线性代数疑问:线性相关/无关向量与齐次方程组Ax=0解集的关系
Hey there! Let's unpack this connection between vector linearity and homogeneous systems—once you see the link between Ax and linear combinations, it'll all click into place.
First, let's clarify what matrix A represents here: Suppose you have a set of vectors v₁, v₂, ..., vₙ (all of the same dimension). If you arrange these vectors as the columns of matrix A, then the product Ax (where x is a column vector [x₁, x₂, ..., xₙ]^T) is exactly the linear combination:x₁v₁ + x₂v₂ + ... + xₙvₙ
That's the key bridge between the two concepts! Now let's answer your questions directly:
i) Solution set for linearly dependent vectors
By definition, a set of vectors is linearly dependent if there exists at least one set of scalars x₁, x₂, ..., xₙ that are not all zero such that their linear combination equals the zero vector.
Translating this to the Ax=0 system: this means there's a non-zero vector x that satisfies the equation. The solution set here is the entire collection of such vectors—including the zero vector (since A*0=0 always holds) and all non-zero vectors that make the linear combination zero.
In linear algebra terms, this is called the non-trivial solution set (or the null space of A, which is a non-zero subspace because it contains more than just the zero vector).
ii) Solution set for linearly independent vectors
Linearly independent vectors have the opposite property: the only way their linear combination equals the zero vector is if all scalars x₁, x₂, ..., xₙ are zero.
For the Ax=0 system, this means the only solution is the zero vector x=0. This is called the trivial solution set—it's just a single element, the zero vector.
To tie this back to your confusion: even without a specific matrix, Ax is just a compact way to write the linear combination of A's column vectors. So whether you're thinking in terms of linear combinations or Ax=0, you're really asking the same question: can we get the zero vector without using all zero weights?
内容的提问来源于stack exchange,提问作者Sifu

