This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Constructing new subspaces
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
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from lingeo import random_good_matrix, column_space_matrix, left_kernel_matrix, kernel_matrix
Let $U$ and $V$ be two subspaces in the same vector space.
In general, the set $U\cup V$ is no more a subspace.
However, we may define the sum of $U$ and $V$ as $U + V = \vspan(U \cup V)$, which is a subspace.
Suppose $\beta_U$ and $\beta_V$ are the bases of $U$ and $V$, respectively.
Then $U + V = \vspan(\beta_U \cup \beta_V)$.
However, $\beta_U \cup \beta_V$ is not necessarily independent.
Suppose $\beta_U$ and $\beta_V$ are finite.
Let $A_U$ and $A_V$ be the matrix whose columns are vectors in $U$ and $V$, respectively.
Then the $\beta_C$ corresponding to $\begin{bmatrix} A_U & A_V \end{bmatrix}$ is a basis of $U + V$.
On the other hand, the intersection $U \cap V$ is indeed a subspace.
Suppose $\beta_U$ and $\beta_V$ are the bases of $U$ and $V$, respectively.
Then $(\beta_U \cap \beta_V)$ is linearly independent but not necessarily spans $U\cap V$.
(Even worse, it is quite possible that $U \cap V = \emptyset$.)
Suppose we are able to matrices $B_U$ and $B_V$ such that $U = \ker(B_U)$ and $V = \ker(B_V)$.
Then $U \cap V$ is the kernel of $\begin{bmatrix} B_U \\ B_V \end{bmatrix}$.
By the expanding lemma, one may find a basis $\beta_\cap$ of $U\cap V$,
expand it to a basis $\beta_U$ of $U$,
expand it to a basis $\beta_V$ of $V$,
and show that $\beta_\cup = \beta_U \cup \beta_V$ is a basis of $U + V$.
Therefore, we have
$$\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V).$$
Suppose $V_1, \ldots, V_k$ are some subspaces in the same vector space.
We say $\{V_1, \ldots, V_k\}$ is linearly independent if
the only choice of $\bv_1\in V_1, \ldots, \bv_k\in V_k$ satisfying
$$\bv_1 + \cdots + \bv_k = \bzero$$
is $\bv_1 = \cdots = \bv_k = \bzero$.
The following two statments are equivalents:
However, even if $V_1,V_2,V_3$ mutually have trivial intersections, $\{ V_1, V_2, V_3 \}$ might not be linearly independent.
In the case that $\{V_1,\ldots, V_k\}$ is linearly independent,
we call the subspace $V_1 + \cdots + V_k$ the direct sum of them and
use the notation $V_1 \oplus \cdots \oplus V_k$ instead to emphasize the linearly independence.
執行以下程式碼。
令 $\bu_1, \bu_2$ 為 $A_U$ 的各行向量、
$\bv_1, \bv_2$ 為 $A_V$ 的各行向量。
令 $U = \Col(A_U)$ 且 $V = \Col(A_V)$。
已知 $R$ 為 $\begin{bmatrix} A_U & A_V \end{bmatrix}$ 的最簡階梯形式矩陣。
Run the code below. Let $\bu_1, \bu_2$ be the columns of $A_I$ and $\bv_1, \bv_2$ the columns of $A_V$. Let $U = \Col(A_U)$ and $V = \Col(A_V)$. Suppose $R$ is the reduced echelon form of $\begin{bmatrix} A_U & A_V \end{bmatrix}$.
### code
set_random_seed(0)
print_ans = False
m,n,r = 4,3,3
A = random_good_matrix(m,n,r)
AU = A[:,:2]
AV = A[:,1:]
AUAV = AU.augment(AV, subdivide=True)
print("[ A_U | A_V ] =")
show(AUAV)
print("R =")
show(AUAV.rref())
BU = left_kernel_matrix(AU)
BV = left_kernel_matrix(AV)
BUBV = BU.stack(BV, subdivide=True)
if print_ans:
print("dim( U + V ) =", r)
print("basis of U + V = columns of")
print(column_space_matrix(AUAV))
print("dim( U cap V) =", 4 - r)
print("basis of U cap V = columns of")
print(kernel_matrix(BUBV))
令
$$ A_U = \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ 2 & 1 \\ 2 & 1 \\ \end{bmatrix}, A_V = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 2 \\ 2 & 1 \\ \end{bmatrix} $$且 $U = \Col(A_U)$ 且 $V = \Col(A_V)$。
Let
$$ A_U = \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ 2 & 1 \\ 2 & 1 \\ \end{bmatrix}, A_V = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 2 \\ 2 & 1 \\ \end{bmatrix}. $$Let $U = \Col(A_U)$ and $V = \Col(A_V)$.
找出 $B_U$ 和 $B_V$
使得 $U = \ker(B_U)$ 且 $V = \ker(B_V)$。
Find $B_U$ and $B_V$ such that $U = \ker(B_U)$ and $V = \ker(B_V)$.
若 $U$ 和 $V$ 為兩個子空間。
證明 $U + V$ 為一個子空間。
Let $U$ and $V$ be subspaces. Show that $U + V$ is a subspace.
若 $U$ 和 $V$ 為兩個子空間。
證明 $U \cap V$ 為一個子空間。
Let $U$ and $V$ be subspaces. Show that $U \cap V$ is a subspace.
若 $U$ 和 $V$ 為兩個子空間。
證明
Let $U$ and $V$ be subspaces. Show that
$$ \dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V). $$令
$$ A_U = \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ 2 & 1 \\ 2 & 1 \\ \end{bmatrix}, A_V = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 2 \\ 2 & 1 \\ \end{bmatrix} $$且 $U = \Col(A_U)$ 且 $V = \Col(A_V)$。
Let
$$ A_U = \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ 2 & 1 \\ 2 & 1 \\ \end{bmatrix}, A_V = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 2 \\ 2 & 1 \\ \end{bmatrix}. $$Let $U = \Col(A_U)$ and $V = \Col(A_V)$.
找出一組非零向量 $\bu\in U$ 及 $\bv\in V$
使得 $\bu + \bv = \bzero$。
藉此說明 $\{U, V\}$ 不線性獨立。
Find nonzero vectors $\bu\in U$ and $\bv\in V$ such that $\bu + \bv = \bzero$. This shows that $\{U, V\}$ is not linearly independent.
令 $V_1$ 和 $V_2$ 為任意的兩個子空間。
證明以下敘述等價：
Let $V_1$ and $V_2$ be subspaces. Show that the following are equivalent:
對於每個 $i = 1,\ldots, k$，令 $U_i = \vspan(\{\bu_i\})$。
證明以下敘述等價：
For each $i = 1,\ldots, k$, let $U_i = \vspan(\{\bu_i\})$. Show that the following are equivalent:
令
$$ V_1 = \vspan\left(\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}\right), V_2 = \vspan\left(\left\{\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\right), V_3 = \vspan\left(\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}\right). $$說明對任意相異的 $i$ 和 $j$ 都有 $V_i\cap V_j = \emptyset$﹐
但是 $\{V_1,V_2,V_3\}$ 並不線性獨立。
Let
$$ V_1 = \vspan\left(\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}\right), V_2 = \vspan\left(\left\{\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\right), V_3 = \vspan\left(\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}\right). $$Verify that $V_i\cap V_j = \emptyset$ for any distinct $i$ and $j$, but $\{V_1, V_2, V_3\}$ is not linearly independent.
若 $\{\bu_1,\ldots,\bu_6\}$ 線性獨立。
令 $V_1 = \{\bu_1, \bu_2\}$、
$V_2 = \{\bu_3, \bu_4\}$ 且
$V_3 = \{\bu_5, \bu_6\}$。
證明 $\{V_1,V_2,V_3\}$ 線性獨立。
（實際上把一群線性獨立的向量分成任意堆﹐
則每堆生成出來的空間
全部合在一起會是線性獨立的。）
Let $\{\bu_1,\ldots,\bu_6\}$ be a linearly independent set. Let $V_1 = \{\bu_1, \bu_2\}$, $V_2 = \{\bu_3, \bu_4\}$, and $V_3 = \{\bu_5, \bu_6\}$. Show that $\{V_1,V_2,V_3\}$ is linearly independent. (In fact, for any partition of a linearly independent set, the collection of the spanning set of its parts is linearly independent.)