求证：满足T²=0的线性变换T:ℝⁿ→ℝⁿ的秩r≤n/2

Got it, let's walk through this proof step by step—no fancy jargon, just clear reasoning using linear algebra basics:

Key Background #

First, let's recap a few essential definitions and theorems we'll rely on:

• Rank of $T$: The dimension of the image (range) of $T$, written $r = \dim(\text{Im}(T))$. This is the number of linearly independent columns in any matrix representation of $T$.
• Nullity of $T$: The dimension of the kernel (null space) of $T$, which is the set of all vectors $\mathbf{x}$ where $T(\mathbf{x}) = \mathbf{0}$.
• Rank-Nullity Theorem: For any linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$, this theorem tells us:
  $$\dim(\text{Im}(T)) + \dim(\text{Ker}(T)) = n$$

Step 1: What does $T^2 = 0$ mean? #

The condition $T^2 = 0$ (where $T^2$ is $T \circ T$, applying $T$ twice) implies that for every vector $\mathbf{x} \in \mathbb{R}^n$, $T(T(\mathbf{x})) = \mathbf{0}$.

In other words: any vector that's in the image of $T$ (i.e., any vector $\mathbf{y} = T(\mathbf{x})$ for some $\mathbf{x}$) gets mapped to the zero vector when we apply $T$ to it. That means the entire image space of $T$ is contained within the kernel of $T$:
$$\text{Im}(T) \subseteq \text{Ker}(T)$$

Step 2: Use subspace dimension inequality #

If one subspace is contained within another, its dimension can't be larger. So from $\text{Im}(T) \subseteq \text{Ker}(T)$, we get:
$$\dim(\text{Im}(T)) \leq \dim(\text{Ker}(T))$$
Substituting the rank $r$ for $\dim(\text{Im}(T))$, this becomes:
$$r \leq \dim(\text{Ker}(T))$$

Step 3: Apply Rank-Nullity Theorem #

From the Rank-Nullity Theorem, we know $\dim(\text{Ker}(T)) = n - r$. Let's substitute that into our inequality:
$$r \leq n - r$$

Now rearrange the terms to solve for $r$:

• Add $r$ to both sides: $2r \leq n$
• Divide by 2: $r \leq \frac{n}{2}$

Wrap-Up #

That's it! The critical insight here is recognizing that $T^2=0$ forces the image of $T$ to live entirely inside its kernel. Combining that with the Rank-Nullity Theorem gives us the upper bound on the rank.

内容的提问来源于stack exchange，提问作者nandita mukherjee