This note is in the spirit of our paper on Robert Sheckley’s answerer for two orthogonal projections [4]. The meta theorem of that paper is that the two projections theorem of Halmos is something like Robert Sheckley’s answerer: no question about the \(W^*\)- and \(C^*\)-algebras generated by two orthogonal projections will go unanswered, provided the question is not foolish. The norm attainment problem asks whether for a given bounded linear operator A there is a unit vector x such that \(\Vert A x\Vert = \Vert A\Vert \). In this generality, a useful answer is not available. Here we pose the question for the case where A is a function of two orthogonal projections or a function of a single skew projection and its adjoint. (Skew projections are frequently also referred to as oblique projections.)

Let P and Q be orthogonal projections acting on a real or complex Hilbert space \({\mathcal {H}}\). According to Halmos’ “Two projections theorem” (see [8] and consult [3, 10] for the history and more on the subject), there is a representation of \({\mathcal {H}}\) as an orthogonal sum

$$\begin{aligned} {\mathcal {H}} = M_{00}\oplus M_{01}\oplus M_{10}\oplus M_{11}\oplus M\oplus M \end{aligned}$$
(1)

(the last two summands have the same dimension and are thus identified via an appropriate unitary similarity) with respect to which

$$\begin{aligned} P = (1,1,0,0)\oplus \begin{bmatrix}I &{} 0 \\ 0 &{} 0\end{bmatrix} \end{aligned}$$
(2)

and

$$\begin{aligned} Q = (1,0,1,0)\oplus \begin{bmatrix}I-H &{} \sqrt{H(I-H)}\, \\ \sqrt{H(I-H)} &{} H\,\end{bmatrix}. \end{aligned}$$
(3)

Here and below we use the string \((a_{00},a_{01},a_{00},a_{11})\) as an abbreviation for \(a_{00}I_{M_{00}}\oplus a_{01}I_{M_{01}}\oplus a_{10}I_{M_{10}}\oplus a_{11}I_{M_{11}}\), while the blocks of the matrix component in (2),(3) are operators on M. The selfadjoint operator H has the spectrum \(\sigma (H)\subseteq [0,1]\), with 0, 1 not being its eigenvalues. In particular, \(\min \sigma (H)<1\). Note that \(M=\{0\}\) if and only if P and Q commute. The other four subspaces \(M_{ij}\) also may or may not be actually present in (1); we will let \(\Lambda =\{ (i,j):\dim M_{ij}\ne 0\}\).

According to the Giles-Kummer theorem (see [7] or [3, Theorem 7.1]), the von Neumann algebra \(W^*(P,Q)\) generated by P and Q consists of the operators A admitting the representation

$$\begin{aligned} A = (a_{00},a_{01},a_{00},a_{11})\oplus \begin{bmatrix}\phi _{00}(H) &{} \phi _{01}(H) \\ \phi _{10}(H) &{} \phi _{11}(H)\end{bmatrix} \end{aligned}$$
(4)

with respect to the decomposition (1) of \({\mathcal {H}}\). Here \(a_{ij}\) are arbitrary complex numbers and \(\phi _{ij}\) are (also arbitrary) functions in \(L^\infty (\sigma (H))\) with respect to the spectral measure of H. We will sometimes write the rightmost summand in (4) as \(\Phi _A(H)\).

The norms of operators from \(W^*(P,Q)\) were computed in [11, Theorem 10], see also [3, Theorem 7.9]. Namely, for A as in (4),

$$\begin{aligned} \left\| A\right\| =\max \left\{ \max _{(j,k)\in \Lambda }\left| a_{jk}\right| ,\max _{x\in \sigma (H)} \sqrt{\frac{\phi (x)+\sqrt{\phi (x)^2-4\left| \omega (x)\right| ^2}}{2}} \right\} . \end{aligned}$$
(5)

Here \(\phi =\sum _{i,j=0,1}\left| \phi _{ij}\right| ^2\) is the square of the Frobenius norm of \(\Phi _A\) and \(\omega =\det \Phi _A=\phi _{00}\phi _{11}-\phi _{01}\phi _{10}\). Note that always \(\phi (x)^2 -4|\omega (x)|^2 \ge 0\).

The question we are addressing here is: when is \(\left\| A\right\| \) attained, i.e., when does there exist a unit vector \(x\in {\mathcal {H}}\) such that \(\left\| Ax\right\| =\left\| A\right\| \)? We will call A a norm attaining operator and write \(A\in {\mathcal {N}}\) if this happens to be the case.

For our purposes it is useful to recall that formula (5) was derived in [11] from the fact that

$$\begin{aligned} \lambda _{\max }:=\max _{x\in \sigma (H)}\frac{\phi (x)+\sqrt{\phi (x)^2-4\left| \omega (x)\right| ^2}}{2} \end{aligned}$$

is the right endpoint of the spectrum \(\sigma \left( \Phi _{A^*A}(H)\right) \). Since for every operator X acting on \({\mathcal {H}}\), we have \(X\in {\mathcal {N}}\) if and only if \(X^*X\in {\mathcal {N}}\) if and only if the right endpoint of \(\sigma (X^*X)\) is its eigenvalue (see [6] and [9]), we just need to figure out when \(\lambda _{\max }\) is (or is not) an eigenvalue of \(A^*A\).

To this end, recall that for operators in \(W^*(P,Q)\), the description of their kernels is also known ([3, Theorem 7.5] or [11, Theorem 1]). There is no need to include its exact form here. A consequence which is important for us reads as follows.

FormalPara Lemma 1

The kernel of \(\Phi _A(H)\) is non-trivial if and only if the spectral measure of \(\{x \in \sigma (H): \omega (x)=0\}\) is non-zero.

When applied to \(A-\lambda I\) in place of A, Lemma 1 immediately yields the following.

FormalPara Lemma 2

Let A be given by (4). Then \(\lambda \in {\mathbb {C}}\) is an eigenvalue of A if and only if the spectral measure of the set \(\{x\in \sigma (H):\lambda ^2-{\text {trace}}\Phi _A(x)\lambda +\det \Phi _A(x)=0\}\) is non-zero.

Applying Lemma 2 to \(A^*A\) and taking into account that \({\text {trace}}\Phi _{A^*A}=\phi \) and \(\det \Phi _{A^*A}=\left| \omega \right| ^2\), we see that the eigenvalues \(\lambda \) of \(A^*A\) are characterized by the property that the spectral measure of the set \(\{x\in \sigma (H):\lambda ^2-\phi (x)\lambda +\left| \omega (x)\right| ^2=0\}\) is non-zero. In particular, \(\lambda _{\max }\) is an eigenvalue of \(A^*A\) (and not just a point of its spectrum) if and only if the spectral measure of the set of all \(x\in \sigma (H)\) on which the function

$$\begin{aligned} \psi (x):=\phi (x)+\sqrt{\phi (x)^2-4\left| \omega (x)\right| ^2} \end{aligned}$$
(6)

attains its maximum value is non-zero. Denoting this set by \(\Sigma (A)\), we arrive at the following conclusion.

FormalPara Theorem 3

Let A be the operator given by (4). Then \(A\in {\mathcal {N}}\) if and only if either (i) \(\max _{(j,k)\in \Lambda }\left| a_{jk}\right| \ge \sqrt{\lambda _{\max }}\) or (ii) \(\max _{(j,k)\in \Lambda }\left| a_{jk}\right| <\sqrt{\lambda _{\max }}\) and the spectral measure of \(\Sigma (A)\) is non-zero.

Let now T be a skew projection on \({\mathcal {H}}\). We assume that T is genuinely skew, which is equivalent to the requirement that \(\Vert T\Vert >1\). Denote by \(P=P_{\mathrm{Ran}\,T}\) the orthogonal projection onto the range of T and by \(Q=P_{\mathrm{Ker}\,T}\) the orthogonal projection onto the kernel of T. Afriat [1] (see also [3, Proposition 1.6]) showed that then \(\Vert PQ\Vert < 1\) and

$$\begin{aligned} T =(I-PQ)^{-1}P(I-PQ). \end{aligned}$$
(7)

Moreover, in (1) then \(M_{00}=M_{11}=\{0\}\), while \(M_{01}\) and \(M_{10}\) may or may not be present. Omitting the \(M_{00}\) and \(M_{11}\) terms in (2) and (3), we obtain

$$\begin{aligned} I-PQ= (1,1)\oplus \begin{bmatrix}H &{} -\sqrt{H(I-H)}\, \\ 0 &{} I \,\end{bmatrix}, \end{aligned}$$

and since \(I-PQ\) is invertible, so also must be H. Formula (7) then gives

$$\begin{aligned} T= (1,0)\oplus \begin{bmatrix}I &{} -\sqrt{H^{-1}(I-H)}\, \\ 0 &{} 0 \,\end{bmatrix} =(1,0)\oplus \begin{bmatrix}I &{} -\sqrt{H^{-1}-I}\, \\ 0 &{} 0 \,\end{bmatrix}. \end{aligned}$$
(8)
FormalPara Corollary 4

A skew projection attains its norm if and only if \(\min \sigma (H)\) is an eigenvalue of the respective operator H.

FormalPara Proof

Indeed, for \(A=T\), we get from (8) that \(\phi (x)=x^{-1}\), \(\omega (x)=0\), and hence \(\psi (x)=2x^{-1}\), which is a monotonically decreasing function. It follows that \(\Sigma (T)\) is the singleton \(\{\min \sigma (H)\}\). Since \(\left\| T\right\| >1\), part (i) of Theorem 3 is irrelevant and the assertion follows from part (ii) of that theorem. \(\square \)

From (2) and (3) and the equalities \(M_{00}=M_{11}=\{0\}\), we infer that

$$\begin{aligned} PQP=(0,0) \oplus \begin{bmatrix}I-H &{} 0\, \\ 0 &{} 0 \,\end{bmatrix}. \end{aligned}$$

Thus, if \(A=T\), we see that the respective operator H in Corollary 4 is

$$\begin{aligned} H=(I-PQP)|\mathrm{Ran}\,P = I|\mathrm{Ran}\, T-P_{\mathrm{Ran}\,T}P_{\mathrm{Ker}\,T}|\mathrm{Ran}\,T. \end{aligned}$$

We conclude with some examples. The authors of [2] recently proved that a skew projection T is in \({\mathcal {N}}\) if and only if the selfadjoint operator \(T+T^*-I\) belongs to \({\mathcal {N}}\). The operator \(T+T^*-I\) appeared in [5] and is therefore called the Buckholtz operator in [2]. The following is an extension of this result.

FormalPara Example 5

Let T be a skew projection. Then the following are equivalent:

  1. (i)

    \(T \in {\mathcal {N}}\),

  2. (ii)

    \(T +\alpha T^* +\beta I \in {\mathcal {N}}\) for some \(\alpha ,\beta \in {\mathbb {R}}\),

  3. (iii)

    \(T +\alpha T^* +\beta I \in {\mathcal {N}}\) for all \(\alpha ,\beta \in {\mathbb {R}}\).

FormalPara Proof

Fix \(\alpha ,\beta \in {\mathbb {R}}\). We have to show that \(T \in {\mathcal {N}} \Longleftrightarrow T+\alpha T^*+\beta I \in {\mathcal {N}}\). According to (8),

$$\begin{aligned} A:=T+\alpha T^*+\beta I= (1+\alpha +\beta ,\beta )\oplus \begin{bmatrix}(1+\alpha +\beta )I &{} -\sqrt{H^{-1}-I} \\ -\alpha \sqrt{H^{-1}-I} &{} \beta I \end{bmatrix}. \end{aligned}$$

Abbreviating \(1+\alpha +\beta \) to s and \(x^{-1}-1\) to f(x), we obtain

$$\begin{aligned} \phi (x)= & {} (1+\alpha +\beta )^2+\beta ^2+(1+\alpha ^2)(x^{-1}-1)=s^2+\beta ^2+(1+\alpha ^2)f(x),\\ \omega (x)= & {} (1+\alpha +\beta )\beta -\alpha (x^{-1}-1)=s\beta -\alpha f(x),\\ \psi (x)= & {} s^2+\beta ^2+(1+\alpha ^2)f(x)\\&+\sqrt{(s^2+\beta ^2+(1+\alpha ^2)f(x))^2-4(s\beta -\alpha f(x))^2}. \end{aligned}$$

The term under the square root equals

$$\begin{aligned} (s^2-\beta ^2)^2+2[(s^2+\beta ^2)(1+\alpha ^2)+4s\beta \alpha ]f(x)+(1-\alpha ^2)^2 f(x)^2. \end{aligned}$$

The function f(x) is monotonically decreasing and non-negative. Since \(2|s\beta | \le s^2+\beta ^2\) and \(2|\alpha |\le 1+\alpha ^2\), we have \((s^2+\beta ^2)(1+\alpha ^2)+4s\beta \alpha \ge 0\). Consequently, the term under the square root and therefore also \(\psi (x)\) are monotonically decreasing, which implies that \(\Sigma (A)=\{\min \sigma (H)\}\). Using that \(\min \sigma (H) <1\), we get

$$\begin{aligned} \lambda _{\max }\ge & {} \frac{s^2+\beta ^2+(1+\alpha ^2)f(\min \sigma (H))+|s^2-\beta ^2|}{2}\\> & {} \frac{s^2+\beta ^2+|s^2-\beta ^2|}{2}=\max (s^2,\beta ^2), \end{aligned}$$

and Theorem  3(ii) implies that \(A \in {\mathcal {N}}\) if and only if the spectral measure of the singleton \(\{\min \sigma (H)\}\) is positive, that is, if and only if \(\min \sigma (H)\) is an eigenvalue of H. This together with Corollary 4 shows that \(A \in {\mathcal {N}} \Longleftrightarrow T \in {\mathcal {N}}\), as desired. \(\square \)

FormalPara Example 6

Let T be a skew projection. Put \(T^{(2)}=TT^*\), \(T^{(3)}=TT^*T\), and more generally, \(T^{(m)}:=\underbrace{TT^*TT^*T\cdots }_{m}\). Then, for each m,

$$\begin{aligned} T^{(m)} \in {\mathcal {N}} \Longleftrightarrow T \in {\mathcal {N}}. \end{aligned}$$
FormalPara Proof

From (8), we infer that if \(m=2k\) is even, then

$$\begin{aligned} T^{(m)}=(TT^*)^k= (1,0)\oplus \begin{bmatrix}H^{-k} &{} 0 \\ 0 &{} 0 \,\end{bmatrix}. \end{aligned}$$

This implies that \(\psi (x)=2x^{-2k}\) and thus \(\Sigma (T^{(m)})=\{\min \sigma (H)\}\). Taking into account that \(\min \sigma (H) <1\), we obtain as above from Theorem 3(ii) and Corollary 4 that \(T^{(m)} \in {\mathcal {N}} \Longleftrightarrow T\in {\mathcal {N}}\). Finally, since \(A\in {\mathcal {N}} \Longleftrightarrow AA^*\in {\mathcal {N}}\), we conclude that

$$\begin{aligned} T^{(2k+1)} \in {\mathcal {N}} \Longleftrightarrow T^{(4k+2)} \in {\mathcal {N}} \Longleftrightarrow T \in {\mathcal {N}}, \end{aligned}$$

which gives the assertion in the case of odd m. \(\square \)

FormalPara Example 7

Let \(\{\omega _n\}_{n=1}^\infty \) be a sequence of positive real numbers that converge monotonically to zero and let T be the skew projection on \(\ell ^2({\mathbb {N}})\) defined by the infinite matrix

$$\begin{aligned} T={\text {diag}}\left\{ \begin{bmatrix}1 &{} -\omega _n \\ 0 &{} 0 \end{bmatrix}_{n=1}^\infty \right\} . \end{aligned}$$

Then \(T \in {\mathcal {N}}\). Put \(A=TT^*+T^*T-T-T^*-I\). If \(\omega _n=1/n\), then \(A \notin {\mathcal {N}}\), but if \(\omega _n=2/n\), then \(A \in {\mathcal {N}}\).

FormalPara Proof

It is clear that \(T \in {\mathcal {N}}\): the norm is attained at the vector

$$\begin{aligned} (1/\sqrt{1+\omega _1^2}, -\omega _1/\sqrt{1+\omega _1^2}, 0,0, \ldots )^\top \in \ell ^2({\mathbb {N}}). \end{aligned}$$

To treat the operator A, we employ Theorem 3. We have \(M_{ij}=\{0\}\) for all i, j and may therefore write

$$\begin{aligned} T=\begin{bmatrix}I &{} -\sqrt{H^{-1}-I}\, \\ 0 &{} 0 \,\end{bmatrix} \quad \text{ with }\quad H={\text {diag}}(x_n)_{n=1}^\infty , \end{aligned}$$

where \(\sqrt{x_n^{-1}-1}=\omega _n\), that is, \(x_n=1/(1+\omega _n^2)\). Straightforward computation gives

$$\begin{aligned} A=\begin{bmatrix}H^{-1}-2I &{} 0\, \\ 0 &{} H^{-1}-2I \,\end{bmatrix}. \end{aligned}$$

Thus, \(\psi (x)=2(2-x^{-1})^2\). If \(\omega _n=1/n\), then

$$\begin{aligned} \sigma (H)=\left\{ \frac{n^2}{n^2+1}: n \in {\mathbb {N}}\right\} \cup \{1\} =\left\{ \frac{1}{2}, \frac{4}{5},\frac{9}{10}, \ldots \right\} \cup \{1\}. \end{aligned}$$

The function \(\psi (x)\) takes its maximum on \(\sigma (H)\) at \(x=1\), with \(\lambda _{\max }=\psi (1)/2=1\). Hence \(\Sigma (A)=\{1\}\), and as 1 is not an eigenvalue of H, Theorem 3(ii) implies that \(A \notin {\mathcal {N}}\). If \(\omega _n=2/n\), we have

$$\begin{aligned}\sigma (H)=\left\{ \frac{n^2}{n^2+4}: n \in {\mathbb {N}}\right\} \cup \{1\} =\left\{ \frac{1}{5}, \frac{1}{2},\frac{9}{13}, \ldots \right\} \cup \{1\}.\end{aligned}$$

This time \(\psi (x)\) assumes its maximum at \(x=1/5\), the value of the maximum being \(\lambda _{\max }=\psi (1/5)/2=3.24\). It follows that \(\Sigma (A)=\{1/5\}\), and since 1/5 is an eigenvalue of H, we deduce from Theorem 3(ii) that \(A \in {\mathcal {N}}\). \(\square \)

The last example can be elaborated to great extent. However, we leave it with Israel M. Gelfand: “Explain this to me on a simple example; the difficult example I will be able to do on my own.” (http://www.israelmgelfand.com/edu_work.html)