Verify Eigenpairs

Question

In this challenge, you will be given a square matrix A, a vector v, and a scalar λ. You will be required to determine if (λ, v) is an eigenpair corresponding to A; that is, whether or not Av = λv.

Dot Product

The dot product of two vectors is the sum of element-wise multiplication. For example, the dot product of the following two vectors is:

(1, 2, 3) * (4, 5, 6) = 1*4 + 2*5 + 3*6 = 32

Note that the dot product is only defined between two vectors of the same length.

Matrix-Vector Multiplication

A matrix is a 2D grid of values. An m x n matrix has m rows and n columns. We can imagine an m x n matrix as m vectors of length n (if we take the rows).

Matrix-Vector multiplication is defined between an m x n matrix and a size-n vector. If we multiply an m x n matrix and a size-n vector, we obtain a size-m vector. The i-th value in the result vector is the dot product of the i-th row of the matrix and the original vector.

Example

        1 2 3 4 5
Let A = 3 4 5 6 7
        5 6 7 8 9

        1
        3
Let v = 5
        7
        9

If we multiply the matrix and the vector Av = x, we get the following:

x₁ = A^T₁ * v /* AT1 means the first row of A; A1 would be the first column */ = (1,2,3,4,5) * (1,3,5,7,9) = 1*1 + 2*3 + 3*5 + 4*7 + 5*9 = 1+6+15+28+45 = 95

x₂ = A^T₂ * v = (3,4,5,6,7) * (1,3,5,7,9) = 3*1 + 4*3 + 5*5 + 6*7 + 7*9 = 3+12+25+42+63 = 145

x₃ = A^T₃ * v = (5,6,7,8,9) * (1,3,5,7,9) = 5*1 + 6*3 + 7*5 + 8*7 + 9*9 = 5+18+35+56+81 = 195

So, we get Av = x = (95, 145, 195).

Scalar Multiplication

Multiplication of a scalar (a single number) and a vector is simply element-wise multiplication. For example, 3 * (1, 2, 3) = (3, 6, 9). It's fairly straightforward.

Eigenvalues and Eigenvectors

Given the matrix A, we say that λ is an eigenvalue corresponding to v and v is an eigenvector corresponding to λ if and only if Av = λv. (Where Av is matrix-vector multiplication and λv is scalar multiplication).

(λ, v) is an eigenpair.

Challenge Specifications

Input

Input will consist of a matrix, a vector, and a scalar. These can be taken in any order in any reasonable format.

Output

Output will be a truthy/falsy value; truthy if and only if the scalar and the vector are an eigenpair with the matrix specified.

Rules

• Standard loopholes apply
• If a built-in for verifying an eigenpair exists in your language, you may not use it.
• You may assume that all numbers are integers

Test Cases

 MATRIX  VECTOR  EIGENVALUE
 2 -3 -1    3
 1 -2 -1    1    1    ->    TRUE
 1 -3  0    0

 2 -3 -1    1
 1 -2 -1    1    -2   ->    TRUE
 1 -3  0    1

 1  6  3    1
 0 -2  0    0    4    ->    TRUE
 3  6  1    1

 1  0 -1    2
-1  1  1    1    7    ->    FALSE
 1  0  0    0

-4 3    1    
 2 1    2    2    ->    TRUE

2    1    2    ->    TRUE

I will add a 4x4 later.

Unreadable Test Cases that are easier for testing

Answer 1

Mathematica, 10 bytes

#2.#==#3#&

Takes input like {vector, matrix, scalar} and returns a boolean.

Answer 2

Jelly, 5 bytes

æ.⁵⁼×

This is a triadic, full program.

Try it online!

How it works

æ.⁵⁼×  Main link
       Left argument:  v (eigenvector)
       Right argument: λ (eigenvalue)
       Third argument: A (matrix)

  ⁵    Third; yield A.
æ.     Take the dot product of v and A, yielding Av.
    ×  Multiply v and λ component by component, yielding λv.
   ⁼   Test the results to the left and to the right for equality.

Answer 3

MATL, 7 bytes

*i2GY*=

Inputs in order: l,v,A.

Explanation:

*  % implicitly get l and v, multiply.
i  % get A
2G % get second input, i.e., v again
Y* % perform matrix multiplication
=  % test equality of both multiplications

Surprisingly long answer, if you ask me, mostly because I needed a way to get all the input correctly. I do not think that less than 5 bytes is possible, but it would be cool if someone found a 5 or 6 byte solution.

Basically, this calculates l*v==A*v.

Answer 4

CJam, 15 bytes

q~W$f.*::+@@f*=

Takes input in the form vector scalar matrix.

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Explanation

q~               e# Read and eval the input
  W$             e# Copy the bottom most value (the vector)
    f.*::+       e# Perform element-wise multiplication with each row of the matrix, then
                 e#   sum the results of each (ie dot product with each row) 
          @@     e# Move the resulting vector to the bottom of the stack
            f*   e# Element-wise multiplication of the scalar and the vector
              =  e# Check if the two vectors are equal

Answer 5

MATLAB, 16 bytes

@(A,v,l)A*v==v*l

Rather trivial answer. Defines an anonymous function taking the inputs, and calculates element-wise equality of the resulting vectors. A single zero in a logical array makes an array falsey in MATLAB.

Answer 6

MATLAB, 38 bytes

function r=f(m,v,s);r=isequal(m*v,s*v)

Returns 1 or 0.

MATLAB, 30 bytes

function r=f(m,v,s);r=m*v==s*v

Returns

1
1
1

as a truthy value. A falsy value is a similar vector with any or all values 0 instead of 1.

Answer 7

C++, 225 203 bytes

Thanks to @Cort Ammon and @Julian Wolf for saving 22 bytes!

#import<vector>
#define F(v,i)for(i=0;i<v.size();++i)
using V=std::vector<float>;int f(std::vector<V>m,V v,float s){V p;int i,j;F(m,i){p.push_back(0);F(v,j)p[i]+=v[j]*m[i][j];}F(v,i)v[i]*=s;return v==p;}

Try it online!

Answer 8

Julia, 17 bytes

(a,b,c)->a*b==b*c

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Answer 9

Python 2.7, 33 bytes

f=lambda m,s,e:all(m.dot(s)==e*s)

input: m=matrix, s=scalar, e=eigenvalue. M and s are numpy arrays

Answer 10

Python 3, 96 70 bytes

No builtins for matrix-vector or scalar-vector multiplication!

lambda A,L,v:all(L*y==sum(i*j for i,j in zip(x,v))for x,y in zip(A,v))

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-26 bytes by using zip thanks to @LeakyNun!

Answer 11

05AB1E, 11 bytes

vy²*O})²³*Q

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vy²*O})     # Vectorized product-sum.
       ²³*  # Vector * scalar.
          Q # Equivalence.

Answer 12

R, 30 25 bytes

s=pryr::f(all(a%*%v==λ*v))

Anonymous function, fairly straightforward. Returns TRUE or FALSE.

Answer 13

oK, 12 bytes

{y~z%+/y*+x}

This is a function, it takes in [matrix;vector;scalar].

This does not work in k for the same reasons that 3.0~3 gives 0 as a result.

---

The following works in k, with 14 bytes:

{(y*z)~+/y*+x}

Answer 14

Axiom, 27 bytes

f(a,b,c)==(a*b=c*b)@Boolean

exercises

(17) -> m:=matrix[[2,-3,-1],[1,-2,-1],[1,-3,0] ]; v:=matrix[[3],[1],[0]];
(18) -> f(m,v,1)
   (18)  true

(19) -> m:=matrix[[2,-3,-1],[1,-2,-1],[1,-3,0] ]; v:=matrix[[1],[1],[1]];
(20) -> f(m,v,-2)
   (20)  true

(21) -> m:=matrix[[1,6,3],[0,-2,0],[3,6,1] ]; v:=matrix[[1],[0],[1]];
(22) -> f(m,v,4)
   (22)  true

(23) -> m:=matrix[[1,0,-1],[-1,1,1],[1,0,0] ]; v:=matrix[[2],[1],[0]];
(24) -> f(m,v,7)
   (24)  false

(25) -> m:=matrix[[-4,3],[2,1] ]; v:=matrix[[1],[2]];
(26) -> f(m,v,2)
   (26)  true

(27) -> f(2,1,2)
   (27)  true

Answer 15

Python, 26 bytes

lambda a,b,c:c*b==a.dot(b)

a and b are numpy arrays, c is an integer.

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Answer 16

Clojure, 60 bytes

#(=(set(map(fn[a v](apply -(* v %3)(map * a %2)))% %2))#{0})

This checks that all deltas are zero, thus collapsing into the set of zero. Calling example:

(def f #(=(set(map(fn[a v](apply -(* v %3)(map * a %2)))% %2))#{0}))
(f [[1 6 3][0 -2 0][3 6 1]] [1 0 1] 4)