矩阵可逆性与秩：如何不使用行列式证明A可逆当且仅当满秩？

Hi there! Great question—going back to basics without determinants really helps solidify why invertibility and full rank are equivalent. Let's work through this step by step for an ( n \times n ) matrix over any field (like ( \mathbb{R} ) or ( \mathbb{C} )):

Key Definitions First #

• Invertible matrix: An ( n \times n ) matrix ( A ) is invertible if there exists an ( n \times n ) matrix ( B ) such that ( AB = I_n ) and ( BA = I_n ), where ( I_n ) is the ( n \times n ) identity matrix.
• Full rank: An ( n \times n ) matrix has full rank if its column rank (the dimension of its column space) equals ( n ) (note: row rank always equals column rank, so we can use either interchangeably).

1. Necessity: If ( A ) is invertible, then ( A ) has full rank #

Suppose ( A ) is invertible, so we have some matrix ( B ) where ( AB = BA = I_n ). Let's look at the column vectors of ( A ): ( \mathbf{a}_1, \mathbf{a}_2, ..., \mathbf{a}_n ). We need to show these vectors are linearly independent (since ( n ) linearly independent vectors in ( \mathbb{F}^n ) form a basis, meaning column rank is ( n )).

Take any linear combination of the columns that equals the zero vector:
$$c_1\mathbf{a}_1 + c_2\mathbf{a}_2 + ... + c_n\mathbf{a}_n = \mathbf{0}$$
This is equivalent to ( A\mathbf{c} = \mathbf{0} ), where ( \mathbf{c} = [c_1, c_2, ..., c_n]^T ). Multiply both sides on the left by ( B ):
$$B(A\mathbf{c}) = B\mathbf{0} \implies (BA)\mathbf{c} = \mathbf{0} \implies I_n\mathbf{c} = \mathbf{0} \implies \mathbf{c} = \mathbf{0}$$
The only solution is the trivial combination (all ( c_i = 0 )), so the columns of ( A ) are linearly independent. Thus, the column rank of ( A ) is ( n )—it has full rank.

2. Sufficiency: If ( A ) has full rank, then ( A ) is invertible #

If ( A ) has full rank, its column vectors ( \mathbf{a}_1, ..., \mathbf{a}_n ) form a basis for ( \mathbb{F}^n ) (since we have ( n ) linearly independent vectors in an ( n )-dimensional space). This means every vector in ( \mathbb{F}^n ) can be written uniquely as a linear combination of ( A )'s columns.

In particular, each standard basis vector ( \mathbf{e}_i ) (with 1 in the ( i )-th position and 0 elsewhere) can be expressed as:
$$\mathbf{e}i = d{1i}\mathbf{a}1 + d{2i}\mathbf{a}2 + ... + d{ni}\mathbf{a}_n$$
This is exactly ( A\mathbf{b}i = \mathbf{e}i ), where ( \mathbf{b}i = [d{1i}, d{2i}, ..., d{ni}]^T ). Now build matrix ( B ) whose columns are ( \mathbf{b}_1, \mathbf{b}_2, ..., \mathbf{b}_n ). By matrix multiplication rules, the ( i )-th column of ( AB ) is ( A\mathbf{b}_i = \mathbf{e}_i ), so ( AB = I_n ).

Now we need to confirm ( BA = I_n ). Let's use linear transformations: let ( T: \mathbb{F}^n \to \mathbb{F}^n ) be the linear map defined by ( T(\mathbf{x}) = A\mathbf{x} ). Since ( A ) has full rank, ( T ) is injective (one-to-one): if ( T(\mathbf{x}) = T(\mathbf{y}) ), then ( A(\mathbf{x}-\mathbf{y}) = \mathbf{0} ). Linear independence of ( A )'s columns implies ( \mathbf{x}-\mathbf{y} = \mathbf{0} ), so ( \mathbf{x} = \mathbf{y} ).

In finite-dimensional vector spaces, injective linear maps are automatically surjective (onto) because the dimension of the image equals the rank of ( A ) (which is ( n )), so the image is all of ( \mathbb{F}^n ). Thus ( T ) is bijective, meaning it has an inverse map ( T^{-1} ) such that ( T(T^{-1}(\mathbf{x})) = \mathbf{x} ) and ( T^{-1}(T(\mathbf{x})) = \mathbf{x} ) for all ( \mathbf{x} ).

Let ( B ) be the matrix representing ( T^{-1} ), so ( T^{-1}(\mathbf{x}) = B\mathbf{x} ). Then:

• ( T(T^{-1}(\mathbf{x})) = A(B\mathbf{x}) = (AB)\mathbf{x} = I_n\mathbf{x} = \mathbf{x} ) (we already knew ( AB = I_n ))
• ( T^{-1}(T(\mathbf{x})) = B(A\mathbf{x}) = (BA)\mathbf{x} = I_n\mathbf{x} = \mathbf{x} ) for all ( \mathbf{x} ), which forces ( BA = I_n )

Therefore, ( B ) is the inverse of ( A ), so ( A ) is invertible.

Wrapping Up #

We've proven both directions using only linear independence, basis properties, and linear transformations—no determinants involved. This equivalence is core to linear algebra because it links the algebraic property of having an inverse to the structural property of spanning the entire space with columns (or rows).

内容的提问来源于stack exchange，提问作者MasaJuno