What Are the Properties of Linear Transformations and Subspaces?

[Ted123]

Homework Statement

Suppose $\phi : V \to W$ is a linear transformation over a field $\mathbb{K}$ .

(a) (i) Define the kernel of $\phi, \; Ker(\phi)$. [2 marks]

(ii) Show that $\phi$ is 1-1 if and only if $Ker ( \phi )=\{\bf{0}\}$. [4 marks]

(b) Suppose X is a subspace of W. Define the set $U_X$ by

$U_X = \{ v\in V : \phi (v) \in X \}$ .

Show that $U_X$ is a subspace containing $Ker ( \phi )$. [7 marks]

(c) Suppose that Y is another subspace of W. Show that $U_X \cap U_Y = U_{X \cap Y}$. [4 marks]

(d) What does it mean to say that W is the direct sum of two subspaces X and Y (written $W=X \oplus Y$ )? [2 marks]

(e) Show that if $W=X \oplus Y$ then $U_X \cap U_Y = Ker(\phi)$. [6 marks]

Total [25 marks]

The Attempt at a Solution

Done (a)

For (b), $\phi(\bf{0})=\bf{0} \in X$ since X is a subspace, so $\bf{0}\in U_X$ and $U_X$ is non-empty.

Suppose $u,v\in U_X$ .

Then $\phi(u+v) = \phi(u) + \phi(v) \in X$ since X is a subspace, so $U_X$ is closed under vector addition.

Suppose $\alpha \in \mathbb{K}$ and $v\in U_X$.

Then $\phi(\alpha v) = \alpha \phi(v) \in X$ since X is a subspace.

Hence $U_X$ is a subspace containing $\bf{0}$ and hence $Ker( \phi )$ .

For (c) is this alright?

$U_X \cap U_Y = \{ v\in V : \phi(v) \in X \} \cap \{ v\in V : \phi(v) \in Y \}$

$= \{ v\in V : \phi(v) \in X\;\text{and}\;\phi(v)\in Y \}$

$= \{ v\in V : \phi(v) \in X \cap Y \} = U_{X\cap Y}$

For (d), the direct sum $W=X \oplus Y$ means $W=X+Y$ and $X\cap Y = \{ \bf {0}\}$

For (e) is this alright?

If $W=X \oplus Y$ then $X\cap Y = \{ \bf {0}\}$ . So

$U_X \cap U_Y = U_{X \cap Y}$ (by part (c))

$= U_{\{\bf{0}\}} = \{v\in V : \phi(v)\in \{\bf{0}\} \} = \{v\in V : \phi(v) = \bf{0} \} = Ker( \phi )$

 

[mighty2000]

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Your solution looks good overall. Here are a few minor suggestions:

- In part (a), you could also mention that \phi is 1-1 if and only if \text{Ker}(\phi)=\{\bf{0}\} because this means that the only element in the kernel is the zero vector, which implies that \phi is injective.
- In part (c), you could add a little more explanation as to why U_X \cap U_Y = \{ v\in V : \phi(v) \in X \} \cap \{ v\in V : \phi(v) \in Y \} = \{ v\in V : \phi(v) \in X\;\text{and}\;\phi(v)\in Y \}.
- In part (d), you could clarify that X+Y is the sum of the subspaces X and Y, and X\cap Y = \{\bf{0}\} means that the only element in common between X and Y is the zero vector.
- In part (e), you could mention that U_X \cap U_Y = U_{X \cap Y} because X\cap Y = \{\bf{0}\} implies that any vector that satisfies both \phi(v)\in X and \phi(v)\in Y must also satisfy \phi(v)\in \{\bf{0}\}.