cdot\Im(\vec{a}_1)\\ &&\phantom{xxx}\color{blue}{\Re(x\cdot y ) = \Re( x)\cdot \Re(y)-\Im(x)\cdot\Im(y)} \end{array}\]In the same manner, using the equality \(\Im(x\cdot y ) = \Re( \mu)\cdot \Im(y)+\Im(x)\cdot\Re(y)\), we can prove that \[L_A (\Im\vec{a}_1) =\Re( \mu)\cdot \Im(\vec{a}_1)+\Im(\mu)\cdot\Re(\vec{a}_1)\]The matrix of the restriction of #A# to the span #\linspan{\Re\vec{a}_1,\Im \vec{a}_1}# relative to #\basis{\Re\vec{a}_1,\Im \vec{a}_1}# hence is equal to #B#.
Next, we calculate the image of the real and imaginary part of #\vec{a}_j# for #j\gt 1#:
\[\begin{array}{rcl} L_A (\Re\vec{a}_j) &=& \Re(A \,\vec{a}_j)\\ &&\phantom{xxx}\color{blue}{\text{ like above for }j = 1}\\ &=& \Re(\mu \vec{a}_j+\vec{a}_{j-1})\\ &&\phantom{xxx}\color{blue}{\text{the matrix of } A\text{ relative to }\alpha\text{ is a Jordan block }}\\ &=&\Re( \mu)\cdot \Re(\vec{a}_j)-\Im(\mu)\cdot\Im(\vec{a}_j)+\Re(\vec{a}_{j-1})\\ &&\phantom{xxx}\color{blue}{\Re(x\cdot y ) = \Re( x)\cdot \Re(y)-\Im(x)\cdot\Im(y)} \end{array}\]In the same manner, using the equality \(\Im(x\cdot y ) = \Re( \mu)\cdot \Im(y)+\Im(x)\cdot\Re(y)\), we can prove that \[L_A (\Im\vec{a}_j) =\Re( \mu)\cdot \Im(\vec{a}_j)+\Im(\mu)\cdot\Re(\vec{a}_j)+\Re(\vec{a}_{j-1})\]The two columns of the matrix of the restriction of #A# to #\beta# corresponding with #\Re\vec{a}_j# en #\Im \vec{a}_j# thus form the submatrix \[ \matrix{0\\ \vdots\\ 0\\ I_2\\ B\\ 0\\ \vdots \\ 0} \]with #j-2# zero matrices above #I_2#. With this we have proven that the matrix of #L_A# relative to #\beta# looks like as indicated.
Assume #V = \mathbb{R}^2# and #L=L_A:V\to V# for #A=\matrix{0&-1\\ 1&0}# with characteristic polynomial #p_L(x) = x^2+1#. The roots of #p_L(x)# are #\ii# and #-\ii#, and #L# has the diagonal matrix with these two values on the main diagonal relative to the basis of #\alpha = \basis{\rv{\ii,1},\rv{-\ii,1}}# for #\mathbb{C}^2#. According to the theorem #A# is conjugate to each of the two real Jordan blocks \[J_{\ii,1} = \matrix{0&1\\ -1&0}\phantom{xx}\text{ and }\phantom{xx}J_{-\ii,1} = \matrix{0&-1\\ 1&0}\]The second one is equal to #A#. This shows that the Jordan block is not uniquely determined by the involved eigenvalue, but almost: the only freedom is the choice of the order: #\mu,\overline{\mu}# or #\overline{\mu},\mu#. For example, if we require that we only work with #J_{\mu,\ell}# for #\Im\mu\ge0#, then the Jordan block is not uniquely determined.
Instead of #B# we can, for a given non-real eigenvalue #\mu#, also choose the matrix \[\matrix{0&c\\ 1&b}\phantom{ xx } \text{ with }\phantom{xx}b = \mu+\overline{\mu} \phantom{ xx } \text{and }\phantom{xx} c =- \mu\cdot\overline{\mu}\]This #(2\times2)#-matrix is not dependent on the order between #\mu# and #\overline{\mu}#. It is the companion-matrix of the characteristic polynomial #x^2-b\cdot x -c# of #B#.
The real #(2\times2)#-matrix #B# is the matrix relative to the basis #\basis{1,\ii}# for #\mathbb{C}#, seen as real vector space, of the linear map "complex multiplication by #\mu#".
With this we arrive at the end result of the search for a certain most simple/unique form for square matrices within a give conjugation class:
Each real square matrix #A# is conjugate to a matrix which has Jordan blocks #J_{\lambda,k}# along the main diagonal, where #\lambda# is a real or complex root of the characteristic polynomial of #A# with #\Re\lambda\ge0#. Conversely, if the eigen values #\lambda# with #\Re\lambda\ge0# are given and the sizes #k# of all occurring Jordan blocks #J_{\lambda,k}#, then the conjugation class of #A# is uniquely determined.
In particular two of the same real square matrices with the same size are conjugate if and only if they are conjugate as complex matrices.
A square matrix is conjugate to a real matrix if and only if the number of Jordan blocks #j_{\lambda,k}# of size #k# are equal to #j_{\overline\lambda,k}# for each eigenvalue #\lambda# and each natural number #k#.
Above we already discussed how a square matrix can be conjugate to the Jordan normal form: First use the direct sum decomposition in generalized eigenspaces for each complex root #\lambda# of the characteristic polynomial with #\Im\lambda\ge0#. Then apply the Jordan decomposition of the theorem The Jordan decomposition at one eigenvalue. If #\lambda# is non-real, then the theorem above indicates how Jordan blocks of the complex linear map can be used to find the real Jordan blocks.
To prove the unicity, we have to reduce that two conjugate Jordan normal forms are only conjugate if the sizes the numbers #j_{\lambda,i}# of Jordan blocks a eigenvalue #\lambda# of size #i# are equal for all #\lambda# and #i#. In that case the Jordan normal forms are equal, apart from exchanging the Jordan blocks. But an exchange of Jordan blocks comes down to an exchange of the order of the vectors in a basis for the involved invariant subspaces. Hence, the two Jordan normal forms are conjugate by means of a permutation matrix.
The last theorem follows directly from the proof of the theorem above: if the number of Jordan Blocks at eigenvalue #\lambda# for each #i# are equal to the one of #\overline{\lambda}# then the matrix is conjugate to the matrix relative to the basis on which the real Jordan normal form can be found. Conversely, if a matrix is real, then the numbers of complex Jordan blocks at given size #i# must be equal for #\lambda# and #\overline{\lambda}#, for example because complex conjugate transforms the one Jordan block at a non real eigenvalue into another Jordan block at the complex conjugate eigenvalue of the same size.
Earlier, with the theorem Criterium for real conjugacy, we have given a direct proof of the fact that, if two real #(n\times n)#-matrices are conjugate over #\mathbb{C}# they are also conjugate over #\mathbb{R}#.
There is one difficulty when finding the Jordan normal form: factorizing the characteristic polynomial in linear factors. This problem is bypassed with the companion matrix. For this we only need to track down the factors occurring twice in the characteristic polynomial. This is possible with the help of the gcd as we have discussed earlier.
To describe a companion matrix we will take a look at the vector space #P_{n-1}# of polynomials in #x# of a degree smaller than #n#, we take a monic polynomial #f(x)# in #x# of degree #n#, and study a linear map #L_x:P_{n-1}\to P_{n-1}# which adds to #g(x)# the remainder of division of #x\cdot g(x)# by #f(x)#. The companion matrix at #f(x)# is the #(n\times n)#-matrix of the linear map #L_x# relative to the basis #\basis{1,x,\ldots,x^{n-1}}#.
The first #n-1# columns of the companion matrix are the vectors #e_2,\ldots,e_{n}#. In the #i#-th row of the last column we have the opposite of the coefficient of #x^{i-1}# in #f(x)#. If for example #f(x) = x^2+1#, then the companion matrix of #f(x)# is equal to #\matrix{0&-1\\1&0}#.
Consider the matrix \[ A = \matrix{-1 & 0 & 6 & 3 & 3 & 0 & 0 & -3 \\ 10 & 1 & -8 & -5 & -19 & 2 & -7 & -5 \\ -3 & 3 & -4 & 0 & 0 & -3 & 0 & 6 \\ -4 & -10 & 3 & -4 & 8 & 6 & 4 & -6 \\ 2 & -5 & 2 & -1 & -6 & 6 & -3 & -16 \\ 3 & -19 & -10 & -16 & -16 & 16 & -5 & -28 \\ 10 & 20 & -2 & 4 & -2 & -15 & 1 & 36 \\ -5 & -9 & -1 & -4 & 3 & 6 & 1 & -11 \\ } \] The characteristic polynomial of the matrix is equal to #{\left(x^2+2 x+10\right)^4}#. Therefore, the eigenvalues of #A# are #-3\, \complexi-1# and #3\, \complexi-1#. Calculation of the ranks of the powers of #A- (-3\, \complexi-1)\cdot I_8# provides the following information
\[\begin{array}{l|r}
\text{matrix}\phantom{x}&\phantom{x}\text{rank}\\
\hline
{A- (-3\, \complexi-1)\cdot I_8} &6\\
(A- (-3\, \complexi-1)\cdot I_8)^2 & 5\\
(A- (-3\, \complexi-1)\cdot I_8)^3 &4\\
\end{array}\] Denote by #B# the matrix of multiplication by #-3\, \complexi-1# with respect to the basis #\basis{1,\ii}#, so #B = \matrix{-1 & 3 \\ -3 & -1 \\ }#. Which of the matrices below is a Jordan normal form of #A#?
#\text{Jordan normal form for }A=# #\matrix{B & {I_2} & 0 & 0 \\ 0 & B & {I_2} & 0 \\ 0 & 0 & B & 0 \\ 0 & 0 & 0 & B \\ }#
To determine the sizes of the complex Jordan blocks of #A# for eigenvalue #-3\, \complexi-1#, we first calculate the dimensions of #\ker{(A-(-3\, \complexi-1)\cdot I_8)^{\ell}}#. If #\ell\ge 4#, then this dimension is equal to #4# because the kernel coincides with the generalized eigenspace, the dimension of which is equal to #4#, which in turn is the multiplicity of the root #-3\, \complexi-1# of the characteristic polynomial. According to the
Rank-nullity theorem for linear maps the dimension #e_\ell=\dim{\ker{(A-(-3\, \complexi-1)\cdot I_8)^\ell}}# of the kernel #(A-(-3\, \complexi-1)\cdot I_8)^\ell# equals #8# minus the rank of the map. Therefore, the information about the rank given in the question gives the following values for #e_\ell#:
\[\begin{array}{l|c}\ell\phantom{i}&e_\ell\\
\hline
1& 2\\
2&3\\
3&4\\
\hline
\end{array}\] According to theorem
The Jordan normal form for a single eigenvalue, we can determine #j_r#, the number of Jordan blocks of size #r# of #A# for eigenvalue #-3\, \complexi-1#, as follows as a function of #e_{\ell}# (not all steps are needed if the dimension #4# of the generalized eigenspace is reached after a smaller number of steps):
\[\begin{array}{rclclcl}
j_4 &=&
e_4-e_3&=&4-4 &=&0\\
j_3 &=&
2e_3-e_4-e_2&=&2\cdot 4-4-3&=&1\\
j_2 &=&2e_2-e_3-e_1&=&2\cdot 3-4-2 &=&0\\
j_1 &=&
2e_1-e_2-e_0&=&2\cdot 2-3-0&=&1\\
\end{array}\] A complex Jordan block corresponds to the real Jordan block of double the size. Consequently, #j_r# is the number of real Jordan blocks of #A# of size #2r#. This means that #A# has a Jordan normal form equal to \[\matrix{B & {I_2} & 0 & 0 \\ 0 & B & {I_2} & 0 \\ 0 & 0 & B & 0 \\ 0 & 0 & 0 & B \\ }\]
A basis with respect to which #A# assumes the indicated Jordan normal form, is made up of the columns of the matrix \[T = \matrix{0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ -1 & 1 & 2 & 2 & -1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & -1 \\ 1 & -1 & -2 & -1 & 0 & -2 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & -1 & 0 & 1 & 1 \\ -1 & -1 & 1 & 1 & 1 & -2 & -1 & -1 \\ 1 & 0 & -1 & -1 & 0 & 0 & 0 & 0 \\ }\] Because \[T^{-1} = \matrix{0 & 2 & 2 & 2 & 2 & -1 & 1 & 3 \\ 1 & -3 & -3 & -4 & -4 & 3 & -1 & -3 \\ -1 & 1 & 2 & 2 & 3 & -1 & 1 & 1 \\ 1 & 1 & 0 & 0 & -1 & 0 & 0 & 1 \\ 1 & 0 & -1 & -1 & -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 & -1 & 0 & 1 \\ 0 & 1 & -1 & 0 & 0 & -1 & 0 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 & 0 & -1 \\ }\] conjugation of #A# to the Jordan form can be realized by #T# in the sense that
\[ T^{-1} A T = \matrix{-1 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\ -3 & -1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 3 & 1 & 0 & 0 & 0 \\ 0 & 0 & -3 & -1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & -3 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & -3 & -1 \\ }=\matrix{B & {I_2} & 0 & 0 \\ 0 & B & {I_2} & 0 \\ 0 & 0 & B & 0 \\ 0 & 0 & 0 & B \\ }\]