mathematics

Schur product theorem

Schur product theorem is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Schur product theorem rather than just read about it. In short: In mathematics, particularly in linear algebra, the Schur product theorem states that the Hadamard product of two positive definite matrices is also a positive definite matrix. The result is named after Issai Schur (Schur 1911, p. 14, Theorem VII) (note that Schur signed as J.

Key takeaways

Schur product theorem belongs to mathematics; place it in that map before memorising details.
Learn the definition first, then one example that makes the definition concrete.
Connect Schur product theorem to a quantity you can measure, compute or draw — that is where exam questions come from.
Reproduce the core statement of Schur product theorem from memory before moving on to harder problems.

Reference excerpt

In mathematics, particularly in linear algebra, the Schur product theorem states that the Hadamard product of two positive definite matrices is also a positive definite matrix. The result is named after Issai Schur (Schur 1911, p. 14, Theorem VII) (note that Schur signed as J. Schur in Journal für die reine und angewandte Mathematik.) The converse of the theorem holds in the following sense: if M {\displaystyle M} is a symmetric matrix and the Hadamard product M ∘ N {\displaystyle M\circ N} is positive definite for all positive definite matrices N {\displaystyle N} , then M {\displaystyle M} itself is positive definite.

Proof

Proof using the trace formula For any matrices M {\displaystyle M} and N {\displaystyle N} , the Hadamard product M ∘ N {\displaystyle M\circ N} considered as a bilinear form acts on vectors a , b {\displaystyle a,b} as

a ∗ ( M ∘ N ) b = tr ⁡ ( M T diag ⁡ ( a ∗ ) N diag ⁡ ( b ) ) {\displaystyle a^{*}(M\circ N)b=\operatorname {tr} \left(M^{\textsf {T}}\operatorname {diag} \left(a^{*}\right)N\operatorname {diag} (b)\right)}

where tr {\displaystyle \operatorname {tr} } is the matrix trace and diag ⁡ ( a ) {\displaystyle \operatorname {diag} (a)} is the diagonal matrix having as diagonal entries the elements of a {\displaystyle a} . Suppose M {\displaystyle M} and N {\displaystyle N} are positive definite, and so Hermitian. We can consider their square-roots M 1 2 {\displaystyle M^{\frac {1}{2}}} and N 1 2 {\displaystyle N^{\frac {1}{2}}} , which are also Hermitian, and write

tr ⁡ ( M T diag ⁡ ( a ∗ ) N diag ⁡ ( b ) ) = tr ⁡ ( M ¯ 1 2 M ¯ 1 2 diag ⁡ ( a ∗ ) N 1 2 N 1 2 diag ⁡ ( b ) ) = tr ⁡ ( M ¯ 1 2 diag ⁡ ( a ∗ ) N 1 2 N 1 2 diag ⁡ ( b ) M ¯ 1 2 ) {\displaystyle \operatorname {tr} \left(M^{\textsf {T}}\operatorname {diag} \left(a^{*}\right)N\operatorname {diag} (b)\right)=\operatorname {tr} \left({\overline {M}}^{\frac {1}{2}}{\overline {M}}^{\frac {1}{2}}\operatorname {diag} \left(a^{*}\right)N^{\frac {1}{2}}N^{\frac {1}{2}}\operatorname {diag} (b)\right)=\operatorname {tr} \left({\overline {M}}^{\frac {1}{2}}\operatorname {diag} \left(a^{*}\right)N^{\frac {1}{2}}N^{\frac {1}{2}}\operatorname {diag} (b){\overline {M}}^{\frac {1}{2}}\right)}

… excerpt ends here. Continue reading the full article.

Worked examples

Example 1 — a first encounter with Schur product theorem

Start with the simplest possible case. Write down what Schur product theorem claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Schur product theorem before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Schur product theorem ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Schur product theorem

In research: Schur product theorem appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry: Engineering practice reuses Schur product theorem in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom: Schur product theorem is common in secondary-school and first-year university syllabi. It links to neighbouring topics Issai Schur, Matrix theory, Theorems in linear algebra, so understanding it makes those chapters shorter.
In everyday life: Look for Schur product theorem outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

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How to study Schur product theorem in 20 minutes

Read the reference excerpt below once, without taking notes.
Close the page and write down what Schur product theorem means in your own words.
Compare your version with the excerpt and mark what you missed.
Work through the three examples above with pen and paper.
Explain Schur product theorem out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Schur product theorem in simple terms?

Why does Schur product theorem matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Schur product theorem?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Schur product theorem.