从零空间和左零空间推导矩阵形式与值域的技术咨询

Hey there! Let's break this down step by step—you're right that null spaces tie to columns and rows, so we'll use core linear algebra properties like the rank-nullity theorem and orthogonality to build out matrix $A$ and its range.

第一步：确定矩阵$A$的维度和秩 #

First, let's lock in the basics using the given null spaces:

• The null space $N(A)$ is spanned by $\begin{bmatrix}1 \ 0 \ -1\end{bmatrix}$, which is a 3-dimensional column vector. This tells us $A$ has 3 columns (since null space vectors match the column count of the matrix).
• The left null space $N(A^T)$ is spanned by two linearly independent vectors: $\begin{bmatrix}1&1&1&1\end{bmatrix}$ and $\begin{bmatrix}1&1&-1&-1\end{bmatrix}$. This means $dim(N(A^T))=2$, so $A$ has 4 rows (left null space dimension = number of rows - rank of $A$).

Using the rank-nullity theorem to confirm the rank:

• For $A$ (a 4×3 matrix):
  • $rank(A) = \text{number of columns} - dim(N(A)) = 3 - 1 = 2$
  • Also, $rank(A) = \text{number of rows} - dim(N(A^T)) = 4 - 2 = 2$
    Both calculations align, so we know $A$ is a 4×3 matrix with rank 2.

第二步：推导矩阵$A$的具体形式 #

We'll use orthogonal relationships between null spaces and row/column spaces to constrain $A$:

约束1：行空间与$N(A)$正交 #

A key linear algebra rule: the row space $R(A^T)$ is the orthogonal complement of the null space $N(A)$. Since $N(A)$ is spanned by $\begin{bmatrix}1 \0 \-1\end{bmatrix}$, any row vector $\begin{bmatrix}x_1&x_2&x_3\end{bmatrix}$ in $R(A^T)$ must satisfy:
$$x_11 + x_20 + x_3*(-1) = 0 \implies x_1 = x_3$$
In plain terms, every row of $A$ must look like $\begin{bmatrix}a&b&a\end{bmatrix}$ where $a,b$ are real numbers.

约束2：列空间与$N(A^T)$正交 #

Another rule: the column space $R(A)$ is the orthogonal complement of the left null space $N(A^T)$. The two basis vectors of $N(A^T)$ mean every column $\begin{bmatrix}c_1\c_2\c_3\c_4\end{bmatrix}$ of $A$ must satisfy:

1. $c_1 + c_2 + c_3 + c_4 = 0$
2. $c_1 + c_2 - c_3 - c_4 = 0$

Solving these equations:

• Adding them gives $2(c_1 + c_2) = 0 \implies c_1 = -c_2$
• Subtracting them gives $2(c_3 + c_4) = 0 \implies c_3 = -c_4$
  So every column of $A$ must look like $\begin{bmatrix}-k\k\-l\l\end{bmatrix}$ where $k,l$ are real numbers.

结合约束得到$A$的最终形式 #

Putting the row and column constraints together, we can write $A$ explicitly as:

[ -p  -r  -p ]
[  p   r   p ]
[ -q  -s  -q ]
[  q   s   q ]

Here, $p,q,r,s$ are real numbers, and $(p,q)$ must be linearly independent of $(r,s)$—this ensures $A$ has rank 2 (if they were proportional, the rank would drop to 1, which contradicts our earlier conclusion).

As a concrete example, set $p=1,q=0,r=0,s=1$, and you get a valid $A$:

[ -1  0  -1 ]
[  1  0   1 ]
[  0 -1   0 ]
[  0  1   0 ]

You can verify this matrix has the exact null spaces given in your question!

第三步：确定$A$的值域（列空间） #

The range $R(A)$ is the subspace spanned by $A$'s columns. From the column constraints, every vector in $R(A)$ satisfies two conditions: $x_1 = -x_2$ and $x_3 = -x_4$.

Conversely, any 4-dimensional vector that meets these two conditions can be written as a linear combination of $A$'s columns. For example, $\begin{bmatrix}-a-b\a+b\-c-d\c+d\end{bmatrix}$ can be expressed as $a*\begin{bmatrix}-1\1\0\0\end{bmatrix} + b*\begin{bmatrix}0\0\-1\1\end{bmatrix}$.

So the range of $A$ is the 2-dimensional subspace of $\mathbb{R}^4$ consisting of all vectors where the first entry is the negative of the second, and the third entry is the negative of the fourth.

备注：内容来源于stack exchange，提问作者Adri Rove