零行矩阵与方阵相乘的列零化及类单位矩阵特例求证

Great question! Let's break this down with clear definitions, a formal proof, and a concrete example to make the logic totally transparent.

First, Let's Define the "Quasi-Identity Matrix" I #

You're referring to an $n \times n$ diagonal matrix where each diagonal entry $I_{ii}$ is either 1 or 0 (all off-diagonal entries are 0). Think of it as the identity matrix with some diagonal 1s replaced by 0s. A key property of this matrix:

I's k-th row is a zero row if and only if I's k-th column is a zero column
Since it's diagonal, the k-th row only has a non-zero entry at the k-th position (the diagonal), and the same goes for the k-th column. So if one is zero, the other must be too.

Formal Proof of the Conclusion #

We want to show: If I's k-th row is a zero row, then the k-th column of the product $AI$ (A is any $n \times n$ matrix) is a zero column.

1. Recall matrix multiplication rules: For any two matrices $A$ and $I$, the j-th column of $AI$ is equal to matrix $A$ multiplied by the j-th column of $I$ (treated as a column vector).
2. Leverage I's diagonal property: Since I's k-th row is a zero row, I's k-th column must be a zero vector (denoted $\mathbf{0}$, where all entries are 0).
3. Matrix times zero vector is zero vector: When we multiply $A$ by the zero vector $\mathbf{0}$, every entry in the resulting column vector is the dot product of a row of $A$ with $\mathbf{0}$. For any row $i$ of $A$, this dot product is:
   $$
   \sum_{t=1}^n A_{it} \times 0 = 0
   $$
   Every entry in the column is 0, so the k-th column of $AI$ is a zero column.

Concrete Example to Verify #

Let's take $n=3$, with I having a zero row (and column) at position 2:
$$
I = \begin{pmatrix}
1 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 1
\end{pmatrix}
$$
And an arbitrary 3x3 matrix A:
$$
A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} \
a_{21} & a_{22} & a_{23} \
a_{31} & a_{32} & a_{33}
\end{pmatrix}
$$
Calculating $AI$ gives:
$$
AI = \begin{pmatrix}
a_{11} & 0 & a_{13} \
a_{21} & 0 & a_{23} \
a_{31} & 0 & a_{33}
\end{pmatrix}
$$
You can clearly see the 2nd column is entirely zero, matching our conclusion.

Why This Differs from General Zero-Row Matrices #

Your initial note about "zero-row matrices making product columns zero" only holds for matrices where zero rows correspond to zero columns (like this diagonal quasi-identity matrix). For a general matrix with a zero row but non-zero column, multiplying it on the right won't produce a zero column—this special case works because of I's diagonal symmetry.

内容的提问来源于stack exchange，提问作者user12888