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Let $A \in \operatorname{Mat}(n\times n, \mathbb{R})$ 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:

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

$\in \operatorname{Mat}(n\times n, \mathbb{R})$,

and $B_j = $

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

I already determined for $2\times 2$-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.

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  • $\begingroup$ There is no Jordan Canonical Form over $\Bbb R$, and the matrix you wrote down is no JCF if there are any blocks $B_j$. $\endgroup$ Feb 20, 2015 at 15:33
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    $\begingroup$ @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. $\endgroup$
    – user1551
    Apr 24, 2015 at 5:14
  • $\begingroup$ Rant. Why do people (even Hoffman and Kunze!) keep calling Jordan form "cannonical"? It's a normal form, not a cannonical form! $\endgroup$
    – user1551
    Apr 24, 2015 at 5:17
  • $\begingroup$ @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.) $\endgroup$ Apr 24, 2015 at 6:58
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    $\begingroup$ @MarcvanLeeuwen Ha ha ha! It's "canon", of course. I've no idea why I typed double-n in the previous comment. Anyway, the uniqueness is indeed my point, and a "real Jordan form" means this. I didn't read the question carefully and didn't realise that the matrix in the OP is required to be diagonalisable. So, a real JNF is not "exactly" the matrix structure in the OP, but that structure is a special case of a real JNF. $\endgroup$
    – user1551
    Apr 24, 2015 at 7:18

1 Answer 1

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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.

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  • $\begingroup$ Very nice and complete answer! Do you have a simple reference for it? Also, I think $S$ diagonalizable is enough and you don’t need the eigenvalues to be distinct: for real eigenvalues with high geometric multiplicity, note that out of $k$ L.I. complex vectors we can always extract $k$ L.I. real vectors. Also I guess a basis for a non-real eigenspace can have it’s real part L.I. if we multiply each vector by a convenient complex scalar. Is this correct? $\endgroup$
    – user334639
    Apr 28, 2019 at 15:12
  • $\begingroup$ @user334639 indeed the assumption that the eigenvalues and complex eigenvalues be distinct can be replaced by the assumption the matrix is complex diagonalizable. For LI, it is nice to use the real and imaginary parts of a complex eigenvector to produce a pair of real LI vectors. This must happen for a genuinely complex eigenvalue ($\alpha+i \beta$ with $\beta \neq 0$). As I mentioned, Abraham's Linear and Multilinear Algebra at least mentions the real Jordan form. I also have considerable more discussion in my current set of lecture notes: supermath.info/LinearNotes2019.pdf $\endgroup$ Apr 28, 2019 at 21:40
  • $\begingroup$ These notes are great. Are we allowed to use them for teaching? I suggest adding a copyright statement saying what people are allowed do with it. Axler Done Right gives a long proof for the general form of a normal operator. It seems that the proof cannot be shortened because this "split real and imaginary parts of eigenvector" trick (can I call it a trick?) doen't ensure that the eigenvectors can be made orthogonal in general. Right? $\endgroup$
    – user334639
    Apr 29, 2019 at 23:13
  • $\begingroup$ PS: for real eigenvalues with multiplicity, I guess you still need a lemma saying that k complex linearly independent vectors “contain” k real linearly independent vectors after you split real and imaginary parts, and still need some more work if you want to show that the Jordan blocks of the real eigenvalues coincide. $\endgroup$
    – user334639
    Apr 30, 2019 at 1:08
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    $\begingroup$ @AleTolcachier I don't think I have one. I have a physical copy of the text, I think it is from around 1970 or so. I think you can buy it for pretty cheap used. It's a short book. $\endgroup$ Apr 17, 2020 at 3:17

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