## 11.10 *Remarks and exercises*

### 11.10.1 Operator decompositions

Analogously to how we can factor polynomials into linear parts, or factor numbers into prime divisors, we can “factor” matrices into smaller components.
Doing so often helps us to better understand the geometry of the situation: we might be able to understand the transformation described by a single matrix as “some reflection, followed by some rotation, followed by some scaling”.
For us, one specific use of such a “factorisation” (known formally as an **operator decomposition**) is in better understanding various operator norms, as we explain in 11.10.2.

Here are three operator decompositions that are particularly useful in quantum information theory.
The second is for arbitrary operators between Hilbert spaces, the first and third are for *normal endomorphisms* (i.e. normal operators from one Hilbert space to itself).

**Spectral decomposition.**Recall Section 4.5: the spectral theorem tells us that every*normal*operatorA\in\mathcal{B}(\mathcal{H}) can be expressed as a linear combination of projections onto pairwise orthogonal subspaces. We write the spectral decomposition ofA asA = \sum_k \lambda_k |v_k\rangle\langle v_k| where\lambda_k are the eigenvalues ofA , with corresponding eigenvectors|v_k\rangle , which form an orthonormal basis in\mathcal{H} .In matrix notation, we can write this as

A = UDU^\dagger whereD is the diagonal matrix whose diagonal entries are the eigenvalues\lambda_k , and whereU is the unitary matrix whose columns are the eigenvectors|v_k\rangle .**Singular value decomposition (SVD).**We have already mentioned the SVD in 5.9.11 when discussing the Schmidt decomposition, but we recall the details here. Consider*any*(non-zero) operatorA\in\mathcal{B}(\mathcal{H},\mathcal{H}') . From this, we can construct two positive semi-definite operators:A^\dagger A\in\mathcal{B}(\mathcal{H}) andAA^\dagger\in\mathcal{B}(\mathcal{H}') . These are both normal, and so we can apply the spectral decomposition to both. In particular, if we denote the eigenvalues ofA^\dagger A by\lambda_k , and the corresponding eigenvectors by|v_k\rangle , then we see that the vectors|u_k\rangle \coloneqq \frac{1}{\sqrt{\lambda_k}}A|v_k\rangle form an orthonormal system in\mathcal{H}' (and are, in fact, eigenvectors ofAA^\dagger ), since\begin{aligned} \langle u_i|u_j\rangle &= \frac{1}{\sqrt{\lambda_i}\sqrt{\lambda_j}} \langle v_i|A^\dagger A|v_j\rangle \\&= \frac{\lambda_j}{\sqrt{\lambda_i}\sqrt{\lambda_j}} \langle v_i|v_j\rangle \\&= \delta_{ij}. \end{aligned} We define the**singular values**s_k ofA to be the square roots of the eigenvalues ofA^\dagger A , i.e.s_k^2=\lambda_k . These singular values satisfyA|v_k\rangle=s_k|u_k\rangle by construction, and so we can writeA = \sum_k s_k|u_k\rangle\langle v_k| which we call the**singular value decomposition**(or**SVD**). This decomposition holds for arbitrary (non-zero) operators as opposed to just normal ones, and also for operators between two different Hilbert spaces as opposed to just endomorphisms. In words, this decomposition says that, givenA , we can find orthonormal bases of\mathcal{H} and\mathcal{H}' such thatA maps thek -th basis vector of\mathcal{H} to a non-negative multiple of thek -th basis vector of\mathcal{H}' (and sends any left over basis vectors to0 , if\dim\mathcal{H}>\dim\mathcal{H}' ).In matrix notation, we can write this as

A = U\sqrt{D}V^\dagger whereD is the diagonal matrix of eigenvalues (and so\sqrt{D} is the diagonal matrix of*singular*values), and bothU andV are unitary.Geometrically, we are decomposing any linear transformation into a composition of a rotation or reflection

V^\dagger , followed by a scaling by the singular values\sqrt{D} , followed by another rotation or reflectionU . This maps the unit sphere in\mathcal{H} onto an ellipsoid in\mathcal{H}' , and the singular values ofA are exactly the lengths of the semi-axes of this ellipsoid.**Polar decomposition.**LetA\in\mathcal{B}(\mathcal{H}) be a normal arbitrary operator. Since it is an endomorphism, it is represented by a square matrix. Forgetting thatA is normal for a moment, we know that its SVD takes the form\begin{aligned} A &= U\sqrt{A^\dagger A} \\&= \sqrt{AA^\dagger}U \end{aligned} where the unitary matrixU connects the two eigenbases:U=\sum_k|u_k\rangle\langle v_k| . We shall return to this unitaryU shortly.

Since **modulus** as
**polar decomposition**

If we decompose the eigenvalues of

### 11.10.2 More operator norms

We have already seen, all the way back in 1.10.2^{198}, how the Euclidean norm (from which we get the Euclidean distance) is the special case ** \ell^p-norms**), where

You might recall that we named the Cauchy–Schwartz inequality as arguably the most useful inequality in analysis.
Well it turns out that it is actually the special case

**Hölder’s inequality**.
Let ^{199}

We will come back to the relevance of these

Throughout, let

**Spectral norm.**This one is so frequently used that it is often simply called the**operator**norm and denoted simply by\|\cdot\| . It is the maximum length of the vectorA|v\rangle over all possible normalised vectors|v\rangle\in\mathcal{H} , i.e.\|A\| \coloneqq \max_{|v\rangle\in S_{\mathcal{H}}^1}\Big\{ |A|v\rangle| \Big\} (whereS_{\mathcal{H}}^1 is the unit sphere in\mathcal{H} , i.e. the set of vectors of norm1 ). From this definition, one can actually show that the norm is given by the largest singular value:\|A\| = \max_k s_k. **Trace norm.**This is given by the sum of the singular values ofA , i.e.\|A\|_{\operatorname{tr}} \coloneqq \sum_k s_k but note that we can rewrite this using the polar decomposition (from Section 11.10.1) as simply\|A\|_{\operatorname{tr}} = \operatorname{tr}|A|. **Frobenius norm.**We have mentioned a few times how inner products give rise to norms, and you might remember that we introduced an inner product on\mathcal{B}(\mathcal{H}) a while ago: the Hilbert–Schmidt norm^{200}\begin{aligned} (A|B) &\coloneqq \operatorname{tr}A^\dagger B \\&= \sum_{i,j} A_{ij}^\star B_{ji}. \end{aligned} The Frobenius norm is the norm induced by this inner product, i.e.\begin{aligned} \|A\|_F &\coloneqq \sqrt{(A|A)} \\&= \sqrt{\operatorname{tr}(A^\dagger A)} \\&= \sqrt{\sum_{i,j}|A_{ij}|^2}. \end{aligned}

Let’s study the relation between the operator norm and the trace norm first.
By definition, we see that
**Hölder’s inequality for matrices**.
To derive this inequality, we can use the SVD of

We can actually use this inequality to recover either the operator or the trace norm, by maximising: for any fixed **variational** characterisation of the trace norm which is very useful at times:

One final special case to point out is what happens if

Now we finally return to the relevance of **Schatten p-norms**.
For

### 11.10.3 Fidelity in a trace norm inequality

There is a useful inequality involving the trace norm:

Let

Writing ^{201}

One particularly nice application of this inequality arises when we take

### 11.10.4 Hamming distance

Show that the Hamming distance (defined in Section 11.1) is indeed a metric.

### 11.10.5 Operator norm

Prove the following properties of the operator norm:

\|A\otimes B\|=\|A\|\|B\| for any operatorsA andB - If
A is normal, then\|A^\dagger\|=\|A\| - If
U is unitary, then\|U\|=1 - If
P\neq0 is an orthogonal projector, then\|P\|=1 .

Using the singular value decomposition^{202}, or otherwise, prove that the operator norm has the following two properties for any operators

**Unitary invariance:**\|UAV\|=\|A\| for any unitariesU andV **Sub-multiplicativity:**\|AB\|\leqslant\|A\|\|B\| .

Recall that we say that

- Prove that, if
V approximatesU with precision\varepsilon , thenV^{-1} approximatesU^{-1} with the same precision\varepsilon .

Using the Cauchy–Schwartz inequality, or otherwise, prove the following, for any vector

|\langle\psi|A^\dagger B|\psi\rangle|\leqslant\|A\|\|B\| .

### 11.10.6 Tolerance and precision

Suppose we wish to implement a quantum circuit consisting of gates

We want our approximate circuit to be within some **tolerance**

How small must

*Hint: recall that |p_U-p_V|\leqslant 2\|U-V\|.*

### 11.10.7 Statistical distance and a special event

- Show that, if
p andq are probability distributions on the same sample space\Omega , thend(p,q) = \max_{A\subseteq\Omega}\{|p(A)-q(A)|\}. - By definition, the above maximum is realised for some specific subset
A\subseteq\Omega , i.e. there exists some event (described by the set of outcomesA ) that is optimal in distinguishingp fromq . What is this event?

### 11.10.8 Joint probability distributions

If we simultaneously sample *two* random variables from the same probability space, then we obtain a **joint distribution**:
**marginals**

So let *Hint:*

### 11.10.9 Distinguishability and the trace distance

Say we have a physical system which is been prepared in one of two states (say, *single* measurement can distinguish between the two preparations with probability *at most*

How does this probability change if the states

\rho_0 and\rho_1 are*not*equally liked, but instead sent with some predetermined probabilitiesp_0 andp_1 , respectively?Suppose that you are given one randomly selected qubit from a pair in the state

^{203}|\psi\rangle = \frac{1}{\sqrt{2}}\left( |0\rangle\otimes\left( \sqrt{\frac23}|0\rangle - \sqrt{\frac13}|1\rangle \right) + |1\rangle\otimes\left( \sqrt{\frac23}|0\rangle + \sqrt{\frac13}|1\rangle \right) \right). What is the maximal probability with which we can determine which qubit (either the first or the second) we were given?

Think how far you’ve come since then!↩︎

The technical statement of Hölder’s inequality is really for something known as a

**measure space**, wherev andw are*functions*, but the details aren’t relevant for us here.↩︎Here we drop the factor of

\frac{1}{2} that we sometimes included for simplifying certain calculations.↩︎This is the non-commutative version of the identity

a^2-b^2=(a+b)(a-b) .↩︎