{"id":37663,"date":"2026-01-14T14:24:20","date_gmt":"2026-01-14T14:24:20","guid":{"rendered":"https:\/\/mp.moonpreneur.com\/math-corner\/?p=37663"},"modified":"2026-02-25T09:15:04","modified_gmt":"2026-02-25T09:15:04","slug":"sherman-morrison-woodbury-identity","status":"publish","type":"post","link":"https:\/\/mp.moonpreneur.com\/math-corner\/sherman-morrison-woodbury-identity\/","title":{"rendered":"Application & Proof of the Sherman-Morrison-Woodbury Identity"},"content":{"rendered":"\t\t
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The Sherman-Morrison-Woodbury identity<\/b><\/h2>

\\(\\displaystyle (A + UV^{T})^{-1} = A^{-1} – A^{-1}U\\left(I + V^{T}A^{-1}U\\right)^{-1}V^{T}A^{-1}\\)<\/p>

Have you ever looked at a massive maths problem and thought, “There has to be a faster way”? If you\u2019re preparing for the SATs, you\u2019re already training your brain to spot patterns and solve equations efficiently. But what happens when the problems get even bigger\u2014like the ones used to build Artificial Intelligence? In this blog, we are going to explore and prove <\/span>\u00a0<\/b>the Sherman-Morrison-Woodbury identity.<\/span><\/p>

Recommended Reading:<\/b> How to Derive and Use the Quadratic Formula (With Examples)<\/span><\/a><\/p>

The Problem: When Calculations Get Too Big<\/b><\/span><\/h3>

In the world of technology, we often work with “matrices”\u2014grids of numbers. Imagine a <\/span>huge matrix (<\/b>A<\/span><\/i>)<\/b> that is 100×100 or even larger. To find the “inverse” of a matrix of that size from scratch is a massive task for a computer, requiring a huge number of operations (known as <\/span>O(<\/span><\/i>n<\/span><\/i>\u00b3)<\/span><\/i>.<\/span><\/i><\/p>

Now, imagine you have already solved that huge matrix, but then <\/span>a little bit of new information is added to it<\/b> (represented by smaller matrices <\/span>U<\/span><\/i> and <\/span>V<\/span><\/i>). If you start the whole calculation over again, you\u2019re wasting time and energy.<\/span><\/p>

For a more detailed walkthrough, you can watch this video: <\/strong><\/span><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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The Solution: The Woodbury Shortcut<\/b><\/span><\/h3>

The Sherman Morrison Woodbury Identity is a way to calculate the inverse of that new, updated matrix <\/span>without starting from scratch.<\/p>

How it works:<\/b><\/p>

\u00a0If you already know the inverse of your original big matrix <\/span>\\(\\displaystyle xA^{-1}y\\)
<\/span>\u00a0you can use it to find the new answer by only doing “small” multiplications with the new data (<\/span>U<\/span><\/i> and <\/span>V<\/span><\/i>),<\/span><\/p>