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Suppose $\mathcal{V}$ be a subspaces of $\mathbb{R}^n$, Suppose $P: \mathbb{R}^n\to\mathbb{R}^n$ be a projection, then I need to prove the following

$$ \mathcal{V}\cap\text {im } P=P^{-1}\mathcal{V}\cap\text{im }P$$

Suppose $x\in \mathcal{V}\cap\text {im } P\Rightarrow x\in\mathcal{V}\text { and } \exists y\in \text{ im } P\ni x=Py\in \mathcal{V}\Rightarrow Px=Py=x=P^{-1}x\in P^{-1}\mathcal{V}\Rightarrow x\in P^{-1}\mathcal{V}\cap im P$

now, suppose $x\in P^{-1}\mathcal{V}\cap im P$ so $Py=x$

I am not able to prove please help.

  • 0
    If $P$ is a projection, it is not invertible in general... – 2017-01-21
  • 0
    I think $ P^{-1} $ is meant to be the reverse image here, not the inverse function – 2017-01-21

1 Answers 1

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I would go and try to prove both inclusions separately. So let's focus on $$ (\mathcal{V}\cap\text {im } P) \subset (P^{-1}\mathcal{V}\cap\text{im }P) $$ first.

$$ x \in (\mathcal{V}\cap\text {im } P) \Rightarrow x \in V \land x \in {im } P $$ $$ \Rightarrow x \in V \land \exists y \in \mathbb{R}^n : Py=x $$ By the definition of a projection we get: $$ \Rightarrow x \in V \land Py = Px = x $$ $$ \Rightarrow x \in (P^{-1}\mathcal{V}\cap\text{im }P) $$

Now for the second part I'll give you the following hint because P is a projection we know: $$ x \in \text {im } P \iff Px = x $$ I hope that will help you a bit.