Today we're gonna take care of the bane of first-year linear algebra students: matrix definiteness! Apparently this doesn't yet have a challenge so here we go:


  • A \$n\times n\$ symmetric Matrix \$A\$ in any convenient format (you may also of course only take the upper or the lower part of the matrix)
  • Optionally: the size of the matrix \$n\$

What to do?

The challenge is simple: Given a real-valued matrix \$n\times n\$ Matrix decide whether it is positive definite by outputting a truthy value if so and a falsey value if not.

You may assume your built-ins to actually work precisely and thus don't have to account for numerical issues which could lead to the wrong behaviour if the strategy / code "provably" should yield the correct result.

Who wins?

This is , so the shortest code in bytes (per-language) wins!

What is a positive-definite Matrix anyways?

There are apparently 6 equivalent formulations of when a symmetric matrix is positive-definite. I shall reproduce the three easier ones and reference you to Wikipedia for the more complex ones.

  • If \$\forall v\in\mathbb R^n\setminus \{0\}: v^T Av>0\$ then \$A\$ is positive-definite.
    This can be re-formulated as:
    If for every non-zero vector \$v\$ the (standard) dot product of \$v\$ and \$Av\$ is positive then \$A\$ is positive-definite.
  • Let \$\lambda_i\quad i\in\{1,\ldots,n\}\$ be the eigenvalues of \$A\$, if now \$\forall i\in\{1,\ldots,n\}:\lambda_i>0\$ (that is all eigenvalues are positive) then \$A\$ is positive-definite.
    If you don't know what eigenvalues are I suggest you use your favourite search engine to find out, because the explanation (and the needed computation strategies) is too long to be contained in this post.
  • If the Cholesky-Decomposition of \$A\$ exists, i.e. there exists a lower-triangular matrix \$L\$ such that \$LL^T=A\$ then \$A\$ is positive-definite. Note that this is equivalent to early-returning "false" if at any point the computation of the root during the algorithm fails due to a negative argument.


For truthy output






For falsey output

(at least one eigenvalue is 0 / positive semi-definite) \begin{pmatrix}3&-2&2\\-2&4&0\\2&0&2\end{pmatrix}

(eigenvalues have different signs / indefinite) \begin{pmatrix}1&0&0\\0&-1&0\\0&0&1\end{pmatrix}

(all eigenvalues smaller than 0 / negative definite) \begin{pmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}

(all eigenvalues smaller than 0 / negative definite) \begin{pmatrix}-2&3&0\\3&-5&0\\0&0&-1\end{pmatrix}

(all eigenvalues smaller than 0 / negative definite) \begin{pmatrix}-7.15&-2.45\\-2.45&-9.37\end{pmatrix}

(three positive, one negative eigenvalue / indefinite) \begin{pmatrix}7.15&2.45&1.23&3.5\\2.45&9.37&2.71&3.14\\1.23&2.71&0&6.2\\3.5&3.14&6.2&0.56\end{pmatrix}

  • \$\begingroup\$ This challenge was sandboxed. \$\endgroup\$ – SEJPM Sep 23 '18 at 11:48
  • \$\begingroup\$ You need to provide a better definition of what we're supposed to be looking for rather than assuming we can all read mathematical notation (or all know what an "eigenvalue" is). A worked example would be useful too. \$\endgroup\$ – Shaggy Sep 23 '18 at 15:53
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    \$\begingroup\$ @Shaggy I think the challenge is better without all the background to clog it up. There are many existing explanations of what an eigenvalue is elsewhere, and this post is already really large. \$\endgroup\$ – Ad Hoc Garf Hunter Sep 23 '18 at 16:37
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    \$\begingroup\$ The challenge would have been nicer hadn't you restrict the input to symmetric matrices. \$\endgroup\$ – polfosol ఠ_ఠ Sep 24 '18 at 15:55
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    \$\begingroup\$ I meant just checking for the sign of eigenvalues is also boring. Different tastes I know ;) \$\endgroup\$ – polfosol ఠ_ఠ Sep 24 '18 at 16:11

12 Answers 12


C, 108 bytes

-1 byte thanks to Logern
-3 bytes thanks to ceilingcat

f(M,n,i)double**M;{for(i=n*n;i--;)M[i/n][i%n]-=M[n][i%n]*M[i/n][n]/M[n][n];return M[n][n]>0&(!n||f(M,n-1));}

Try it online!

Performs Gaussian elimination and checks whether all diagonal elements are positive (Sylvester's criterion). Argument n is the size of the matrix minus one.

  • \$\begingroup\$ Perhaps save a character with float instead of double? \$\endgroup\$ – Jens Sep 23 '18 at 15:10
  • \$\begingroup\$ 106 bytes - Try it online! \$\endgroup\$ – Logern Sep 23 '18 at 15:35
  • \$\begingroup\$ You can shave another character if you drop i=0 in the for loop, make the recursive call f(M,n-1,0) and the initial call with 0 as the third arg. \$\endgroup\$ – Jens Sep 23 '18 at 19:36
  • \$\begingroup\$ @Jens 1. Using floats instead of doubles can quickly lead to noticable rounding errors, so I don't think the one byte saved is worth it. 2. Initializing a variable via an additional argument looks like cheating to me. \$\endgroup\$ – nwellnhof Sep 24 '18 at 17:42
  • \$\begingroup\$ @Logern I refuse to use the "omit the return statement" trick in my C answers. But thanks for the other byte saved. \$\endgroup\$ – nwellnhof Sep 24 '18 at 17:45

MATLAB/Octave, 19 17 12 bytes


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The function eig provides the eigenvalues in ascending order, so if the first eigenvalue is greater than zero, the other ones are too.

  • \$\begingroup\$ You can drop the f= at the beginning - anonymous functions are generally accepted as answers. \$\endgroup\$ – Delfad0r Sep 23 '18 at 15:05
  • \$\begingroup\$ Thanks for the tip! \$\endgroup\$ – Daniel Turizo Sep 23 '18 at 15:11
  • \$\begingroup\$ Even if its a vector? Interesting \$\endgroup\$ – Daniel Turizo Sep 23 '18 at 15:17
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    \$\begingroup\$ +1. I've added a link for trying it online. Hope you don't mind. Note that it is also proving that the output values despite being arrays do count as the correct "truthy" or "falsey" values as per the link @Delfad0r posted. \$\endgroup\$ – Tom Carpenter Sep 23 '18 at 17:03
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    \$\begingroup\$ Having said that, it fails for the first "falsey" test case on TIO. I'm guessing due to a precision issue - one of the Eigen values comes out as 8.9219e-17 rather than 0. \$\endgroup\$ – Tom Carpenter Sep 23 '18 at 17:05

Jelly, 11 10 bytes


Uses Sylvester's criterion.

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How it works

ṖṖ€$ƬÆḊṂ>0  Main link. Argument: M (matrix)

   $Ƭ       Do the following until a fixed point is encountered.
Ṗ             Pop; remove the last row of the matrix.
 Ṗ€           Pop each; remove the last entry of each row.
     ÆḊ     Take the determinants of the resulting minors.
       Ṃ    Take the minimum.
        >0  Test if the least determinant is positive, i.e., if all determinants are.

R, 29 bytes


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Alternative using cholesky :

R, 34 33 bytes


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-1 byte thanks to @Giuseppe


Haskell, 56 bytes


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Basically a port of nwellnhof's answer. Performs gaussian elimination and checks whether the elements on the main diagonal are positive.

Fails the first falsey output because of rounding errors, but it would theoretically work with infinite precision. Thanks to Curtis Bechtel's suggestion, now the outputs are all correct.

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    \$\begingroup\$ you can add inputs :: [[[Rational]]] to get correct answers \$\endgroup\$ – Curtis Bechtel Sep 23 '18 at 20:38

Wolfram Language (Mathematica), 20 bytes


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  • \$\begingroup\$ Should 4th test case be False? \$\endgroup\$ – tsh Sep 23 '18 at 14:48
  • \$\begingroup\$ @tsh Fixed, I am dumb! \$\endgroup\$ – Mr. Xcoder Sep 23 '18 at 14:57
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    \$\begingroup\$ Funny how Mathematica has a builtin for this, but its name is longer than your solution. \$\endgroup\$ – Federico Poloni Sep 23 '18 at 15:04
  • \$\begingroup\$ @FedericoPoloni: Wouldn't a solution using NullSpace or MatrixRank be even shorter? If Null space is zero then the matrix is positive definite. \$\endgroup\$ – Phil H Sep 24 '18 at 15:41
  • \$\begingroup\$ @PhilH No, I'm afraid that doesn't work by itself. For instance, the second falsey example (diagonal matrix with (1,-1,1)) has rank 3, but is not positive definite. \$\endgroup\$ – Federico Poloni Sep 24 '18 at 16:39

MATL, 4 bytes


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I've noticed that for the [3 -2 2; -2 4 0; 2 0 2] test case, MATL calculates the 0 eigenvalue with a very small inaccuracy, computing something as low as about \$10^{-18}\$. This therefore fails the first falsy test case due to precision issues. Thanks to Luis Mendo for pointing out that a non-empty array is truthy iff all its entries differ from 0, saving 1 byte.

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    \$\begingroup\$ @LuisMendo Thanks, I learned something new about MATL today! \$\endgroup\$ – Mr. Xcoder Sep 26 '18 at 21:53
  • \$\begingroup\$ My pleasure :-) Here's a better explanation by Suever. I forgot to say that for complex-valued arrays only the real part is compared against zero. So [1 2 3j] is falsey \$\endgroup\$ – Luis Mendo Sep 26 '18 at 22:04

Mathematica, 29 28 bytes


Definition 2.


Julia 1.0, 28 bytes

using LinearAlgebra

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Julia 0.6, 8 bytes


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MATL, 6 bytes

It is possible to do it using even less bytes, @Mr. Xcoder managed to find a 5 byte MATL answer!



Yv     compute eigenvalues
  X<   take the minimum
    0> check whether it is greather than zero

Try it online!

  • \$\begingroup\$ Fails on the first falsy test case. See my deleted answer. \$\endgroup\$ – Mr. Xcoder Sep 25 '18 at 6:59
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    \$\begingroup\$ @Mr.Xcoder Oh, you even submitted an answer before me. I think you should undelete your answer as it just depends on rounding issues. (I think you can expect answers to use limited precision arithmetic - I think only the CAS languages here use exact computations.) \$\endgroup\$ – flawr Sep 25 '18 at 18:02
  • \$\begingroup\$ Following your advice, I've undeleted it. \$\endgroup\$ – Mr. Xcoder Sep 25 '18 at 21:18

Maple, 33 bytes

(i.e. my 2 cents)

  • \$\begingroup\$ Hello and welcome to PPCG; I am unfamiliar with Maple, though is the newline necessary? \$\endgroup\$ – Jonathan Frech Sep 24 '18 at 16:06
  • \$\begingroup\$ @JonathanFrech Hello and thanks. No it's not. I didn't count it btw. \$\endgroup\$ – polfosol ఠ_ఠ Sep 24 '18 at 16:09
  • \$\begingroup\$ To me your current byte count seems to reflect the newline character. \$\endgroup\$ – Jonathan Frech Sep 24 '18 at 16:16
  • \$\begingroup\$ @JonathanFrech Duh, my bad \$\endgroup\$ – polfosol ఠ_ఠ Sep 24 '18 at 16:18
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    \$\begingroup\$ Well ... now your code and your byte count disagree. \$\endgroup\$ – Jonathan Frech Sep 24 '18 at 17:07

JavaScript (ES6),  99 95  88 bytes

Same method as nwellnhof's answer. Returns \$0\$ or \$1\$.


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