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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.


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

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


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 $\{0\} \subseteq H$


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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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