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I will prove that for $T:H\rightarrow H$, A bounded linear operator on a Hilbert space H, $ker(T)$ and $range(T)$ are subspaces of $H$. Is this valid? I will appreciate any corrections that you all may point out.

$Definition:$

$Range(T)=\{Tx:x\in H\}$

$Ker(T) = \{x\in H: Tx=0\} = T^{-1}(\{0_h\})$

$Proof:$

Let $w_1$ and $w_2$ $\in$ $range(T)$ and $\lambda_1$, $\lambda_2 \in R$.

Then there are vectors $v_1, v_2 \in H$ with $Tv_1=w_1$ and $Tv_2=w_2$.

Since $H$ is a Vector/Hilbert Space, $\lambda_1v_1+\lambda_2v_2 \in H$.

Since $T$ is linear on $v$, we have

$T(\lambda_1v_1+\lambda_2v_2) = \lambda_1Tv_1+\lambda_2Tv_2 = \lambda_1w_1+\lambda_2w_2$

Thus, $\lambda_1w_1+\lambda_2w_2 \in Range(T)$

This shows that $Range(T)$ is a vector/hilbert space of $H$

Next, to show $ker(T)$ is a subspace of $H$, we must first show that $T^{-1}(U)$ is a subspace of $H$ where $U\subseteq H$.

Let $v_1$ and $v_2 \in$ $T^{-1}(U)$ and $\lambda_1$, $\lambda_2 \in R$.

Then $Tv_1$ and $Tv_2 \in U$

Since $U\subseteq H$, $\lambda_1Tv_1+\lambda_2Tv_2 \in U$

Since $T$ is Linear on $H$, we have

$T(\lambda_1v_1+\lambda_2v_2) =\lambda_1Tv_1+\lambda_2Tv_2 \in U$

Thus, $\lambda_1v_1+\lambda_2v_2 \in T^{-1}(U)$. This shows that $T^{-1}(U) \subseteq H$

From this, since $ker(T)=T^{-1}(\{0\}$, then the one-point set $\{0\} \subseteq H$

$Q.E.D.$

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1 Answer 1

up vote 1 down vote accepted

Your range proof is almost correct, you've done the esential part. You have proved that the range is stable under linear combinations. You also need to observe that $0=T(0)$ belongs to it (one does not call the empty set a subspace). Now this is a vector subspace.

Your kernel proof needs a precision. Your more general statement holds if $U$ is a vector subspace of $H$. Otherwise $T^{-1}(U)$ could fail to be a subspace. As an example, take $f(x)=x$ on $\mathbb{C}$ and $U=\{1\}$. Then $T^{-1}(U)=\{1\}$ is not a subspace. Now with the further assumption that $U$ is a subspace your steps are correct. Again, you have proven that $T^{-1}(U)$ is stable under linear combinations. It remains to note that $T(0)=0\in U$, hence $0$ belongs to $T^{-1}(U)$.

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Julien, thank you for your response. It does fail on the complex field and my proof is dealing only in the real analysis. As for the rest, I will take note of T(0) –  User69127 Mar 20 '13 at 3:47
    
@Zuniga For any Hilbert space, real or complex, and any linear operator $T$. Take any set $U$ which does not contain $0$. Then $0$ does not belong to $T^{-1}(U)$. But every subspace must contain $0$. So $T^{-1}(U)$ is not a subspace. This counterexample works for every vector space over any field. –  1015 Mar 20 '13 at 3:50
    
Ah yes you're right! A very nice counterexample. –  User69127 Mar 20 '13 at 3:58

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