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.