证明含零特征值的n阶方阵A不可逆并解释零特征值意义

Alright, let's work through this problem step by step—first the formal proof, then we'll unpack what a 0 eigenvalue really means and why it's the key to this result.

反证法证明过程 #

Let's use a proof by contradiction here—it's perfect for this kind of "prove it can't be true" problem:

1. 假设A可逆：Suppose we claim ( A ) is invertible, so there exists an inverse matrix ( A^{-1} ) such that ( A^{-1}A = I ) (the ( n \times n ) identity matrix).
2. 利用特征值的核心定义：Since 0 is an eigenvalue of ( A ), there must be a non-zero vector ( v ) (called an eigenvector for 0) that satisfies:
   [
   Av = 0v = \mathbf{0}
   ]
   Important note: Eigenvectors are always non-zero by definition—this isn't just a technicality, it's what makes the contradiction work later.
3. 计算 ( A^{-1}Av ) 从两个方向推导：
   • Left side: If we first compute ( Av = \mathbf{0} ), then multiply by ( A^{-1} ), we get ( A^{-1}(Av) = A^{-1}\mathbf{0} = \mathbf{0} ) (multiplying any matrix by the zero vector always gives the zero vector).
   • Right side: Using associativity of matrix multiplication, ( (A^{-1}A)v = Iv = v ) (the identity matrix leaves any vector unchanged).
4. 导出矛盾：Putting these two results together, we get ( v = \mathbf{0} )—but this directly contradicts the fact that ( v ) is a non-zero eigenvector. Our initial assumption that ( A ) is invertible must be wrong.

0作为特征值的含义及作用 #

Now let's break down why a 0 eigenvalue is such a critical piece of this puzzle:

• Basic meaning: For a square matrix ( A ), an eigenvalue ( \lambda ) means there's a non-zero vector ( v ) where ( Av = \lambda v ). When ( \lambda = 0 ), this simplifies to ( Av = \mathbf{0} ).
• What this tells us about ( A ): This equation means ( v ) is a non-zero vector in the null space (or kernel) of ( A )—the set of all vectors that get mapped to the zero vector when multiplied by ( A ).
• Direct link to invertibility: A square matrix is invertible if and only if its null space contains only the zero vector (i.e., the only solution to ( Av = \mathbf{0} ) is ( v = \mathbf{0} )). Since we have a non-zero ( v ) satisfying ( Av = \mathbf{0} ), ( A ) can't be invertible.
• Rank-nullity theorem perspective: For an ( n \times n ) matrix, the rank-nullity theorem states ( \text{rank}(A) + \text{dim(null}(A)) = n ). If ( \text{dim(null}(A)) \geq 1 ) (because we have that non-zero ( v )), then ( \text{rank}(A) \leq n-1 )—so ( A ) doesn't have full rank, which is another necessary condition for a square matrix to be invertible.

内容的提问来源于stack exchange，提问作者Elena