矩阵概率与秩的关系：2×2二元元素矩阵秩的概率计算

Hey there, let's break this down step by step for you:

First, let's confirm the basics: we have a total of (2^4 = 16) matrices in set X (each of the 4 elements has 2 choices: 1 or 2). Now let's tackle each rank case, plus your specific questions:

Rank 0 #

Rank 0 matrices are zero matrices (all elements are 0), but our matrix elements can only be 1 or 2—there are no such matrices in X.

So to answer your direct questions:

• The probability of rank 0 is 0, not 1/16.
• The all-1 matrix has a rank of 1 (its rows are identical and linearly dependent), so it doesn't qualify as rank 0 at all. There are no matrices in X that meet the rank 0 criteria, so the "only all-1 matrix" claim is also incorrect.

Rank 1 #

A 2×2 matrix has rank 1 if its rows (or columns) are linearly dependent, which is equivalent to its determinant being 0 ((ad = bc) for matrix (\begin{bmatrix}a&b\c&d\end{bmatrix})).

Counting these valid matrices:

• All elements identical: (\begin{bmatrix}1&1\1&1\end{bmatrix}), (\begin{bmatrix}2&2\2&2\end{bmatrix}) (2 matrices)
• Rows are scalar multiples (but not identical): (\begin{bmatrix}1&1\2&2\end{bmatrix}), (\begin{bmatrix}2&2\1&1\end{bmatrix}), (\begin{bmatrix}1&2\1&2\end{bmatrix}), (\begin{bmatrix}2&1\2&1\end{bmatrix}) (4 matrices)

Total rank 1 matrices: 6. Probability: (\frac{6}{16} = \frac{3}{8}).

Rank 2 #

A matrix has rank 2 if its determinant is non-zero ((ad \neq bc)). We can calculate this by subtracting the rank 0 and rank 1 counts from the total:

Total rank 2 matrices: (16 - 0 - 6 = 10). Probability: (\frac{10}{16} = \frac{5}{8}).

内容的提问来源于stack exchange，提问作者Mike W.