This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Diagonalization
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
We know that an $n\times n$ matrix is diagonalizable if and only if there is a basis of $\mathbb{R}^n$ composed of eigenvectors.
We also know how to find the eigenvalues through the characteristic polynomial.
Indeed, that is all we need to perform the diagonalization of a matrix.
Let $A$ be an $n\times n$ matrix.
Here are the steps for its diagonalization:
That is, $[f_A]_\beta^\beta = D$ is a diagonal matrix.
Equivalently, $D = Q^{-1}AQ$, where $Q$ is the matrix whose columns are the vectors in $\beta$.
If $\beta = \{\bv_1, \ldots, \bv_n\}$ is a basis of $\mathbb{R}^n$ such $\bv_i$ is an eigenvector of $A$ with respect to $\lambda_i$ for all $i$.
Then the equalities $A\bv_i = \lambda_i\bv_i$ is equivalent to
The above process might not able to finish for a few different reasons:
Here is an example of diagonalizing
$$ A = \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix}. $$Solve for the eigenvalues:
The characteristic polynomial of $A$ is
$$ p_A(x) = \det\begin{bmatrix} 2 - x & 3 \\ 3 & 2 - x \end{bmatrix} = x^2 - 4x - 5, $$which as roots $-1, 5$.
Solve for the eigenvectors:
For $\lambda = 5$, calculate the basis of $\ker(A - 5I)$ and get
$$ \ker(A - 5I) = \ker \begin{bmatrix} -3 & 3 \\ 3 & -3 \end{bmatrix} = \vspan\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}. $$For $\lambda = -1$, calculate the basis of $\ker(A + I)$ and get
$$ \ker(A + I) = \ker \begin{bmatrix} 3 & 3 \\ 3 & 3 \end{bmatrix} = \vspan\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}. $$By setting
$$ \beta = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\} \text{ and } Q = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, $$we get
$$ [f_A]_\beta^\beta = Q^{-1}AQ = \begin{bmatrix} 5 & 0 \\ 0 & -1 \end{bmatrix}. $$### code
set_random_seed(0)
print_ans = False
n = 3
spec = random_int_list(n, 3)
D = diagonal_matrix(spec)
Q = random_good_matrix(n,n,n,2)
A = Q * D * Q.inverse()
pretty_print(LatexExpr("A ="), A)
pA = (-1)^n * A.charpoly()
print("characteristic polyomial =", pA)
print(" =", factor(pA))
if print_ans:
print("eigenvalues are:" + ", ".join("%s"%val for val in spec))
print("corresponding eigenvectors are:")
for i in range(n):
pretty_print(LatexExpr(r"\lambda ="), spec[i], ", eigenvector =", Q[:,i])
pretty_print(LatexExpr("Q ="), Q)
求出 $Q$ 使得 $D = Q^{-1}AQ$ 是一個對角矩陣。
Find a matrix $Q$ such that $D = Q^{-1}AQ$ is a diagonal matrix.
將以下矩陣 $A$ 對角化。
（求出可逆矩陣 $Q$ 和對角矩陣 $D$ 使得 $D = Q^{-1}AQ$。）
Diagonalize the following matrices $A$. (That is, find an invertible matrix $Q$ and a diagonal matrix $D$ such that $D = Q^{-1}AQ$.)
將以下矩陣 $A$ 對角化。
（求出可逆矩陣 $Q$ 和對角矩陣 $D$ 使得 $D = Q^{-1}AQ$。）
Diagonalize the following matrices $A$. (That is, find an invertible matrix $Q$ and a diagonal matrix $D$ such that $D = Q^{-1}AQ$.)
將以下矩陣 $A$ 對角化，
並說明 $f_A$ 的作用。
Diagonalize the following matrices $A$ and describe the effect of $f_A$.
令
$$ A = \begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix}. $$說明 $A$ 矩陣只有一個特徵值 $3$（重根三次），
但 $\ker(A - 3I)$ 的維度只有 $1$，
找不到足夠的特徵向量將 $A$ 對角化。
Let
$$ A = \begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix}. $$Show that $A$ only has one eigenvalue $3$ (repeated three times), but $\ker(A - 3I)$ has dimension $1$ and we do not have enough independent eigenvectors to diagonalize $A$.
令
$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$並令 $\zeta = e^{\frac{2\pi}{5}i}$ 為 $1$ 的五次方根。
這個練習告訴我們，如果運氣很好有找到所有的特徵向量，則不見得要解特徵多項式也可以找得到 $\spec(A)$。
Let
$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$and let $\zeta = e^{\frac{2\pi}{5}i}$ be the fifth roots of $1$.
Through the following exercises, we will see that $\spec(A)$ can also be found by the eigenvectors (if we know them by luck) rather than the characteristic polynomial.
令
$$ Q = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & \zeta & \zeta^2 & \zeta^3 & \zeta^4 \\ 1 & \zeta^2 & \zeta^4 & \zeta^6 & \zeta^8 \\ 1 & \zeta^3 & \zeta^6 & \zeta^9 & \zeta^{12} \\ 1 & \zeta^4 & \zeta^8 & \zeta^{12} & \zeta^{16} \\ \end{bmatrix}. $$驗證 $Q^*Q = 5I_5$，
因此 $Q$ 可逆且其行向量集合是一組 $\mathbb{C}^5$ 的基底。
（這裡 $Q^*$ 的意思是將 $Q$ 轉置後再逐項取共軛。）
Let
$$ Q = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & \zeta & \zeta^2 & \zeta^3 & \zeta^4 \\ 1 & \zeta^2 & \zeta^4 & \zeta^6 & \zeta^8 \\ 1 & \zeta^3 & \zeta^6 & \zeta^9 & \zeta^{12} \\ 1 & \zeta^4 & \zeta^8 & \zeta^{12} & \zeta^{16} \\ \end{bmatrix}. $$Verify that $Q^*Q = 5I_5$. Therefore, $Q$ is invertible and it columns form a basis of $\mathbb{C}^5$.
(Here $Q^*$ is the conjugate transpose of $Q$.)
說明 $Q$ 的每個行向量都是 $A$ 的特徵向量，並找出對應的特徵值。
利用這點將 $A$ 對角化。
Show that each column of $Q$ is an eigenvector of $A$ and find the corresponding eigenvalue. Use them to diagonalize $A$.