Talk:Singular value decomposition

Dagger versus asterisk

In the entire article, it is not clear to me whether $V^{*}$ means the conjugate transpose of $V$, or just the complex conjugate. I guess it doesn't really matter, because if $V$ is unitary, so are the conjugate of $V$ and the conjugate transpose of $V$. But in any case, I'm confused. In the definition, should we add something like " ..., and $V^{*}$ is the complex conjugate of $V$ " ? — Preceding unsigned comment added by Barbireau (talk • contribs) 13:16, 1 April 2020 (UTC)

Hello I'm another user adding that agrees, but is unsure for what V* is, also maybe a voting measure or addable box for comments in discussions would be beneficial! — Preceding unsigned comment added by DaltSalt (talk • contribs) 11:20, 12 May 2020 (UTC)

Look of Sigma-symbol representing the matrix holding the singular values

I have just replaced every (capital) Sigma representing the matrix holding the singular values that I could find in the article (apart from captions) with the version of the first appearance in the text that also matches the height of the unitary matrix symbols frequently used next to it. (And what many ways there are to code it, and miscode it for that matter; it took me over half an hour to figure out how to code it geared to the surroundings where it appears.) But at least, if I succeeded, there is now unity of notational appearance in this respect.Redav (talk) 13:26, 23 May 2020 (UTC)

Error in Figure

There is an error in the figure. The sigma's should align with the U vectors. Here is an example of a correct figure http://i.stack.imgur.com/IM6Fn.png — Preceding unsigned comment added by 73.12.205.109 (talk) 19:40, 4 February 2021 (UTC)

Crucial error in first sentence?

The opening sentence seems to claim that only normal matrices are diagonalizable. That claim is false, right? What's the fix? Or is there a reliable source that can explain it to me? Mgnbar (talk) 00:39, 27 July 2021 (UTC)

The claim is talking about the special eigendecomposition of a normal matrix where one can say that the matrix is unitarily similar to a diagonal matrix, not merely similar. Normal matrices are (definitionally) the only ones that can be decomposed in this way. I tried to make the lead more correct, but the wording is a bit clunky. Fawly (talk) 01:19, 27 July 2021 (UTC)

Thank you. I understand the intent much better now. I have further broken the big, complicated sentence into small sentences, hopefully for the better. Mgnbar (talk) 12:03, 27 July 2021 (UTC)

Is Lagrange multiplier equation in existence proof Based on variational characterization correct?

currently it says:

$\nabla \sigma =\nabla \mathbf {u} ^{\textsf {T}}\mathbf {M} \mathbf {v} -\lambda _{1}\cdot \nabla \mathbf {u} ^{\textsf {T}}\mathbf {u} -\lambda _{2}\cdot \nabla \mathbf {v} ^{\textsf {T}}\mathbf {v}$.

I don't quite know what this means, I would think it should be

$\nabla \mathbf {u} ^{\textsf {T}}\mathbf {M} \mathbf {v} -\lambda _{1}\cdot \nabla \mathbf {u} ^{\textsf {T}}\mathbf {u} -\lambda _{2}\cdot \nabla \mathbf {v} ^{\textsf {T}}\mathbf {v} =0$

instead to match the eigenvalue version as well as the following equations. 2001:56A:F98E:2400:6C3F:8EA:5F2:540A (talk) 21:23, 6 November 2022 (UTC)

History - reference to proof by Eckart and Young

The article claims

The first proof of the singular value decomposition for rectangular and complex matrices seems to be by Carl Eckart and Gale J. Young in 1936;

but their paper states up front:

The solution of the problem is much simplified by an appeal to two theorems which are generalizations of well known theorems on square matrices.[here is given a citation to (Courant and Hilbert 1924) getting SVD for square matrices] They will not be proven here.

Theorem I. For any real matrix a, two orthogonal matrices u and U can be found so that lambda = uaU' is a real diagonal matrix with no negative elements.

This is their statement of the purely algebraic fact about SVD (the problem they are considering is about approximating matrices, which is not what the main article is about), and not proved! The usual statement is given immediately following the theorem in equation (10). It might be that Eckart and Young first stated SVD for non-square matrices, but false to say they gave a proof. It's also not true that they considered the complex case; the application they have in mind is purely real, and nowhere do they mention complex entries. 121.44.213.90 (talk) 07:43, 19 July 2023 (UTC)

Dead link in ref 4

Reference 4 link is dead to me. Try https://people.wou.edu/~beavers/Talks/Willamette1106.pdf SAM26 (talk) 20:34, 20 September 2025 (UTC)

I put this link in, although it's not a great source. Tito Omburo (talk) 20:40, 20 September 2025 (UTC)

Excessive lead section

@JavBol, the new lead section here is getting ridiculously overstuffed. The lead section can have some brief concrete material for context, but is not a good place to put long examples, derivations, or more than one or two mathematical expressions.

Remember, Wikipedia is not a textbook.

See MOS:LEAD and MOS:MATH#Article introduction. –jacobolus (t) 01:19, 28 June 2026 (UTC)

Yes, the lede as it currently stands is egregiously overstuffed. Most (or perhaps all) of the recently added content should be moved elsewhere in the article (or perhaps deleted). Mgnbar (talk) 13:28, 28 June 2026 (UTC)

@JavBol To be honest I don't think the example is very illuminating. Maybe put it in an appendix at the bottom or something. But as is this seems distracting for readers and it's not clear what they get out of it. –jacobolus (t) 04:38, 29 June 2026 (UTC)

I agree that the lede was overstuffed before the new parts of the example were moved down to #Example.

But the example legitimately shows why $3,{\sqrt {5}},2,0$ are singular values, $u_{1},...,u_{4}$ are left-singular vectors, and $v_{1},...,v_{4}$ are right-singular vectors of the matrix $M,$ and why $v_{5}$ can pad the matrix $V.$

PS: I've checked on wolframalpha.com that $Mv_{1}=3u_{1},~Mv_{2}={\sqrt {5}}u_{2},~Mv_{3}=2u_{3},~Mv_{4}=0u_{4}=(0~0~0~0)^{\mathsf {T}},~Mv_{5}=(0~0~0~0)^{\mathsf {T}},$

that ⁠$M^{*}u_{1}=3v_{1}$⁠:

{{1,0,0,0,2},{0,0,3,0,0},{0,0,0,0,0},{0,2,0,0,0}}^T × {{0},{1},{0},{0}} gives 3{{0},{0},{1},{0},{0}},

that ⁠$M^{*}u_{2}={\sqrt {5}}v_{2}$⁠:

{{1,0,0,0,2},{0,0,3,0,0},{0,0,0,0,0},{0,2,0,0,0}}^T × {{1},{0},{0},{0}} gives {{1},{0},{0},{0},{2}}, i.e. (√5){{√(0.2)},{0},{0},{0},{√(0.8)}},

that ⁠$M^{*}u_{3}=2v_{3}$⁠:

{{1,0,0,0,2},{0,0,3,0,0},{0,0,0,0,0},{0,2,0,0,0}}^T × {{0},{0},{0},{1}} gives 2{{0},{1},{0},{0},{0}},

and that ⁠$M^{*}u_{4}=0v_{4}=(0~0~0~0~0)^{\mathsf {T}}$⁠:

{{1,0,0,0,2},{0,0,3,0,0},{0,0,0,0,0},{0,2,0,0,0}}^T × {{0},{0},{1},{0}} gives {{0},{0},{0},{0},{0}}.

—JavBol (talk) 16:34, 1 July 2026 (UTC)

It doesn't show "why" anything, or give much explanation at all;. It just states the results of several calculations repeated a few times in a giant block of numbers. I don't think most readers are going to get much out of it, and it's a pretty big distraction to load at the front of the article. –jacobolus (t) 17:03, 1 July 2026 (UTC)

Even if you want to include an example like this, the current version is absurdly redundant. The same information can be conveyed more clearly and much more concisely, e.g.

Example

For example, the ⁠$4\times 5$⁠ matrix ⁠$\mathbf {M}$⁠ below can be decomposed as ⁠$\mathbf {U} \mathbf {\Sigma } \mathbf {V} ^{*}$⁠: $${\begin{aligned}\mathbf {M} &={\begin{bmatrix}1&0&0&0&2\\0&0&3&0&0\\0&0&0&0&0\\0&2&0&0&0\end{bmatrix}}\\&={\begin{bmatrix}\color {RubineRed}0&\color {Mulberry}1&\color {RoyalBlue}0&\color {CadetBlue}0\\\color {RubineRed}1&\color {Mulberry}0&\color {RoyalBlue}0&\color {CadetBlue}0\\\color {RubineRed}0&\color {Mulberry}0&\color {RoyalBlue}0&\color {CadetBlue}1\\\color {RubineRed}0&\color {Mulberry}0&\color {RoyalBlue}1&\color {CadetBlue}0\end{bmatrix}}{\begin{bmatrix}\color {RubineRed}3&\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Mulberry}{\sqrt {5}}&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {RoyalBlue}2&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {CadetBlue}0&\color {Gray}0\end{bmatrix}}{\begin{bmatrix}\color {RubineRed}0&\color {RubineRed}0&\color {RubineRed}1&\color {RubineRed}0&\color {RubineRed}0\\\color {Mulberry}{\tfrac {1}{\sqrt {5}}}&\color {Mulberry}0&\color {Mulberry}0&\color {Mulberry}0&\color {Mulberry}{\tfrac {2}{\sqrt {5}}}\\\color {RoyalBlue}0&\color {RoyalBlue}1&\color {RoyalBlue}0&\color {RoyalBlue}0&\color {RoyalBlue}0\\\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}1&\color {CadetBlue}0\\\color {CadetBlue}{\tfrac {2}{\sqrt {5}}}&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}-{\tfrac {1}{\sqrt {5}}}\end{bmatrix}}\\&=\mathbf {U} \mathbf {\Sigma } \mathbf {V} ^{*}.\end{aligned}}$$

The singular values of ⁠$\mathbf {M}$⁠ are the diagonal entries of ⁠$\mathbf {\Sigma }$⁠: $\{3,{\sqrt {5}},2,0\}$. The corresponding left- and right-singular vectors are the columns of ⁠$\mathbf {U}$⁠ and rows of ⁠$\mathbf {V} ^{*}$⁠, respectively.

The matrices ⁠$\mathbf {U}$⁠ and ⁠$\mathbf {V} ^{*}$⁠ are unitary (and as real-valued matrices, orthogonal), meaning ⁠$\mathbf {U} \mathbf {U} ^{*}=\mathbf {I} _{4}$⁠ and ⁠$\mathbf {V} \mathbf {V} ^{*}=\mathbf {I} _{5}$⁠, where ⁠$\mathbf {I} _{k}$⁠ is the ⁠$k\times k$⁠ identity matrix.

The matrix ⁠$\mathbf {M}$⁠ has rank ⁠$3$⁠, so there are only three non-zero singular values. This particular singular value decomposition is not unique: the last two rows of ⁠$\mathbf {V} ^{*}$⁠ do not contribute to the product because they are multiplied by zero, so are substantially arbitrary, and can be replaced with any pair of vectors which are orthogonal to each-other and the rest of the rows and have unit magnitude. Likewise for the last column of ⁠$\mathbf {U}$⁠.

The compact SVD eliminates the superfluous last column from ⁠$\mathbf {U}$⁠, last two rows from ⁠$\mathbf {V} ^{*}$⁠, and last row and two columns from ⁠$\mathbf {\Sigma }$⁠: $$\mathbf {M} ={\begin{bmatrix}\color {RubineRed}0&\color {Mulberry}1&\color {RoyalBlue}0\\\color {RubineRed}1&\color {Mulberry}0&\color {RoyalBlue}0\\\color {RubineRed}0&\color {Mulberry}0&\color {RoyalBlue}0\\\color {RubineRed}0&\color {Mulberry}0&\color {RoyalBlue}1\end{bmatrix}}{\begin{bmatrix}\color {RubineRed}3&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Mulberry}{\sqrt {5}}&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {RoyalBlue}2\\\end{bmatrix}}{\begin{bmatrix}\color {RubineRed}0&\color {RubineRed}0&\color {RubineRed}1&\color {RubineRed}0&\color {RubineRed}0\\\color {Mulberry}{\tfrac {1}{\sqrt {5}}}&\color {Mulberry}0&\color {Mulberry}0&\color {Mulberry}0&\color {Mulberry}{\tfrac {2}{\sqrt {5}}}\\\color {RoyalBlue}0&\color {RoyalBlue}1&\color {RoyalBlue}0&\color {RoyalBlue}0&\color {RoyalBlue}0\end{bmatrix}}.$$

Instead of the product of three matrices ⁠$\mathbf {M} =\mathbf {U} \mathbf {\Sigma } \mathbf {V} ^{*}$⁠, the SVD can be written as a sum of three rank-⁠$1$⁠ matrices, each formed as the outer product of one column ⁠$\mathbf {u} _{j}$⁠ of ⁠$\mathbf {U}$⁠ times the corresponding row ⁠$\mathbf {v} _{j}^{*}$⁠ of ⁠$\mathbf {V} ^{*}$⁠, scaled by the corresponding singular value ⁠$\sigma _{j}$⁠:

$${\begin{aligned}\mathbf {M} &=\sum _{j}\sigma _{j}\mathbf {u} _{j}\mathbf {v} _{j}^{*}={\color {RubineRed}3\mathbf {u} _{1}\mathbf {v} _{1}^{*}}+{\color {Mulberry}{\sqrt {5}}\mathbf {u} _{2}\mathbf {v} _{2}^{*}}+{\color {RoyalBlue}2\mathbf {u} _{3}\mathbf {v} _{3}^{*}}\\&={\begin{bmatrix}0&0&0&0&0\\0&0&3&0&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}}+{\begin{bmatrix}1&0&0&0&2\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}}+{\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&2&0&0&0\end{bmatrix}}\end{aligned}}$$

---

I think you'd get a more instructive example by picking a matrix which (a) was less sparse, and (b) didn't need square roots to express the entries in the decomposition. But I still maintain that such an example is unnecessary and mostly serves as a distraction. There are a bunch of interesting patterns you might want to point out about an example like this if you were a teacher leading an introductory class, or maybe even a textbook author, but are completely out of scope for this type of article. Remember, Wikipedia is not a textbook, and every bit of text we add to an article really needs to pull its weight. I'm probably going to remove the example from the article pending further discussion, but feel free to keep thinking about it and workshopping it. –jacobolus (t) 20:40, 1 July 2026 (UTC)

• These equalities don't show how to calculate the singular values and the corresponding singular vectors, but they do show that ⁠$\{3,u_{1},v_{1}\}$⁠, ⁠$\{{\sqrt {5}},u_{2},v_{2}\}$⁠, ⁠$\{2,u_{3},v_{3}\}$⁠, and ⁠$\{0,u_{4},v_{4}\}$⁠ each satisfy the definition of a singular value and its left- and right-singular vectors for the matrix ⁠$M$⁠, and that $v_{5}$ can pad the matrix $V$ (since the ⁠$5$⁠th column of $\varSigma$ must be ⁠$(0~0~0~0)^{\mathsf {T}}$⁠). Also, these equalities justify that $MV=U\varSigma$ and $M^{*}U=V\varSigma$ (i.e., ⁠$U^{*}M=\varSigma \,V^{*}$⁠).
• This example was already in the article before I started editing it.
• You're bothered by this example probably because you're a high-graduate mathematician, but please have mercy for average scientists, students, high-school pupils… Most readers badly NEED an example! Please, just move it down within the article, or turn it to an appendix…
• You might find a simultaneously less sparse and square-root-free example, but I won't find such a wonder… —JavBol (talk) 23:12, 1 July 2026 (UTC)

Ah, I see. The example before you arrived was also already distracting and not very helpful, but at least wasn't bloated to monstrous proportions and didn't have each bit of information repeated several times. I'll put back an example pending further discussion, but I don't think this particular example is really very instructive or illuminating. –jacobolus (t) 23:18, 1 July 2026 (UTC)

@Jacobolus: Your (mostly legitimate) latest edit on the SVD article also changed several legitimate ⁠$\mathbf {R}$⁠ (matrices) to ⁠$\mathbb {R}$⁠ (set of real numbers). —JavBol (talk) 21:16, 2 July 2026 (UTC)

Oh, sorry for the sloppy mistake. I'll go fix that. –jacobolus (t) 01:24, 3 July 2026 (UTC)

I made a hopefully somewhat better example. $${\begin{aligned}\mathbf {M} &={\begin{bmatrix}~4&~8&~8&~0&~0\\-3&-6&-6&~0&~0\\-4&~4&-2&~0&~0\\~0&~0&~0&-4&~3\end{bmatrix}}\\&={\begin{bmatrix}\color {PineGreen}~{\frac {4}{5}}&\color {BrickRed}~0&\color {BlueViolet}~0&\color {CadetBlue}~{\frac {3}{5}}\\\color {PineGreen}-{\frac {3}{5}}&\color {BrickRed}~0&\color {BlueViolet}~0&\color {CadetBlue}~{\frac {4}{5}}\\\color {PineGreen}~0&\color {BrickRed}~1&\color {BlueViolet}~0&\color {CadetBlue}~0\\\color {PineGreen}~0&\color {BrickRed}~0&\color {BlueViolet}~1&\color {CadetBlue}~0\end{bmatrix}}{\begin{bmatrix}\color {PineGreen}15&\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {BrickRed}6&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {BlueViolet}5&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {CadetBlue}0&\color {Gray}0\end{bmatrix}}{\begin{bmatrix}\color {PineGreen}~{\frac {1}{3}}&\color {PineGreen}~{\frac {2}{3}}&\color {PineGreen}~{\frac {2}{3}}&\color {PineGreen}~0&\color {PineGreen}~0\\\color {BrickRed}-{\frac {2}{3}}&\color {BrickRed}~{\frac {2}{3}}&\color {BrickRed}-{\frac {1}{3}}&\color {BrickRed}~0&\color {BrickRed}~0\\\color {BlueViolet}~0&\color {BlueViolet}~0&\color {BlueViolet}~0&\color {BlueViolet}-{\frac {4}{5}}&\color {BlueViolet}~{\frac {3}{5}}\\\color {CadetBlue}-{\frac {2}{3}}&\color {CadetBlue}-{\frac {1}{3}}&\color {CadetBlue}~{\frac {2}{3}}&\color {CadetBlue}~0&\color {CadetBlue}~0\\\color {CadetBlue}~0&\color {CadetBlue}~0&\color {CadetBlue}~0&\color {CadetBlue}~{\frac {3}{5}}&\color {CadetBlue}~{\frac {4}{5}}\end{bmatrix}}\end{aligned}}$$ instead of $${\begin{aligned}\mathbf {M} &={\begin{bmatrix}1&0&0&0&2\\0&0&3&0&0\\0&0&0&0&0\\0&2&0&0&0\end{bmatrix}}\\&={\begin{bmatrix}\color {PineGreen}0&\color {BrickRed}1&\color {BlueViolet}0&\color {CadetBlue}0\\\color {PineGreen}1&\color {BrickRed}0&\color {BlueViolet}0&\color {CadetBlue}0\\\color {PineGreen}0&\color {BrickRed}0&\color {BlueViolet}0&\color {CadetBlue}1\\\color {PineGreen}0&\color {BrickRed}0&\color {BlueViolet}1&\color {CadetBlue}0\end{bmatrix}}{\begin{bmatrix}\color {PineGreen}3&\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {BrickRed}{\sqrt {5}}&\color {Gray}0&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {BlueViolet}2&\color {Gray}0&\color {Gray}0\\\color {Gray}0&\color {Gray}0&\color {Gray}0&\color {CadetBlue}0&\color {Gray}0\end{bmatrix}}{\begin{bmatrix}\color {PineGreen}0&\color {PineGreen}0&\color {PineGreen}1&\color {PineGreen}0&\color {PineGreen}0\\\color {BrickRed}{\tfrac {1}{\sqrt {5}}}&\color {BrickRed}0&\color {BrickRed}0&\color {BrickRed}0&\color {BrickRed}{\tfrac {2}{\sqrt {5}}}\\\color {BlueViolet}0&\color {BlueViolet}1&\color {BlueViolet}0&\color {BlueViolet}0&\color {BlueViolet}0\\\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}1&\color {CadetBlue}0\\\color {CadetBlue}{\tfrac {2}{\sqrt {5}}}&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}0&\color {CadetBlue}-{\tfrac {1}{\sqrt {5}}}\end{bmatrix}}.\end{aligned}}$$

I was hoping I'd be able to construct one where the specific entries of ⁠$\mathbf {M}$⁠ weren't entirely separate (i.e. having some of the rank 1 parts overlapping), but it's hard to balance that with avoiding too many zeros, keeping the overall matrix size down, and avoiding square roots. The ⁠$4\times 5$⁠ size with rank ⁠$3$⁠ seemed about right, so I stuck with that. –jacobolus (t) 22:09, 3 July 2026 (UTC)

@JavBol, I took out the footnote you just added to the lead about the first ⁠$r$⁠ singular vectors spanning the columns of the matrix. For anyone else reading along, here was its content:

Indeed: Let $\mathbf {A}$ be the $r\times n$ product matrix ⁠$\mathbf {\Sigma } _{r}\mathbf {V} _{r}^{*}$⁠; then ⁠$\mathbf {M} =\mathbf {U} _{r}\mathbf {A}$⁠,
i.e., ⁠$\forall (j,k)\in \{1,\ldots ,m\}\times \{1,\ldots ,n\},~\mathbf {M} _{j,k}=\sum _{\ell =1}^{r}\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}$⁠;
i.e., ⁠$\forall k\in \{1,\ldots ,n\},~{\bigl [}\mathbf {M} _{j,k}{\bigr ]}_{1\leq j\leq m}=\left[\sum _{\ell =1}^{r}\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}\right]_{1\leq j\leq m}$⁠, where ${\bigl [}\mathbf {u} _{j}{\bigr ]}_{1\leq j\leq m}$ denotes the column vector ⁠$\mathbf {u}$⁠;
i.e., denoting by $\mathbf {m} _{k}$ the ⁠$k$⁠-th column of ⁠$\mathbf {M}$⁠,
⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}{\bigl [}\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}{\bigr ]}_{1\leq j\leq m}$⁠ (since sums of coordinates are coordinates of sum),
i.e., ⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}\mathbf {A} _{\ell ,k}{\bigl [}\mathbf {U} _{j,\ell }{\bigr ]}_{1\leq j\leq m}$⁠ (by factoring $\mathbf {A} _{\ell ,k}$ out of the coordinates),
i.e., ⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}\mathbf {A} _{\ell ,k}\mathbf {u} _{\ell }$⁠.

First, the lead is the entirely wrong place for any such explanation. Remember: the lead is a short summary of the content, not a venue for extended mathematical asides. But second, any explanation of this should be done with words and pictures, not with a solid blob of mathematical symbols. Many (most?) readers aren't even going to be able to make sense of this footnote, and the ones who can either already know this or can figure it out on the spot. –jacobolus (t) 18:46, 6 July 2026 (UTC)

If you like we can remove the observation that these vector span the columns and rows of the matrix from the lead; it's not really necessary. –jacobolus (t) 18:55, 6 July 2026 (UTC)

My footnote was brief & had no figure… because it was just a footnote! Anyway, explaining this spanning with words instead of mathematical formulas would probably be less clear, & I don't know what figures could explain it…

Also anyway, with your argument:

Many (most?) readers aren't even going to be able to make sense of this footnote, and the ones who can either already know this or can figure it out on the spot.,

Wikipedia would never give any mathematical proof…! —JavBol (talk) 20:41, 6 July 2026 (UTC)

With some exceptions (especially articles specifically about a theorem or about a particular proof) Wikipedia shouldn't be in the business of providing mathematical proofs. We should provide links to external sources, and try to make explanations that hold together and are comprehensible to readers, but formal proofs of arbitrary mathematical claims in Wikipedia are optional and often (usually?) unhelpful. In a case like this, where the claim is a direct and pretty obvious consequence, a justification in the form of a cumbersome bunch of explicit formalized steps in pure notation without prose explanation is not pulling its weight, even in "just a footnote". –jacobolus (t) 22:55, 6 July 2026 (UTC)

For anyone reading along, I took the footnote out again. Its content was:

Indeed: Let $\mathbf {A}$ be the $r\times n$ product matrix ⁠$\mathbf {\Sigma } _{r}\mathbf {V} _{r}^{*}$⁠; then ⁠$\mathbf {M} =\mathbf {U} _{r}\mathbf {A}$⁠,
i.e., for any ⁠$j\in \{1,\ldots ,m\}$⁠ and any ⁠$k\in \{1,\ldots ,n\},~\mathbf {M} _{j,k}=\sum _{\ell =1}^{r}\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}$⁠;
i.e., for any ⁠$k\in \{1,\ldots ,n\},~{\begin{bmatrix}\mathbf {M} _{1,k}\\\vdots \\\mathbf {M} _{j,k}\\\vdots \\\mathbf {M} _{m,k}\end{bmatrix}}={\begin{bmatrix}\sum _{\ell =1}^{r}\mathbf {U} _{1,\ell }\mathbf {A} _{\ell ,k}\\\vdots \\\sum _{\ell =1}^{r}\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}\\\vdots \\\sum _{\ell =1}^{r}\mathbf {U} _{m,\ell }\mathbf {A} _{\ell ,k}\end{bmatrix}}$⁠;
i.e., denoting by $\mathbf {m} _{k}$ the ⁠$k$⁠-th column of ⁠$\mathbf {M}$⁠,
⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}{\begin{bmatrix}\mathbf {U} _{1,\ell }\mathbf {A} _{\ell ,k}\\\vdots \\\mathbf {U} _{j,\ell }\mathbf {A} _{\ell ,k}\\\vdots \\\mathbf {U} _{m,\ell }\mathbf {A} _{\ell ,k}\end{bmatrix}}$⁠ (since sums of coordinates are coordinates of sum),
i.e., ⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}\mathbf {A} _{\ell ,k}{\begin{bmatrix}\mathbf {U} _{1,\ell }\\\vdots \\\mathbf {U} _{j,\ell }\\\vdots \\\mathbf {U} _{m,\ell }\end{bmatrix}}$⁠ (by factoring $\mathbf {A} _{\ell ,k}$ out of the coordinates),
i.e., ⁠$\mathbf {m} _{k}=\sum _{\ell =1}^{r}\mathbf {A} _{\ell ,k}\mathbf {u} _{\ell }$⁠.

This seems even harder to read than the previous version. –jacobolus (t) 06:52, 7 July 2026 (UTC)

I don't understand revision as of 18:15, 8 July 2026

@Jacobolus:

• I agree with your revision as of 18:23, 8 July 2026, of the article.
• I agree with your restorations of «\right.» instead of «\right\}» (sorry, I thought «\right.» was a typo).
• I don't mind about your changes to Singular value decomposition#Image compression.
• But in your revision as of 18:15, 8 July 2026, of the article, I don't understand your other reversals… Can you explain some of these, please? —JavBol (talk) 21:51, 8 July 2026 (UTC)

Your changes seems worse to me. For example, making multiple small chunks of inline mathematical notation into one giant blob makes it impossible to have a line break, and the word "for" in running text should be just ordinary wikimarkup rather than LaTeX. Changing "unit vectors" to "unit rows" is confusing, since "unit rows" is unexplained unusual terminology. Adding "without changing any column of U" etc. is unnecessarily wordy addition to an already distractingly wordy aside. Changing "a sum" to "the sum" reads worse to me and seems like a completely unnecessary change. You put LaTeX in an image caption, which should be avoided wherever possible (due to a Wikimedia bug). Changing "span" to "linear span" is not necessary; the term "span" is typical in this context and there's no other way that vectors would "span" a space that we'd confuse it for. –jacobolus (t) 23:26, 8 July 2026 (UTC)

I had changed

the SVD of ⁠$\mathbf {M}$⁠ can be written as a sum of ⁠$r$⁠ rank-⁠$1$⁠ matrices

to

the SVD of ⁠$\mathbf {M}$⁠ can be written as the sum of ⁠$r$⁠ rank-⁠$1$⁠ matrices

because in this phrase, the SVD is «determined», so the ⁠$r$⁠ rank-⁠$1$⁠ matrices are «determined», so their sum is «determined». —JavBol (talk) 19:05, 9 July 2026 (UTC)

Yes, but changing the word changes the emphasis. The point is that the product can be written as a sum. –jacobolus (t) 19:19, 9 July 2026 (UTC)

Clarification needed in #«Ky Fan norms»?

In Singular value decomposition#Ky Fan norms, the passage:

One can easily verify the relationship between the Ky Fan ⁠$1$⁠-norm and singular values. Indeed, it is true in general, for a bounded operator ⁠$\mathbf {M}$⁠ on a (possibly infinite-dimensional) Hilbert space, that $$\|\mathbf {M} \|_{\text{op}}={\sqrt {\|\mathbf {M} ^{*}\mathbf {M} \|_{\text{op}}}}.$$

But, in the matrix case, ${\sqrt {\mathbf {M} ^{*}\mathbf {M} }}$ is a normal matrix, so ${\sqrt {\|\mathbf {M} ^{*}\mathbf {M} \|_{\text{op}}}}$ is the largest eigenvalue of ⁠${\sqrt {\mathbf {M} ^{*}\mathbf {M} }}$⁠, i.e. $\|\mathbf {M} \|_{\text{op}}$ is the largest singular value of ⁠$\mathbf {M}$⁠.,

should be clarified to something like:

One can easily verify the relationship between the Ky Fan ⁠$1$⁠-norm and singular values. Indeed, it is true in general, for a bounded operator ⁠$\mathbf {M}$⁠ on a (possibly infinite-dimensional) Hilbert space, that $$\|\mathbf {M} \|_{\text{op}}={\sqrt {\|\mathbf {M} ^{*}\mathbf {M} \|_{\text{op}}}}.$$

But, in the matrix case, $\mathbf {M} ^{*}\mathbf {M}$ is Hermitian positive semi-definite, so it has a unique Hermitian positive semi-definite square root, ⁠${\sqrt {\mathbf {M} ^{*}\mathbf {M} }}$⁠; so ${\sqrt {\|\mathbf {M} ^{*}\mathbf {M} \|_{\text{op}}}}$ is the largest eigenvalue of ⁠${\sqrt {\mathbf {M} ^{*}\mathbf {M} }}$⁠, i.e. $\|\mathbf {M} \|_{\text{op}}$ is the largest singular value of ⁠$\mathbf {M}$⁠.,

shouldn't it?

Also, what about specifying that, using any SVD of ⁠$\mathbf {M}$⁠ as ⁠$\mathbf {U} \mathbf {\Sigma } \mathbf {V} ^{*}$⁠, this (principal) square root ${\sqrt {\mathbf {M} ^{*}\mathbf {M} }}$ can be defined as ⁠$\mathbf {V} {\sqrt {\mathbf {\Sigma } ^{*}\mathbf {\Sigma } }}~\mathbf {V} ^{*}$⁠ ? —JavBol (talk) 01:19, 22 July 2026 (UTC)