求5×4实矩阵A的秩

Hey there! Let's work through this problem step by step—since you're using Hoffman and Kunze, we can lean on a core theorem they cover in depth: the Rank-Nullity Theorem.

First, let's rephrase the problem's key condition to make it clearer: the only solution to the equation (Ax=0) is the zero vector (x=0).

What does this tell us about the null space (kernel) of (A)? The null space of (A) is the set of all vectors (x) that satisfy (Ax=0). Since only the zero vector fits here, the dimension of the null space (nullity(A)) is 0—there's no non-trivial subspace here, just the single zero vector.

Now, the Rank-Nullity Theorem states that for any (m \times n) matrix, the sum of the matrix's rank and its nullity equals the number of columns (n). In formula terms:
rank(A) + nullity(A) = n

For our 5×4 matrix (A), (n=4) (the number of columns). We already established nullity(A) = 0, so substituting these values in:
rank(A) + 0 = 4

That means the rank of (A) must be 4.

Just to confirm this makes sense: the maximum possible rank of a 5×4 matrix is 4 (since rank can't exceed the smaller of the number of rows or columns). Since our null space is trivial, this implies all 4 columns of (A) are linearly independent—so the column space has dimension 4, which is exactly the rank of the matrix. This lines up perfectly with our conclusion.

备注：内容来源于stack exchange，提问作者user1237867