This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Understand the Jordan canonical form
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
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from lingeo import random_int_list, random_good_matrix
Let $\lambda\in\mathbb{C}$ and $n$ an integer.
A Jordan block of $\lambda$ of order $n$ is the $n\times n$ matrix
$$J_{\lambda,n} = \begin{bmatrix}
\lambda & 1 & & \\
& \ddots & \ddots & \\
& & & 1 \\
& & & \lambda \\
\end{bmatrix}.$$
According to the spectrum theorem, for every symmetric matrix $A$, there is an orthonormal basis $\beta$ such that $[f_A]_\beta^\beta$ is orthogonal.
This is not necessarily true for not necessarily symmetric matrices, even we give up the condition of being orthonormal.
Any Jordan block of order at least $2$ is such an example.
Fortunately, we have the folloing canonical form.
Let $A$ be an $n\times n$ matrix.
Then there is a basis $\beta$ of $\mathbb{R}^n$ such that
$$[f_A]_\beta^\beta = J_{\lambda_1,n_1}\oplus \cdots \oplus J_{\lambda_d,n_d},$$
where $\oplus$ is the direct sum of two matrices.
That is, there is an invertible matrix $Q$ such that $Q^{-1}AQ = J_{\lambda_1,n_1}\oplus \cdots \oplus J_{\lambda_d,n_d}$.
We say two $n\times n$ matrices $A$ and $B$ are similar if there is an invertible matrix $Q$ such that $Q^{-1}AQ = B$.
Equivalently, $A$ and $B$ are similar if there is a basis $\beta$ of $\mathbb{R}^n$ such that $[f_A]_\beta^\beta = B$.
It is not trivial how to determine whether two given matrices are similar.
It turns out that two matrices are similar if and only if they have the same Jordan canonical form.
執行以下程式碼。
已知 $A$ 和 $B$ 相似﹐且 $Q^{-1}AQ = B$。
Run the code below. Suppose $A$ and $B$ are similar and $Q^{-1}AQ = B$.
### code
set_random_seed(0)
print_ans = False
n = 2
A = matrix(2, random_int_list(n*n))
Q = random_good_matrix(n,n,n)
B = Q.inverse() * A * Q
print("Qinv * A * Q = B")
pretty_print(Q.inverse(), A, Q, "=", B)
if print_ans:
print("rank =", A.rank())
print("nullity =", A.nullity())
print("determinant =", A.determinant())
print("B (Qinv v) = (Qinv b)")
驗證 $A$ 和 $B$ 有相同的秩、核數、行列式值。
Verify that $A$ and $B$ have the same rank, nullity, and determinant.
令 $A$ 和 $B$ 為兩 $n\times n$ 矩陣。
證明以下敘述等價：
Let $A$ and $B$ be $n\times n$ matrices. Show that the following are equivalent:
令
$$ J = J_{2,3} = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix}. $$Let
$$ J = J_{2,3} = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix}. $$說明若存在一組基底 $\beta = \{ \bv_1,\ldots,\bv_3 \}$ 使得 $[f_J]_\beta^\beta$ 是對角矩陣 $\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$﹐
則對每個 $i = 1,\ldots, 3$ 都有 $J\bv_i = \lambda_i \bv_i$。
Show that if there is a basis $\beta = \{ \bv_1,\ldots,\bv_3 \}$ such that $[f_J]_\beta^\beta$ is a diagonal matrix $\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$, then $J\bv_i = \lambda_i \bv_i$ for each $i = 1, \ldots, 3$.
說明若這樣的基底存在﹐必定是 $\lambda_1 = \cdots = \lambda_3 = 2$﹐
然而 $J$ 不可能和 $2I$ 相似。
Show that if such a basis exists, then it must be $\lambda_1 = \cdots = \lambda_3 = 2$. However, $J$ is impossible to be similar with $2I$.
令 $A$ 為一 $n\times n$ 矩陣。
已知 $\beta$ 為 $\mathbb{R}^n$ 的一組基底使得 $[f_A]_\beta^\beta = J = J_{\lambda,n}$。
將每個 $A\bv_i$ 寫成 $\beta$ 的線性組合。
Let $A$ be an $n\times n$ matrix. Suppose $\beta$ is a basis of $\mathbb{R}^n$ such that $[f_A]_\beta^\beta = J = J_{\lambda,n}$. Write each $A\bv_i$ as a linear combination of $\beta$.
令
$$ A = \begin{bmatrix} 5 & 1 & 0 \\ -7 & -1 & 1 \\ -6 & -2 & 2 \end{bmatrix}. $$令
$\bv_1 = (1, -3, -2)$、
$\bv_2 = (1, -2, -2)$、
$\bv_3 = (2, -5, -3)$、
$\beta = \{ \bv_1, \ldots, \bv_3 \}$。
求 $[f_A]_\beta^\beta$。
Let
$$ A = \begin{bmatrix} 5 & 1 & 0 \\ -7 & -1 & 1 \\ -6 & -2 & 2 \end{bmatrix}. $$Let $\bv_1 = (1, -3, -2)$, $\bv_2 = (1, -2, -2)$, $\bv_3 = (2, -5, -3)$, and $\beta = \{ \bv_1, \ldots, \bv_3 \}$. Find $[f_A]_\beta^\beta$.