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Let $\mathcal{H}$ be a Hilbert space and $H_1 ,H_2$ are two closed subspaces of $\mathcal{H}$. Then $\mathcal{H}=H_1\oplus H_1^\perp=H_2\oplus H_2^\perp.$

My questions:

  1. If $dim(H_1)=dim(H_2)$ is finite, then is $dim(H_1^\perp)=dim(H_2^\perp)$ true ?
  2. If $dim(H_1)=dim(H_2)$ , then does there exist an invertible operator $T :H_1\rightarrow H_2?$

If $\mathcal{H}$ is finite dimensional , both are true . If $\mathcal{H}$ is infinite dimensional, I don't know whether my 1st question is valid or not. Kindly make me clarify.

Thanks in advance.

B.B.S
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  • Assuming that the dimension is defined as $\infty$ if the space is infinite dimensional: The first question is still valid and both dimensions must be equal to the same value (which one?) if $\mathcal{H}$ is infinite dimensional. You can show this by assuming that this was not the case. – jd27 Jun 17 '23 at 17:45

1 Answers1

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The answer is yes to both. For any orthonormal subset $S\subseteq H$ there exists, by Zorn's lemma, a maximal orthonormal subset $T$ containing $S$. A maximal orthonormal subset $T$ of $H$ is an orthonormal basis. To see this, let $W$ be the closure of the linear span of $T$. If $W\ne H$ then $W^\perp \ne \{0\}$, and any vector in $W^\perp$ of norm $1$ extends the orthonormal set $T$ - contradicting its maximality.

All orthonormal bases of $H$ have the same cardinality. See here.

So let $V_1$ be an orthonormal basis of $H_1$ and $V_2$ be an orthonormal basis of $H_2$. Since $\dim H_1 = \dim H_2$, $|V_1| = |V_2|$. $V_1$ and $V_2$ each extend to orthonormal bases $B_1, B_2$ of $H$. Since all orthonormal bases have the same cardinality $|B_1| = |B_2|$, and since $|V_1|=|V_2|$ are finite, by cardinal arithmetic we must have $\kappa=|B_1\setminus V_1|=|B_2\setminus V_2|$. So $H_1^\perp, H_2^\perp$ have the same orthogonal dimension $\kappa$. If $\Pi:B_1\setminus V_1\to B_2\setminus V_2$ is any bijection, then $\Pi$ extends uniquely to a partial isometry of $H_1^\perp$ onto $H_2^\perp$. Any invertible operator $T:H_1\to H_2$ can be extended to an invertible operator on $H$ by acting as $\Pi$ on $H_1^\perp$.

Chad K
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  • In last paragraph , u have mentioned $\pi$ extends uniquely to a unitary mapping .That means if I set $\pi(f_i)=g_j ,$ then $\pi^(g_j)=f_i$ such that $\pi \pi^=I $ and $\pi^*\pi=I$. Am i right ? – B.B.S Jun 18 '23 at 00:56
  • Yes, nearly, but I changed the term to 'partial isometry' instead of 'unitary mapping'. $\Pi^* \Pi = P_{H_1^\perp}$ and $\Pi ,\Pi^* = P_{H_2^\perp}$ where $P_{xx}$ are the orthogonal projections on these subspaces. – Chad K Jun 18 '23 at 04:36