### Invariant subspaces of linear maps: Invariant subspaces

### Real Jordan normal form for non-real eigenvalues

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.

Real Jordan blockLet #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# \[J_{\mu,\frac{n}{2}}=\matrix{B&I_2&0&\cdots &0\\ 0&B&I_2&\cdots& 0\\ \vdots&\ddots&\ddots&\ddots& \vdots\\ 0&\cdots&\ddots&B& I_2\\ 0&\cdots&\cdots&0&B} \phantom{xxx}\text{where } B = \matrix{\Re\mu&\Im\mu\\ -\Im\mu&\Re\mu}\]

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:

Real JordanformEach 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#.

\[\begin{array}{l|r}

\text{matrix}\phantom{x}&\phantom{x}\text{rank}\\

\hline

{A- (-\complexi-1)\cdot I_8} &6\\

(A- (-\complexi-1)\cdot I_8)^2 & 4\\

(A- (-\complexi-1)\cdot I_8)^3 &4\\

\end{array}\] Denote by #B# the matrix of multiplication by #-\complexi-1# with respect to the basis #\basis{1,\ii}#, so #B = \matrix{-1 & 1 \\ -1 & -1 \\ }#. Which of the matrices below is a Jordan normal form of #A#?

To determine the sizes of the complex Jordan blocks of #A# for eigenvalue #-\complexi-1#, we first calculate the dimensions of #\ker{(A-(-\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 #-\complexi-1# of the characteristic polynomial. According to the

*Rank-nullity theorem for linear maps*the dimension #e_\ell=\dim{\ker{(A-(-\complexi-1)\cdot I_8)^\ell}}# of the kernel #(A-(-\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&4\\

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 #-\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 4-4-4&=&0\\

j_2 &=&2e_2-e_3-e_1&=&2\cdot 4-4-2 &=&2\\

j_1 &=&

2e_1-e_2-e_0&=&2\cdot 2-4-0&=&0\\

\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 \[\matrix{B & {I_2} & 0 & 0 \\ 0 & B & 0 & 0 \\ 0 & 0 & B & {I_2} \\ 0 & 0 & 0 & B \\ }\]

\[ T^{-1} A T = \matrix{-1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 \\ }=\matrix{B & {I_2} & 0 & 0 \\ 0 & B & 0 & 0 \\ 0 & 0 & B & {I_2} \\ 0 & 0 & 0 & B \\ }\]

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