Proof for real Jordan canonical form

Let A $\epsilon$ Mat(nxn, $\mathbb R$) be a matrix that is diagonalizable in $\mathbb C$ with k real eigenvalues of algebraic multiplicity 1 and (n-k)/2 pairs of complex-conjugated eigenvalues of algebraic multiplicity 1.

I need to show that the Jordan canonical form of A in $\mathbb R$ is:

$$\begin{matrix} \lambda_1 & \ldots & \ldots & \ldots & \ldots &0\\ 0 & \ddots & \ldots & \ldots & \ldots & 0\\ \vdots & \ldots & \lambda_k & \ldots & \ldots & 0\\ \vdots & \ldots & \ldots & B_1 & \ldots & 0\\ \vdots & \ldots & \ldots & \ldots & \ddots & 0\\ 0 & \ldots & \ldots & \ldots & 0 & B_{(n-k)/2} \end{matrix}$$

$\epsilon$ Mat(nxn, $\mathbb R$),

and B_j =

$$\begin{matrix} a_j & b_j\\ -b_j & a_j \end{matrix}$$

I already determined for 2x2-matrices that we can write AP=PB, where P is a matrix with vectors of the decomposition of an eigenvector of A. But I don't really know how to prove this general case.

Thanks for help.

• There is no Jordan Canonical Form over $\Bbb R$, and the matrix you wrote down is no JCF if there are any blocks $B_j$. – Marc van Leeuwen Feb 20 '15 at 15:33
• @MarcvanLeeuwen That's why the OP mentions real Jordan form in the title. The term "real Jordan form" refers not to a usual Jordan form that happens to have real entries, but exactly the matrix structure mentioned in the OP. The opening sentence of the second paragraph is confusing, though. – user1551 Apr 24 '15 at 5:14
• Rant. Why do people (even Hoffman and Kunze!) keep calling Jordan form "cannonical"? It's a normal form, not a cannonical form! – user1551 Apr 24 '15 at 5:17
• @user1551: Not so, at least not with "cannon" in place of "canon". More seriously, could you be more explicit on what's the subtle distinction between canonical and normal? (I think indeed Jordan normal form is better because the form is not quite unique, with a choice of ordering the eigenvalues; by contrast rational canonical form is fine with me because it is truly unique. But I'm not sure that is your point as well. In any case these forms are matrices, while in neither case the basis on which it is obtained is canonical/unique at all. But that does not de-canonicalise the form.) – Marc van Leeuwen Apr 24 '15 at 6:58
• @user1551: As for your previous comment, I don't see how "exactly the matrix structure mentioned in the OP" could be what is a real Jordan form, since it is limited to real matrices that are diagonalisable over $\Bbb C$; to call something that is limited to the diagonalisable case a Jordan form seems ridiculous to me (and it rubs me when I see people throw hints "use JNF" at diagonalisable situations). Maybe the "real JNF" is something like this? – Marc van Leeuwen Apr 24 '15 at 7:07

Let $A \in \mathbb{R}^{n \times n}$ as you describe. In particular, we assume the complexified map $T(v)=Av$ for $v \in \mathbb{C}^n$ has $k$-fold distinct real eigenvalues. Of course, $T$ over $\mathbb{C}$ has $n$-complex eigenvalues as $\text{det}(T-tI)=0$ has $n$-zeros since $\mathbb{C}$ is algebraically closed. Continuing, we assume the remaining $(n-k)$ eigenvalues are also distinct and complex and arranged in $(n-k)/2$ conjugate pairs which I will denote $\alpha_j \pm i\beta_j$ for $j=1,2,\dots (n-k)/2$. Observe, over $\mathbb{C}^n$ this complexified map has $n$-distinct eigenvalues hence $T$ is diagonalizable. Let $\gamma = \{ v_1,\dots ,v_k, w_1, \bar{w}_1 \dots , w_{(n-k)/2}, \bar{w}_{(n-k)/2} \}$ be the complex eigenbasis which diagonalizes $T$. In particular, for $j=1,\dots , k$ we simply have: $$T(v_j) = \lambda_j v_j$$ then, as $\bar{A}=A$ we have $Aw=(\alpha+i\beta)w$ conjugated yields $A\bar{w}=(\alpha-i\beta)\bar{w}$. This shows the familiar result that, for real matrices $A$, complex eigenvectors come in conjugate pairs with conjugate eigenvalues. This is a beautiful and useful analog of the elementary pairing of complex roots to real polynomial equations. In particular, if $w=a+ib$ where $a,b \in \mathbb{R}^n$ and $\alpha,\beta \in \mathbb{R}$ such that: \begin{align} T(w)=(\alpha+i\beta)w \ \ &\Rightarrow \ \ T(a+ib)=(\alpha+i\beta)(a+ib) \\ &\Rightarrow \ \ T(a)+iT(b)=\alpha a-\beta b + i(\beta a+\alpha b) \\ &\Rightarrow \ \ T(a)= \alpha a-\beta b \ \ \& \ \ T(b)= \beta a+\alpha b \\ \end{align} Hence, we find an interesting real result. If we define $S$ to be the transformation on $\mathbb{R}^n$ with the same matrix $A$ as the complexified $T$ ( in other words, $S$ is the map with $[S]=A$ which we began this story) then $$S(a)= \alpha a-\beta b \ \ \& \ \ S(b)= \beta a+\alpha b$$ Let $W = \text{span}\{ a,b \}$ then $S|_W=S'$ has the following block structure with respect to basis $\omega = \{a,b \}$: (your text might use $[S']_{\omega }^{\omega}$) $$[S']_{\omega \omega} = \left[[S(a)]_{\omega} \ | \ [S(b)]_{\omega}\right] = [[\alpha a-\beta b]_{\omega} \ | \ [\beta a+\alpha b]_{\omega}] = \left[ \begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array} \right]$$ Therefore, to find the block-decomposition requested by the OP we simply use a basis for $\mathbb{R}^n$ assembled from $k$-eigenvectors $v_1, \dots v_k$ followed by the real and imaginary parts of the complex eigenvectors where we choose to make use of the first copy rather than the conjugate in each conjugate pair. In particular, $$\gamma = \{ v_1,\dots , v_k, a_1,b_1, \dots , a_m,b_m \}$$ where $m = (n-k)/2$ and $A(a_j+ib_j)=(\alpha_j+i\beta_j)(a_j+ib_j)$ for each $j=1, \dots , m$. It follows from the discussion given thus far: $$[\gamma]^{-1}A[\gamma] = \left[ \begin{array}{cccccccc} \lambda_1 & & & & & & & \\ & \ddots & & & & & & \\ & & \lambda_k & & & & & \\ & & & \alpha_1 & \beta_1 & & & \\ & & & -\beta_1 & \alpha_1 & & & \\ & & & & & \ddots & & \\ & & & & & & \alpha_m & \beta_m \\ & & & & & & -\beta_m & \alpha_m \end{array}\right]$$ This is known as the real Jordan form of $A$. In the event that the complexified $A$ is not diagonalized the real Jordan form picks up more terms in the super diagonals from the $1$'s in the superdiagonal of the complex Jordan form of $A$. Philosophically, this discussion is at odds with those mathematicians who refuse to think across fields. There is a pun here I intend. One text which has this is Ralph Abraham's classic text Linear and Multilinear Algebra. Most of my other linear algebra texts have little on this topic as it is just cleaner to stick with either $\mathbb{R}$ or $\mathbb{C}$ etc. But, this concept of complexification is exceedingly important to properly understand applications of complex math to real problems and I think we would do well to spend more time on the exposition of this structure.