Kernel subsets of transformations

[dustbin]

Homework Statement

Let $T_1,T_2:ℝ^n\rightarrowℝ^n$ be linear transformations. Show that $\exists S:ℝ^n\rightarrowℝ^n$ s.t. $T_1=S\circ T_2 \Longleftrightarrow kerT_2\subset kerT_1$.

The Attempt at a Solution

$(\Longrightarrow)$ Let $S:ℝ^n\rightarrowℝ^n$ be a linear transformation s.t. $T_1 = S\circ T_2$ and let $\vec{v}\in kerT_2$. Then $S(T_2(\vec{v})) = S(\vec{0}) = \vec{0}$ by linearity. Then $T_1(\vec{v}) = \vec{0}$. Thus $\vec{v}\in kerT_1 \quad \forall\vec{v}\in kerT_2$. Therefore $kerT_2 \subset kerT_1$.

$(\Longleftarrow)$ Suppose that $kerT_2\subset kerT_1$ and choose $S:ℝ^n\rightarrowℝ^n$ s.t. $S$ is linear and $T_1 = S\circ T_2$. Then for $\vec{v}\in kerT_2,\quad T_1(\vec{v}) = S(T_2(\vec{v}) = S(\vec{0}) = \vec{0}.$ Thus there exists such a transformation.

 

[Dick]

Homework Statement

Let $T_1,T_2:ℝ^n\rightarrowℝ^n$ be linear transformations. Show that $\exists S:ℝ^n\rightarrowℝ^n$ s.t. $T_1=S\circ T_2 \Longleftrightarrow kerT_2\subset kerT_1$.

The Attempt at a Solution

$(\Longrightarrow)$ Let $S:ℝ^n\rightarrowℝ^n$ be a linear transformation s.t. $T_1 = S\circ T_2$ and let $\vec{v}\in kerT_2$. Then $S(T_2(\vec{v})) = S(\vec{0}) = \vec{0}$ by linearity. Then $T_1(\vec{v}) = \vec{0}$. Thus $\vec{v}\in kerT_1 \quad \forall\vec{v}\in kerT_2$. Therefore $kerT_2 \subset kerT_1$.

$(\Longleftarrow)$ Suppose that $kerT_2\subset kerT_1$ and choose $S:ℝ^n\rightarrowℝ^n$ s.t. $S$ is linear and $T_1 = S\circ T_2$. Then for $\vec{v}\in kerT_2,\quad T_1(\vec{v}) = S(T_2(\vec{v}) = S(\vec{0}) = \vec{0}.$ Thus there exists such a transformation.

The first part seems ok. For the second, the problem is to show that such an S exists given ker(T2) is contained in ker(T1). Not to assume it exists.