📂Linear Algebra

The Equivalence Condition When the Range of a Linear Transformation is Smaller than the Kernel

Theorem¹

Let $V$ be a vector space, and $T : V \to V$ be a linear transformation. Then, the following holds:

$$T^{2} = T_{0} \iff R(T) \subset N(T)$$

Here, $T_{0}$ is the zero transformation, and $R(T), N(T)$ are the respective range and null space of $T$.

Generalization

Let $U, V, W$ be a vector space, and $T_{1} : U \to V$, $T_{2} : V \to W$ be linear transformations. Then, the following holds:

$$T_{2}T_{1} = T_{0} \iff R(T_{1}) \subset N(T_{2})$$

Explanation

It’s actually an obvious matter if you think about it. The proof method for the generally written theorem is the same.

Meanwhile, a linear transformation that is $T^{2} = T_{0}$ is called Nilpotent.

Proof

$(\Longrightarrow)$

Let’s assume $T^{2} = T_{0}$. Let $T(x) \in R(T)$ $(x\in V)$. Then,

$$T(T(x)) = T^{2}(x) = 0$$

Therefore, by the definition of $N(T)$, $T(x) \in N(T)$ is true. Hence,

$$R(T) \subset N(T)$$

$(\Longleftarrow)$

Let’s assume $R(T) \subset N(T)$. Then, for all $x \in V$, since $T(x) \in R(T) \subset N(T)$, by the definition of $N(T)$,

$$R(T) \subset N(T)$$

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