11.22 Null Space

Let \(V\) be a vector space and \(T \in \mathcal{L}(V, V)\).
1. Prove that \(\text{Null}(T) \subseteq \text{Null}(T^2)\) (\(T^2 = T \circ T\)).
  
  Let \(v \in \text{Null}(T) \implies T(v) = 0\).
  
  Now, \(T^2(v)=T(T(v))=T(0)=0\implies v\in\text{Null}(T^2)\). 2. Prove that \(\text{Null}(T) = \text{Null}(T^2) \iff \text{Null}(T) \cap \text{Range}(T) = \{0\}\).
  
  \(\Rightarrow)\) Let \(v\in\text{Null}(T)\cap\text{Range}(T)\implies T(v)=0\) and \(v = T(w)\) for some \(w \in V\).
  
  Now \(0=T(v)=T(T(w))=T^2(w)\)
  
  \(\implies w\in\text{Null}(T^2)=\text{Null}(T)\implies T(w)=0\implies v=0\).
  
  Therefore, \(\text{Null}(T) \cap \text{Range}(T) = \{0\}\).
  
  \(\Leftarrow\)) By exercise (a), we just need to check that \(\text{Null}(T^2) \subseteq \text{Null}(T)\).
  
  Let \(v\in\text{Null}(T^2)\implies T^2\left(v\right)=0\implies T(v)\in\text{Null}(T)\cap\text{Range}(T)\).
  
  Since \(\text{Null}(T) \cap \text{Range}(T) = \{0\}\), we have \(T(v)=0\implies v\in\text{Null}(T)\).
  
  Therefore, \(\text{Null}(T^2) = \text{Null}(T)\).
Suppose \(V\) is a vector space and \(S, T \in \mathcal{L}(V, V)\) are such that \(\text{Range}(S) \subseteq \text{Null}(T)\). Prove that \((S \circ T)^2 = 0\).

Let \(v \in V\), \((S\circ T)^2(v)=(S\circ T)\circ(S\circ T)(v)=(S\circ T)\circ(S(T(v)))\)

\(= S(T(S(T(v)))) = S(0) = 0\).

Since this is valid for every \(v \in V\), \((S \circ T)^2 = 0\).
Let \(\dim(V) = n\) and \(T \in \mathcal{L}(V, V)\).
1. If \(n\) is odd \(\implies \text{Null}(T) \neq \text{Range}(T)\).
  
  We know that (by the fundamental theorem of linear maps):
  \(\dim(V) = \dim(\text{Null}(T)) + \dim(\text{Range}(T)).\)
  
  Suppose by contradiction that \(\text{Null}(T) = \text{Range}(T)\).
  
  \(\implies n = 2 \dim(\text{Null}(T)) \implies n\) is even. Contradiction!
  
  So, \(\text{Null}(T) \neq \text{Range}(T)\). 2. If \(n\) is even, give an example of \(T \in \mathcal{L}(V, V)\) such that \(\text{Null}(T) = \text{Range}(T)\)
  
  Consider \(T: \mathbb{R}^2 \to \mathbb{R}^2\) ( \(\dim(\mathbb{R}^2) = 2\), even).
  
  By exercise (2), if we want \(\text{Null}(T) = \text{Range}(T)\), we need \(T^2 = 0\).
  
  Consider \(T(x, y) = (y, 0)\).
  
  Clearly \(T\) is a linear map. Now let's see:
  - \(\text{Null}(T) = \{(x, y) : T(x, y) = (0, 0)\} = \{(x, y) : (y, 0) = (0, 0)\}\)
  \(= \{(x, 0) : x \in \mathbb{R}\} = \langle (1, 0) \rangle\).
  - \(\text{Range}(T) = \{T(x, y) : x, y \in \mathbb{R}\} = \{(y, 0) : x, y \in \mathbb{R}\} = \langle (1, 0) \rangle\).
  Therefore, \(\text{Null}(T) = \text{Range}(T)\).
Suppose \(T: \mathbb{F}^4 \to \mathbb{F}^2\) is a linear map with \(\text{Null}(T) = \{(x_1, x_2, x_3, x_4) \in \mathbb{F}^4 : x_1 = 5x_2 \text{ and } x_3 = 7x_4\}\).

Prove that \(T\) is surjective.

We have that \((x_1, x_2, x_3, x_4) \in \text{Null}(T) \iff (x_1, x_2, x_3, x_4) = (5x_2, x_2, 7x_4, x_4)\) \(= x_2(5, 1, 0, 0) + x_4(0, 0, 7, 1)\)

So, \(\text{Null}(T) = \langle (5, 1, 0, 0), (0, 0, 7, 1) \rangle \implies \dim(\text{Null}(T)) = 2\).

We know that (by the fundamental theorem of linear maps): \(\dim(\mathbb{F}^4) = \dim(\text{Null}(T)) + \dim(\text{Range}(T))\)

\(\implies 4 = 2 + \dim(\text{Range}(T))\)\(\implies \dim(\text{Range}(T)) = 2\).

And since \(\text{Range}(T) \subseteq \mathbb{F}^2\) (subspace), with \(\dim(\mathbb{F}^2) = 2 \implies \text{Range}(T) = \mathbb{F}^2\).

Therefore, \(T\) is surjective.
Prove that there does not exist a linear map \(T: \mathbb{F}^5 \to \mathbb{F}^2\) such that
\(\text{Null}(T) = \{(x_1, x_2, x_3, x_4, x_5) : x_1 = 3x_2, x_3 = x_4 = x_5\}\).

We have that \(\text{Null}(T) = \langle (3, 1, 0, 0, 0), (0, 0, 1, 1, 1) \rangle\)\(\implies \dim(\text{Null}(T)) = 2\).

By the dimension theorem, \(\dim(\mathbb{F}^5) = \dim(\text{Null}(T)) + \dim(\text{Range}(T))\)

\(\implies 5 = 2 + \dim(\text{Range}(T))\)\(\implies \dim(\text{Range}(T)) = 3\).

Contradiction, since \(\text{Range}(T) \subseteq \mathbb{F}^2\).

Therefore, such a linear map \(T\) does not exist.
Suppose \(\{v_1, \dots, v_n\} \subseteq V\) is a set of vectors.

Define \(T \in \mathcal{L}(\mathbb{F}^n, V)\) by \(T(\lambda_1, \dots, \lambda_n) = \lambda_1 v_1 + \dots + \lambda_n v_n\).
1. What property of \(T\) corresponds to \(\{v_1, \dots, v_n\}\) spanning \(V\)?
  
  We want \(\{v_1, \dots, v_n\}\) to span \(V\). This means, \(\forall v \in V\), \(\exists \lambda_1, \dots, \lambda_n \in \mathbb{F}\) such that \(v = \lambda_1 v_1 + \dots + \lambda_n v_n = T(\lambda_1, \dots, \lambda_n)\).
  
  So the property we need is: \(T\) is surjective.
  
  If \(T: \mathbb{F}^n \to V\) is surjective, by definition,
  
  \(\forall v \in V, \exists (\lambda_1, \dots, \lambda_n) \in \mathbb{F}^n\) such that \(T(\lambda_1, \dots, \lambda_n) = v\). 2. What property of \(T\) corresponds to \(\{v_1, \dots, v_n\}\) being linearly independent?
  
  We want to have that \(a_1 v_1 + \dots + a_n v_n = 0 \implies a_1 = \dots = a_n = 0\).
  
  So the property we need is: \(T\) is injective.
  
  Remember: \(T\) is injective \(\iff \text{Null}(T) = \{0\}\).
  
  So if \(T\) is injective, \(T(a_1, \dots, a_n) = 0 \iff (a_1, \dots, a_n) = (0, \dots, 0)\),
  
  which corresponds to \(a_1 v_1 + \dots + a_n v_n = 0\).

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