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A basis of a vector space \$V\$ is a set of vectors \$B\$ such that every vector \$\vec v \in V\$ can be uniquely written as a linear combination of the vectors in \$B\$. In other words, let \$B = \{\vec b_1, \dots, \vec b_n\}\$ be a basis of some vector space \$V\$. For every possible \$\vec v \in V\$, we can say that

$$\vec v = \lambda_1 \vec b_1 + \lambda_2 \vec b_2 + \cdots + \lambda_n \vec b_n$$

for some unique real numbers \$\lambda_1, \lambda_2, \dots, \lambda_n\$. Note that this requires the vectors in \$B\$ to be linearly independent (i.e., you cannot write a vector in \$B\$ as a linear combination of other vectors in \$B\$).

For example, let \$V = \mathbb R^2\$ i.e. the set of all 2 dimensional vectors. We can see that \$B = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}\$ is a basis of \$V\$, as any 2 dimensional vector \$\vec v = \begin{pmatrix} x \\ y \end{pmatrix}\$ can be written as

$$\vec v = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}x + \begin{pmatrix} 0 \\ 1 \end{pmatrix}y$$

Furthermore, examine \$B = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix} \right\}\$. As the two vectors are linearly independent, they must form a basis of \$\mathbb R^2\$, and so we can write any 2-dimensional vector as a combination of the two. For example,

$$\begin{pmatrix} 2 \\ -10 \end{pmatrix} = -2\begin{pmatrix} 1 \\ 1 \end{pmatrix} + -4\begin{pmatrix} -1 \\ 2 \end{pmatrix}$$

However, note that if we have \$B = \left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \right\}\$, then this does not form a basis for all 3 dimensional vectors. There exists no such \$\lambda_1, \lambda_2\$ such that

$$\begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\lambda_1 + \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\lambda_2$$

and so \$B\$ is not a basis for \$\begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix}\$.


You are to take a vector \$\vec v \in \mathbb Z^m\$ of \$m\$ integers, and a list of \$n\$ vectors \$B = \{\vec b_1, \dots, \vec b_n\}\$. You should then output two distinct consistent values to indicate whether or not \$B\$ forms a basis for \$\vec v\$ (more precisely, whether \$B\$ forms a basis for some space containing \$\vec v\$). That is, whether or not there exists a unique set of numbers \$\lambda_1, \lambda_2, \dots, \lambda_n\$ such that

$$\vec v = \lambda_1 \vec b_1 + \lambda_2 \vec b_2 + \cdots + \lambda_n \vec b_n$$

You may assume that \$\vec b_i \in \mathbb Z^m\$ for all vectors \$b_i \in B\$ - that is, they all have the same number of elements as \$\vec v\$, and they are all integer vectors.

You may take the inputs in any reasonable, convenient format and method, including a list of lists of \$B\$, or a list of numbers for \$\vec v\$ etc. The output may be any two distinct, consistent values to indicate whether or not \$B\$ is a basis. You may freely choose these values.

This is , so the shortest code in bytes wins

Test cases

B
v
Output (λ1, λ2, ..., λn or reason)

[[8, 1, 2], [-7, -8, -9], [-1, 9, -5]]
[12, 36, -4]
True (1, -1, 3)

[[-9, -3]]
[3, 1]
True (-1/3)

[[1, 0, 1], [0, 1, 2]]
[1, -1, -1]
True (1, -1)

[[1, 0, 1], [0, 1, 2]]
[4, 6, 10]
False (shown above)

[[7, -2], [-10, -6], [-4, 9]]
[1, -21]
False (too many unknowns, an infinite number of solutions exist)

[[-1, 8, 1, 6, 3], [8, -6, -5, -10, 6], [-8, -3, -3, -4, 5], [-6, -6, 0, 3, 9], [2, 2, 0, -1, -3]]
[1, 1, 1, 1, 1]
False ([-6, -6, 0, 3, 9] and [2, 2, 0, -1, -3] are linearly dependent, so any way of writing the sum is not unique)
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  • \$\begingroup\$ Note that "whether or not \$B\$ forms a basis for \$\vec v\$" isn't mathematially correct; vector spaces have bases but individual vectors do not (and replacing \$\vec v\$ by \$\langle\vec v\rangle\$ doesn't make it correct either). By the criteria described, we're being asked whether \$B\$ is linearly independent and simultaneously \$\vec v \in{}\$Span\$B\$. \$\endgroup\$ Dec 3, 2021 at 19:36
  • \$\begingroup\$ @GregMartin Yeah, I'm not too happy with the wording, but I believe that, although it isn't technically correct, it makes enough sense in the context of the challenge. If you have a better way of expressing the concept that \$\vec v$ can be expressed as some linear combination of a subset of \$B\$, feel free to edit. \$\endgroup\$ Dec 3, 2021 at 20:36
  • \$\begingroup\$ made a stab at it. I agree that the challenge itself is described well enough \$\endgroup\$ Dec 3, 2021 at 21:03

7 Answers 7

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Octave, 43 41 39 bytes

@(B,v)(r=rank(B))-rows(B)|r-rank([B;v])

Anonymous function that inputs the set of vectors as a matrix B with each vector in a row, and v as a row vector. The output is true (displayed as 1) if B is not a basis for v, and false (displayed as 0) if it is.

Try it online!

How it works

The set of vectors defined by the rows of B is a basis for v if and only if the rank of the matrix B equals its number of rows (i.e. the rows of B are linearly independent) and the rank of the extended matrix with v as last row is the same (i.e. v is in the linear span of the rows of B).

To save bytes, the code checks if either of those conditions is not satisfied; that is, if the rank of B minus its number of rows is nonzero or if the rank of B minus that of the extended matrix is nonzero.

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3
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JavaScript (ES6), 218 bytes

Expects (matrix)(vector). Returns \$0\$ or \$1\$.

m=>v=>!(g=m=>m.reduce(o=>m.some((r,y)=>r.some((_,x)=>m[y+w]&&(D=m=>+m||m.reduce((s,[v],i)=>v*D(m.map(([,...r])=>r).filter(_=>i--))-s,0))(m.slice(y,y-~w).map(r=>r.slice(x,x-~w)))/r[x+w]))*++w||o,w=0)-w)(m)&!~g([...m,v])

Try it online!

How?

The method is the same as the one used by Luis.

To compute the rank of a matrix, we look for the largest sub-matrix of size \$n\times n\$ whose determinant is not equal to \$0\$. This is most probably not the shortest way of doing it.

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JavaScript (Node.js), 93 bytes

f=B=>1/B[0][p=B.find(t=>t[0]),1]?!!p&&f(B.map(q=>q.map((v,i)=>v*p[0]-q[0]*p[i]).slice(1))):!p

Try it online!

For testcase [[8, 1, 2], [-7, -8, -9], [-1, 9, -5]], [12, 36, -4] it input as f([[8, -7, -1, 12], [1, -8, 9, 36], [2, -9, -5, -4]]). Extra 47 bytes may be needed if we have to input as matrix B and vector V as two parameters.

This function simply reduce the array using Gaussian elimination.

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Pari/GP, 46 bytes

f(a,b)=#a==matrank(a)&&#matintersect(a,Mat(b))

Try it online!

Takes a matrix and a column vector as input.

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2
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R, 61 bytes

Or R>=4.1, 54 bytes by replacing the word function with \.

function(B,v,`+`=Matrix::rankMatrix)+B-nrow(B)|+rbind(B,v)-+B

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Port of Luis Mendo's answer.

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Wolfram Mathematica, 51 41 bytes

Tr[1^#]&@@Solve[i=0;s@++i&/@#.#==#2]===i&

-10 (!) bytes due to @att

Try it online!

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  • \$\begingroup\$ 41 bytes \$\endgroup\$
    – att
    Dec 3, 2021 at 21:15
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Charcoal, 43 bytes

⊞θηW∧θ⌈Φ§θ⁰κ≔EΦθλEκ⁻×μιק§θ⁰ν§κ⌕§θ⁰ιθ⁼θ⟦Eη⁰

Try it online! Link is to verbose version of code. Explanation: Port of @tsh's JavaScript answer.

⊞θη

Append the vector to the basis.

W∧θ⌈Φ§θ⁰κ

Repeat until the basis becomes degenerate...

≔EΦθλEκ⁻×μιק§θ⁰ν§κ⌕§θ⁰ιθ

... perform Gaussian elimination on the basis, dropping the first row.

⁼θ⟦Eη⁰

Check whether the result is a zero vector.

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