Real canonical form

hbghlyj

Real canonical form, Jason Murphy
Suppose $M$ is a real $n \times n$ matrix with real eigenvalues $\left\{\mu_1, \ldots, \mu_a\right\}$ and complex eigenvalues $\left\{\lambda_1, \overline{\lambda_1}, \ldots, \lambda_b, \overline{\lambda_b}\right\}$. Let $\mathcal{B}$ be a basis of generalized eigenvectors of $M$ that puts $M$ into Jordan canonical form. In particular, we can write
\[
\mathcal{B}=\left\{v_j^{\ell}: \begin{array}{c}
j=1, \ldots, a \\
\ell=1, \ldots \ell_j
\end{array}\right\} \cup\left\{w_j^k: \begin{array}{c}
j=1, \ldots, b \\
k=1, \ldots k_j
\end{array}\right\} \cup\left\{\overline{w_j^k}: \begin{array}{l}
j=1, \ldots, b \\
k=1, \ldots k_j
\end{array}\right\},
\]
where the $v_j^{\ell}$ are real, $\ell_j$ is the dimension of the generalized eigenspace for $\mu_j$, and $k_j$ is the dimension of the generalized eigenspace for $\lambda_j$ (so that $\sum_{j=1}^a \ell_j+2 \sum_{j=1}^b k_j=n$ ).
In particular, we have
(1) $\quad M w_j^1=\lambda_j w_j^1, \quad M w_j^{k+1}=\lambda_j w_j^{k+1}+w_j^k \quad$ for $\quad k=1, \ldots, k_j-1$.
Arranging the elements of $\mathcal{B}$ into the columns of a matrix $S$ (in the order suggested above), we can write $M=S J S^{-1}$, where $J$ is the usual block diagonal Jordan form of $M$. We wish to find a real-valued substitute for the standard Jordan normal form.
Let $y_j^k=\operatorname{Re} w_j^k$ and $z_j^k=\operatorname{Im} w_j^k$ for $j=1, \ldots, b$ and $k=1, \ldots, k_j$. Note that
\[
\tilde{\mathcal{B}}=\left\{v_j^{\ell}: \begin{array}{c}
j=1, \ldots, a \\
\ell=1, \ldots \ell_j
\end{array}\right\} \cup\left\{\left(y_j^k, z_j^k\right): \begin{array}{l}
j=1, \ldots, b \\
k=1, \ldots k_j
\end{array}\right\}
\]
forms a basis for $\mathbb{R}^n$. We now arrange the elements of $\tilde{\mathcal{B}}$ into the columns of a matrix $T$ (in the order suggested above) and define $\tilde{J}$ via $M=T \tilde{J} T^{-1}$. Starting from the top left corner of $\tilde{J}$, there will be $a$ Jordan blocks corresponding to the real eigenvalues identical to those appearing in $J$.
We now claim that in fact $\tilde{J}$ is also block diagonal, with $b$ more blocks of a particular form. To see this, we will use (1) to see what relations the $\left(y_j^k, z_j^k\right)$ satisfy. Forgetting the subscripts and superscripts for a moment, we are examining relations of the form
\[
M w=\lambda w+w_0,
\]
with $w_0$ possibly equal to zero. Writing $y=\operatorname{Re} w, z=\operatorname{Im} w, y_0=\operatorname{Re} w_0, z_0=\operatorname{Im} w_0$, we deduce
\[
M y=(\operatorname{Re} \lambda) y-(\operatorname{Im} \lambda) z+y_0, \quad M z=(\operatorname{Im} \lambda) y+(\operatorname{Re} \lambda) z+z_0 .
\]
Using this, one can see that the remaining portion of $\tilde{J}$ will indeed be block diagonal. There will be one block $B_j$ corresponding to each pair $\left(\lambda_j, \overline{\lambda_j}\right)$ of size $2 k_j \times 2 k_j$ of the following form: defining
\[
\Lambda_j:=\left(\begin{array}{cc}
\operatorname{Re} \lambda_j & \operatorname{Im} \lambda_j \\
-\operatorname{Im} \lambda_j & \operatorname{Re} \lambda_j
\end{array}\right), \quad I_2=\left(\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right),
\]
the block $B_j$ will look just like a standard Jordan block with $k_j$ entries along the diagonal, except with each $\lambda_j$ replaced by $\Lambda_j$ and each 1 replaced by $I_2$ (so that the block is of size $2 k_j \times 2 k_j$ ).

hbghlyj

Converting Jordan Normal Form into Real Jordan Form
Note that the eigenvectors are "conjugates" of each other themselves (in the entry-wise sense). This is no accident. In general, we have that
$$Av = \lambda v \implies \overline{Av} = \overline{\lambda} \overline{v} \implies A \overline{v} = \overline{\lambda} \overline{v},$$
which is to say, if $v$ is an eigenvector corresponding to $\lambda$, then $\overline{v}$ is an eigenvector corresponding to $\overline{\lambda}$. This says something very trivial if $v$ and $\lambda$ are real, but something useful when they are complex.

hbghlyj

the block $B_j$ will look just like a standard Jordan block with $k_j$ entries along the diagonal, except with each $\lambda_j$ replaced by $\Lambda_j$ and each 1 replaced by $I_2$ (so that the block is of size $2 k_j \times 2 k_j$ ).

这里有一个$B_j$的示意图:
An operator $T$ on a generalized eigenspace $N(T-λI)$ can be represented by a matrix of the form$$\left[\begin{array}{ccccc}\lambda & 0 & \cdots & \cdots & 0 \\ 1 & \ddots & \ddots & & \vdots \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & 1 & \lambda\end{array}\right]\tag1$$If $\lambda=a+b i \in \mathbb{C} \backslash \mathbb{R}$ is an eigenvalue of an operator $T \in \mathcal{L}\left(\mathbb{R}^n, \mathbb{R}^n\right)$, and $\mathcal{Z}(x, T-\lambda I)$ is one of the cyclic subspaces whose direct sum is $N(T-\lambda I)$, then $\mathcal{Z}(\bar{x}, T-\bar{\lambda} I)$ can be taken to be one of the cyclic subspaces whose direct sum is $N(T-\bar{\lambda} I)$. If we set $k=\operatorname{nil}(x, T-\lambda I)-1$ and $y_j=\operatorname{Re}\left((T-\lambda I)^j x\right)$ and $z_j=\operatorname{Im}\left((T-\lambda I)^j x\right)$ for $j=0, \ldots, k$, then we have $T y_j=a y_j-b z_j+y_{j+1}$ and $T z_j=b y_j+a z_j+z_{j+1}$ for $j=0, \ldots, k-1$, and $T y_k=a y_k-b z_k$ and $T z_k=b y_k+a z_k$. The $2 k+2$ real vectors $\left\{z_0, y_0, \ldots, z_k, y_k\right\}$ span $Z(x, T-\lambda I) \oplus Z(\bar{x}, T-\bar{\lambda} I)$ over $\mathbb{C}$ and also span a $(2 k+2)$-dimensional space over $\mathbb{R}$ that is invariant under $T$. On this real vector space, the action of $T$ can be represented by the matrix
$$\left[\begin{array}{cc|cc|cc|cc|cc}a & -b & 0 & 0 & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\ b & a & 0 & 0 & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\ \hline 1 & 0 & \ddots & \ddots & \ddots & \ddots & & & \vdots & \vdots \\ 0 & 1 & \ddots & \ddots & \ddots & \ddots & & & \vdots & \vdots \\ \hline 0 & 0 & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots & \vdots \\ \hline \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 0 & 0 \\ \hline 0 & 0 & \cdots & \cdots & 0 & 0 & 1 & 0 & a & -b \\ 0 & 0 & \cdots & \cdots & 0 & 0 & 0 & 1 & b & a\end{array}\right]$$
上面的段落和复杂array的代码都是 Mathpix 识别的

hbghlyj

1楼是一般的实矩阵, 下面是对于正交阵的证明, 用的是另一种方法:
2016 linear algebra 7 第6题
(c) Let $T: V \rightarrow V$ be an orthogonal linear transformation of a real inner product space of dimension 3. Show that there is an orthonormal basis $\mathcal{B}$ such that $\sideset{_\mathcal{B}}{_\mathcal{B}}{[T]}$ is block diagonal with blocks $\pm 1$ and $R_\theta$ where $R_\theta$ is a rotation by $\theta$.
Theorem 8.32.
Proof. Let $S=T+T^{-1}=T+T^*$. Then $S^*=\left(T+T^*\right)^*=T^*+T=S$. So $S$ is self-adjoint and has a basis of orthonormal eigenvectors by Theorem 8.21 and thus
$$
V=V_1 \oplus \cdots \oplus V_k
$$
decomposes into orthogonal eigenspaces of $S$ with distinct eigenvalues $\lambda_1, \ldots, \lambda_k\in\Bbb R$.
Note that each $V_i$ is also $T$-invariant as for $v \in V_i$
$$
S(T(v))=T(S(v))=\lambda_i T(v) \text { and so } T(v) \in V_i .
$$
So we may restrict ourselves to $\left.T\right|_{V_i}$.
By definition of $V_i$, for all $v \in V_i$ we have $\left(T+T^{-1}\right) v=\lambda_i v$ and hence $T^2-\lambda_i T+I=0$. Thus the minimal polynomial of $\left.T\right|_{V_i}$ divides $x^2-\lambda_i x+1$ and any eigenvalue of $\left.T\right|_{V_i}$ is a root of it.
If $\lambda_i=\pm 2$, then $(T+I)^2=0$ or $(T-I)^2=0$. Thus the only eigenvalue of $\left.T\right|_{V_i}$ is $-1$ or $+1$, respectively. Since we know $\left.T\right|_{V_i}$ may be diagonalised over $\mathbb{C}$ (Theorem 8.29), we must have $\left.T\right|_{V_i}=+I$ or $-I$.

If $\lambda_i \neq \pm 2$ then $\left.T\right|_{V_i}$ does not have any real eigenvalues as they would have to be $\pm 1$ by Lemma 8.26 (with product $+1$) forcing $\lambda_i=\pm 2$. So $\{v, T(v)\}$ are linearly independent over the reals for $v \neq 0 \in V_i$. Consider the plane $W=\langle v, T(v)\rangle$ spanned by $v$ and $T v$. Then $W$ is $T$-invariant as
$$
v \mapsto T(v), T(v) \mapsto T^2(v)=\lambda_i T(v)-v .
$$
Hence $W^{\perp}$ is also $T$-invariant by Lemma 8.28. Repeating the argument for $\left.T\right|_{W^{\perp}}$ if necessary, we see that $V_i$ splits into 2-dimensional $T$-invariant subspaces. By our Example 8.31, with respect to some orthonormal basis of $W$
$$
\left.T\right|_W=R_\theta=\left[\begin{array}{cc}
\cos (\theta) & -\sin (\theta) \\
\sin (\theta) & \cos (\theta)
\end{array}\right]
$$
for some $θ \ne 0, π$. (Note, the fact that $T_W$ does not have any real eigenvalues implies that $T_W$ is not a reflection and $θ \ne 0, π$.)

---

脚注${ }^4$ One can also deduce this theorem just from our spectral theorem for unitary matrices (Theorem 8.29), by grouping the non-real eigenvalues in complex conjugate pairs $\lambda$ and $\bar{\lambda}$ and taking an orthonormal basis of "real vectors" for each of the two dimensional spaces you get by choosing an eigenvector for $\lambda$ and the conjugate one for $\bar{\lambda}$
指的就是1楼的方法

hbghlyj

1楼是一般的实矩阵, 下面是对于实斜对称矩阵:
反對稱矩陣#譜理論
斜对称矩阵的特征根永远以$±λ$的形式成对出现，因此一个实数斜对称矩阵的非零特征根为纯虚数将会如下：$iλ_1, −iλ_1, iλ_2, −iλ_2, …$，其中 $λ_k$ 是实数。
实斜对称矩阵是正规矩阵（它们与伴随矩阵可交换），因此满足谱定理的条件，它说明任何实斜对称矩阵都可以用一个酉矩阵对角化。由于实斜对称矩阵的特征值是纯虚数，因此无法用实矩阵来对角化。然而，通过正交变换，可以把每一个斜对称矩阵化为2阶块的对角阵。特别地，每一个$2n × 2n$的实斜对称矩阵都可以写成$A = Q Σ Q^T$的形式，其中$Q$是正交矩阵，且：$$\Sigma = \begin{bmatrix}
\begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} &  0 & \cdots & 0 \\
0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} &  & 0 \\
\vdots &  & \ddots & \vdots \\
0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \\
& & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix}
\end{bmatrix}$$对于实数$λ_k$。这个矩阵的非零特征值是$±iλ_k$。在奇数维的情况中，$Σ$总是至少有一个行和一个列全是零。