Construct the Identity Matrix

Question

The challenge is very simple. Given an integer input n, output the n x n identity matrix. The identity matrix is one that has 1s spanning from the top left down to the bottom right. You will write a program or a function that will return or output the identity matrix you constructed. Your output may be a 2D array, or numbers separated by spaces/tabs and newlines.

Example input and output

1: [[1]]
2: [[1, 0], [0, 1]]
3: [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
4: [[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]]
5: [[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1]]

1
===
1

2
===
1 0
0 1

3
===
1 0 0
0 1 0
0 0 1

etc.

This is code-golf, so the shortest code in bytes wins.

Answer 1

MATL, 2 bytes

Xy

A translation of my Octave answer.

Try it online.

A 4 byte version with no built-ins (thanks to Luis Mendo):

:t!=
:     take input n and a generate row array [1,2,...n]
 t    duplicate
  !   zip
   =  thread compare over the result

Answer 2

TI-BASIC, 2 bytes

identity(Ans

Fun fact: The shortest way to get a list {N,N} is dim(identity(N.

Here's the shortest way without the builtin, in 8 bytes:

randM(Ans,Ans)^0

randM( creates a random matrix with entries all integers between -9 and 9 inclusive (that sounds oddly specific because it is). We then take this matrix to the 0th power.

Answer 3

Julia, 9 3 bytes

eye

This is just a built-in function that accepts an integer n and returns an nxn Array{Float64,2} (i.e. a 2D array). Call it like eye(n).

Note that submissions of this form are acceptable per this policy.

Answer 4

APL, 5 bytes

∘.=⍨⍳

This is a monadic function train that accepts an integer on the right and returns the identity matrix.

Try it here

Answer 5

Octave, 10 4 bytes

@eye

Returns an anonymous function that takes a number n and returns the identity matrix.

Answer 6

R, 4 bytes

diag

When given a matrix, diag returns the diagonal of the matrix. However, when given an integer n, diag(n) returns the identity matrix.

Try it online

Answer 7

Python 2, 42 bytes

lambda n:zip(*[iter(([1]+[0]*n)*n)]*n)[:n]

An anonymous function, produces output like [(1, 0, 0), (0, 1, 0), (0, 0, 1)],

First, creates the list ([1]+[0]*n)*n, which for n=3 looks like

[1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0]

Using the zip/iter trick zip(*[iter(_)]*n to make groups of n gives

[(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)]

Note that the 1 comes one index later each time, giving the identity matrix. But, there's an extra all-zero row, which is removed with [:n].

Answer 8

Jelly, 4 bytes

R=€R

Doesn't use a built-in. Try it online!

How it works

R=€R    Main link. Input: n

R       Range; yield [1, ..., n].
   R    Range; yield [1, ..., n].
 =€     Compare each.
        This compares each element of the left list with the right list, so it
        yields [1 = [1, ..., n], ..., n = [1, ..., n]], where comparison is
        performed on the integers.

Answer 9

J, 4 bytes

=@i.

This is a function that takes an integer and returns the matrix.

Answer 10

Haskell, 43 37 bytes

f n=[[0^abs(x-y)|y<-[1..n]]|x<-[1..n]]

Pretty straightforward, though I think one can do better (without a language that already has this function built in, as many have done).

Edit: dropped some bytes thanks to Ørjan Johansen

Answer 11

Pyth, 7 bytes

XRm0Q1Q

Try it online: Demonstration

Creating a matrix of zeros and replacing the diagonal elements with ones.

Answer 12

JavaScript ES6, 68 62 52 bytes

Saved 10 bytes thanks to a neat trick from @Neil

x=>[...Array(x)].map((_,y,x)=>x.map((_,z)=>+(y==z)))

Trying a different approach than @Cᴏɴᴏʀ O'Bʀɪᴇɴ's. Could possibly be improved.

Answer 13

Retina, 25

Credit to @randomra and @Martin for extra golfing.

\B.
 0
+`(.*) 0$
$0¶0 $1

Try it online.

Note this takes input as a unary. If this is not acceptable, then decimal input may be given as follows:

Retina, 34

.+
$0$*1
\B.
 0
+`(.*) 0$
$0¶0 $1

Try it online.

Answer 14

Python 3, 48

Saved 1 byte thanks to sp3000.

I love challenges I can solve in one line. Pretty simple, build a line out of 1 and 0 equal to the length of the int passed in. Outputs as a 2d array. If you wrap the part after the : in '\n'.join(), it'll pretty print it.

lambda x:[[0]*i+[1]+[0]*(x+~i)for i in range(x)]

Answer 15

Haskell, 54 bytes

(#)=replicate
f n=map(\x->x#0++[1]++(n-x-1)#0)[0..n-1]

f returns the identity matrix for input n. Far from optimal.

Answer 16

Lua, 77 75 65 bytes

x,v=z.rep,io.read()for a=1,v do print(x(0,a-1)..'1'..x(0,v-a))end

Well, I'm not sure if lua is the best language for this with the two period concatenation... But hey, there's a shot at it. I'll see if there's any improvements to be made.

EDIT:

I figured something out on accident which I find rather odd, but, it works.

In Lua, everyone knows you have the ability to assign functions to variables. This is one of the more useful CodeGolf features.

This means instead of:

string.sub("50", 1, 1) -- = 5
string.sub("50", 2, 2) -- = 0
string.sub("40", 1, 1) -- = 4
string.sub("40", 2, 2) -- = 0

You can do this:

s = string.sub
s("50", 1, 1) -- = 5
s("50", 2, 2) -- = 0
s("40", 1, 1) -- = 4
s("40", 2, 2) -- = 0

But wait, Lua allows some amount of OOP. So you could potentially even do:

z=""
s = z.sub
s("50", 1, 1) -- = 5
s("50", 2, 2) -- = 0
s("40", 1, 1) -- = 4
s("40", 2, 2) -- = 0

That will work as well and cuts characters.

Now here comes the weird part. You don't even need to assign a string at any point. Simply doing:

s = z.sub
s("50", 1, 1) -- = 5
s("50", 2, 2) -- = 0
s("40", 1, 1) -- = 4
s("40", 2, 2) -- = 0

Will work.

---

So you can visually see the difference, take a look at the golfed results of this:

Using string.sub (88 characters)

string.sub("50", 1, 1)string.sub("50", 2, 2)string.sub("40", 1, 1)string.sub("40", 2, 2)

Assigning string.sub to a variable (65 characters)

s=string.sub s("50", 1, 1)s("50", 2, 2)s("40", 1, 1)s("40", 2, 2)

Assigning string.sub using an OOP approach (64 characters)

z=""s=z.sub s("50", 1, 1)s("50", 2, 2)s("40", 1, 1)s("40", 2, 2)

Assigning string.sub using a.. nil approach? (60 characters)

s=z.sub s("50", 1, 1)s("50", 2, 2)s("40", 1, 1)s("40", 2, 2)

If someone knows why this works, I'd be interested.

Answer 17

C, 59 or 59 56 or 56

Two versions of identical length.

3 bytes saved due to suggestion from anatolyg: (n+1) --> ~n

Iterates i from n*n-1 to zero. Prints a 1 if i%(n+1) is zero, otherwise 0. Then prints a newline if i%n=0 otherwise a space.

i;f(n){for(i=n*n;i--;)printf(i%n?"%d ":"%d\n",!(i%~n));}

i;f(n){for(i=n*n;i--;)printf("%d%c",!(i%~n),i%n?32:10);}

Answer 18

K6, 1 byte

=

= is exactly this

Try it online!

Answer 19

Mata, 4 bytes

I(3)

Output

[symmetric]
       1   2   3
    +-------------+
  1 |  1          |
  2 |  0   1      |
  3 |  0   0   1  |
    +-------------+

Mata is the matrix programming language available within the Stata statistical package. I(n) creates an identity matrix of size n*n

Answer 20

Brain-Flak, 206 170 162 bytes

(([{}])){({}<>(())<><(({})<{({}()(<>)<>)}{}>)>)}{}(({}<><(())>)){({}()<({[()]<({}()<({}<>((((()()()()){}){}){})((()()()()){}){})<>>)>}{})>)<>((()()()()()){})<>}<>

Try it online!

Answer 21

Japt, 14 12 10 bytes

Uo £Z®¥X|0

Test it online! Note: this version has a few extra bytes to pretty-print the output.

Uo £Z®¥X|0    // Implicit: U = input integer
Uo £          // Create the range [0..U). Map each item X and the full array Z to:
    Z®        //  Take the full array Z, and map each item Z to:
      ¥X|0    //   (X == Z) converted to a number. 1 for equal, 0 for non-equal.
              // Implicit: output result of last expression

Answer 22

Python 3.5 with NumPy - 57 49 30 bytes

import numpy
numpy.identity

NumPy.identity takes in an integer, n, and returns a n by n identity matrix. This answer is allowable via this policy.

Answer 23

Pyth, 8 bytes

mmsqdkQQ

Try it here.

Answer 24

Mathematica, 35 Bytes

without using IdentityMatrix

Table[Boole[i==j],{i,1,#},{j,1,#}]&

Answer 25

Javascript, 40

f=
n=>'0'.repeat(n).replace(/./g,"$`1$'\n")

I.oninput=_=>O.innerHTML=f(+I.value)
I.oninput()

<input id=I value=5>
<pre id=O>

Answer 26

R, 50 bytes

f=function(n)matrix(c(1,rep(c(rep(0,n),1),n-1)),n)

Without using diag(), because I started coding without checking if there was a command for that... Also, I like my solution, despite being way too long.

Answer 27

Octave, 15 bytes

Because built-ins are no fun!

@(x)(k=1:x)==k'
@(x)~~diag(1:x)

Try it online!

The first one creates a vector from 1, 2, .. x, and compares it to itself transposed, expanding it to a square matrix where the diagonal elements are 1.

The second one creates a diagonal matrix with 1, 2, .. x is on the diagonal, then negates it twice, to get the diagonal equal 1, 1, .. 1.

Answer 28

Jolf, 7 + 1 = 8 bytes

Replace all ☺s with \x01 and \xad with a soft hyphen, or use the online interpreter. (Turn pretty output on for one byte.)

!☺!X\xad☺j
  !X       eye
    \xad☺   take one argument
         j  (the input)
!☺         get the data
           implicit printing

Probably more interesting, a solution without a builtin is 11 + 1 = 12 bytes

ZXZyjjdP=Sn
  Zyjj       create a matrix of width = height = j (input)
ZX    d      matrix map
        =Sn  if x-marker is y-marker
       P     as a number

Answer 29

JavaScript, 75 74 bytes

n=>eval("for(a=[],i=0;i<n;i++){a[i]=[];for(j=0;j<n;)a[i][j]=+(i==j++)};a")

Can be improved, probably.

If n=0 isn't necessary... 70 bytes

n=>eval("for(a=[i=0];i<n;i++)for(a[i]=[j=0];j<n;)a[i][j]=+(i==j++);a")

Answer 30

K, 7 bytes

t=\:t:!

Take the equality cross product of two vectors containing [0,n).

In action:

  t=\:t:!3
(1 0 0
 0 1 0
 0 0 1)
  t=\:t:!5
(1 0 0 0 0
 0 1 0 0 0
 0 0 1 0 0
 0 0 0 1 0
 0 0 0 0 1)

Answer 31

Java, 60 bytes

n->{int[][]i=new int[n][n];for(;n-->0;)i[n][n]=1;return i;};

Creates a 2D array and replaces elements where row and column are equal with 1.

Answer 32

CJam, 7 bytes

{,_ff=}

This is a code block that pops an integer from the stack and pushes a 2D array in return.

Try it online!

Answer 33

Perl, 39 33 bytes

/$/,say map$`==$_|0,@%for@%=1..<>

Thanks to Ton Hospel for saving 6 bytes

Running with the -E perlrun:

$ echo 3 | perl -E'@%=1..<>;$a=$_,say map{$a==$_|0}@%for@%'
100
010
001

Answer 34

Mathematica, 14 bytes

IdentityMatrix

Test case

IdentityMatrix[4]
(* {{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}} *)

Answer 35

Rust, 59 bytes

|n|{let mut l=vec![vec![0;n];n];for i in 0..n{l[i][i]=1};l}

Defines an anonymous function, creates a vector of vectors of the specified length, sets the diagonals to 1 and returns it. Boring, but shorter than any iterator-based approaches due to .collect overhead.

Alternatively, if returning an iterator of iterators is allowed it can be done in 44 characters:

|n|(0..n).map(|i|(0..n).map(|j|(j==i)as u8))

Answer 36

Matlab, 3 bytes

eye

Does exactly this. eye(n) for an n-by-n matrix.

and also one without eye, 15 bytes:

diag(ones(1,n))

Answer 37

R, 32 31 bytes

function(n)+outer(1:n,1:n,"==")

Try it online!

Thanks to Kirill L. for -1 byte

Alternative to Ivanneke's solution, but obviously far longer than just plain old diag.

Answer 38

R, 28 bytes

function(n,x=1:n)+!x%o%x-x^2

Try it online!

For pure fun, just another attempt at a non-built-in R solution. Uses the shortcut %o% version of the outer function.

Answer 39

05AB1E, 4 bytes

°¨œÙ

Try it online!

Unique permutations of (10^input)[0:-1]

Answer 40

PowerShell, 52 bytes

param($a)1..$a|%{$d++;(1..$a|%{+($d-eq$_)})-join' '}

Since the only built-in .NET class for matrices is ... umm ... inadequate, shall we say, we're just going to construct and output a string representation here...

Takes input $a then executes a double-for-loop over range 1..$a. Each outer loop we increment helper variable $d (the first loop, since $d isn't initialized, turns $d++ into $null + 1 = 1 ... in PowerShell logic it makes sense), and each inner loop is simply an integer-cast with + of an equality between $d and the current element $_. Each inner loop is -joined together with spaces so it prints nicely, and each outer loop causes a newline, so output code is pretty cheap.

Example

PS C:\Tools\Scripts\golfing> .\construct-identity-matrix.ps1 10
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1

Answer 41

Haskell, 27 bytes

import Data.Matrix
identity

Maybe a little bit boring.

Usage example:

Prelude Data.Matrix> identity 4
( 1 0 0 0 )
( 0 1 0 0 )
( 0 0 1 0 )
( 0 0 0 1 )

Answer 42

Ruby, 56 48 bytes

->n{(([1]+[0]*n)*(n-1)+[1]).each_slice(n).to_a}

Explanation:

Concatenates n - 1 sequences of 1 followed by n 0s into an array, and appends 1 to the end. Then slices the result into n arrays of length n and evaluates to an array having the slices as its elements.

Per Alex A., shaved 8 bytes by putting it in the form of a lambda.

Answer 43

Ruby,38 bytes

Returns a 2D array.

->n{(0..n-1).map{|i|s=[0]*n;s[i]=1;s}}

Iterates through each row of the array. For each iteration generates a row of n zeros, then changes one of them to a 1

Usage

f=->n{(0..n-1).map{|i|s=[0]*n;s[i]=1;s}}
p f[5]

[[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1]]

A different approach (string):

Ruby, 47 bytes

->n{(0..n-1).map{|i|s='0 '*n;s[i*2]=?1;puts s}}

for each row, makes a string of '0 'repeated n times, changes one of the 0s to 1, and prints it

Answer 44

PARI/GP, 11 5 bytes

matid

is sufficient, or

n->matid(n)

(11 bytes) as a 'roll-your-own' closure. If you want to work entirely by hand,

n->matrix(n,n,i,j,i==j)

(23 bytes) should suffice.

Answer 45

Perl, 31 24 bytes

Includes +3 for -p (code contains ' so I can't just use the implied -e)

Run with the count on STDIN, e.g.

./diagonal.pl <<< 3

Outputs:

100
010
001

diagonal.pl:

#!/usr/bin/perl -p
$_=0 x$_;s/./$`1$'
/g

Answer 46

PHP, 64 bytes

for(;$i<$argn**2;$m?:$k++)echo$k==$m?:0,($m=++$i%$argn)?" ":"
";

Try it online!

Answer 47

Jelly, 2 bytes

=þ

Try it online!

Answer 48

tinylisp repl, 107 bytes

(d =(q((A R C S)(i C(=(c(e R C)A)R(s C 1)S)A
(d #(q((A R S)(i R(#(c(=()R S S)A)(s R 1)S)A
(d f(q((S)(#()S S

Using the repl saves 6 bytes in implied closing parentheses at the ends of lines. Defines a function f which takes a positive integer and returns the identity matrix of that size as a nested list. Try it online!

Explanation

The function = constructs a row in the matrix. Its arguments are (in order) an Accumulator, the Row number, the current Column number, and the Size of the matrix. If the column is not zero (i C ...), then cons a number to the accumulator (1 if the row and column are equal, 0 otherwise) (c (e R C) A), decrease the column number (s C 1), and recurse (= ...). Otherwise, return the accumulator A.

The function # constructs the matrix as a list of rows. Its arguments are an Accumulator, the current Row number, and the Size of the matrix. If the row number is not zero (i R ...), then generate a row (=()R S S) and cons it to the accumulator (c ... A), decrease the row number (s R 1), and recurse (# ...). Otherwise, return the accumulator A.

Finally, the function f is simply a single-argument wrapper function that calls #.

Answer 49

Clojure, 45 bytes

#(for[i(range %)](assoc(vec(repeat % 0))i 1))

Answer 50

PHP, 49 bytes

function($n){for(;$n--;)$k[$n][$n]=1;return $k;};

There is a bit of a caveat with this answer, and I will delete it or mark it as non-competing if it is deemed invalid.

The array returned by this function technically only has n elements, those being a 1 at every intersection. ie: 0:0 1:1, 2:2, etc

However, not only does this satisfy the definition of the identity matrix as given by the question:

The identity matrix is one that has 1s spanning from the top left down to the bottom right

It also works if used as an identity matrix, as an empty value in an array in PHP is treated as 0 if used arithmetically.

Answer 51

Axiom, 68 bytes

f(n:PI):Matrix INT==matrix([[(i=j=>1;0)for i in 1..n]for j in 1..n])

test and results

(30) -> for i in 1..3 repeat output f(i)
   [1]
   +1  0+
   |    |
   +0  1+
   +1  0  0+
   |       |
   |0  1  0|
   |       |
   +0  0  1+
                                                               Type: Void

Answer 52

Python 2, 52 bytes

y=x=2**input()
while~-x:x/=2;print str(bin(y+x))[3:]

Try it online!

Using binary representation of powers of 2.

Answer 53

Mathematica ,26 bytes

without using IdentityMatrix

Boole[#==#2]&~Array~{#,#}&

Answer 54

PowerShell, 45 bytes

{param($n)0..--$n|%{"$(,0*$_+1+,0*($n-$_))"}}

Try It Online!

Answer 55

Excel VBA, _{55 52} 43 Bytes

Anonymous VBE immediate window function that takes input as number from cell [A1] and outputs the identity matrix to the range of [A1].Resize(n,n)

[A1].Resize([A1],[A1])="=1*(Row()=Column())

-2 bytes thanks to Engineer Toast

Previous Version

n=[A1]:[A1].Resize(n,n)=0:For i=1To n:Cells(i,i)=1:Next

Answer 56

Bash (w/ coreutils and awk), 51 50 bytes

yes 0|head -$((N*N))|xargs -n$N echo|awk {\$NR=1}1

Usage

Set the N environment variable to the matrix size

$ N=5
$ yes 0|head -$((N*N))|xargs -n$N echo|awk {\$NR=1}1
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1

Explanation

• yes 0 generates an endless stream of 0s to stdout;
• head -$((N*N)) takes the first N² lines;
• xargs -n$N echo groups every N lines and outputs them together;
• awk {\$NR=1}1 changes the value of the field whose index matches the line number to a 1 (otherwise it remains 0) and then outputs the whole record.

There may be a shorter way to generate N² lines of 0s, but I haven't found one yet...

Answer 57

Vim, 47 keystrokes

"zy$␘a@q␛"xYoa␛"zPy$qqo␛@"0␛q@x3Gqqr1jlq@x1G2dd

␘ is CtrlX and ␛ is Esc

Explanation

"zy$␘           Save n for use later
a@q␛"xY         Create the macro `x` that repeats the macro `q` n-1 times
oa␛"zPy$        Create an anonymous macro that inserts a character n times
qqo␛@"0␛q@x    Create the macro `q` that adds a new line with n zeroes and run it n times
3G               Go to the top-left corner of the matrix
qqr1jlq@x        Redefine `q` as a macro that replaces the character at the cursor with
                   a one and moves the cursor one step down-right. Run `q` n times
1G2dd            Delete the two lines used in macro defenitions

Answer 58

Tcl, 92 bytes

proc I n {time {incr j;set a "";set i 0;time {lappend a [expr [incr i]==$j]} $n;puts $a} $n}

Try it online!

Answer 59

Husk, 4 bytes

´Ṫ=ḣ

Try it online!

Explanation

´Ṫ=ḣ  -- example input: 2
   ḣ  -- range: [1,2]
´     -- duplicate argument: [1,2] [1,2]
 Ṫ    -- outer product by
  =   -- | equality: [[1==1,1==2],[2==1,2==2]]
      -- : [[1,0],[0,1]]

Answer 60

Gol><>, 10 bytes

IR0PlF}D|;

Try it online!

The output for n = 5 looks like this:

[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]

How it works

IR0PlF}D|;

IR0P       Take input n, then fill the stack with that many 0s
           and then make the top 1; stack = [0 ... 0 1]
    lF..|  Do the following n times...
      }D   Rotate the stack once to the right, and print the whole stack
         ; Terminate

Answer 61

05AB1E, 6 bytes

Lã€Ësô

Try it online.

Explanation:

L         # List in range [1,n]
          #  i.e. 3 → [1,2,3]
 ã        # Cartesian power of this list
          #  i.e. [1,2,3] → [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
  €Ë      # Check for every pair if they are equal or not
          #  i.e. [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
          #   → [1,0,0,0,1,0,0,0,1]
    sô    # Split it into parts equal to the input
          #  i.e. [1,0,0,0,1,0,0,0,1] and 3 → [[1,0,0],[0,1,0],[0,0,1]]

Answer 62

Perl 6,  41 37  35 bytes

{$/=1..$_;$/.map: {$/.map: {+($^a==$_)}}} # 41 bytes
{$/=1..$_;$/.map: {$/.map: +(*==$_)}}     # 37 bytes
{$_=1..$^a;.map: {.map: +(*==$^a)}}       # 35 bytes

This outputs a list of lists.

Usage:

# give it a lexical name
my &identity-matrix = {…}

# format it so that it is readable
sub readable ( @_ ) { @_.join: "\n" }

say readable identity-matrix 10

1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1

Answer 63

, 3 chars / 6 bytes

Мƕï

Try it here (Firefox only).

I'm finding a lot of these 3-char solutions. It's just a builtin.

Bonus solution, 11 chars / 22 bytes

⩥ïⓜãĉ⇝+($≔a

Try it here (Firefox only).

The code in the interpreter link uses some more bytes to pretty-print output.

Explanation

⩥ïⓜãĉ⇝+($≔a // implicit: ï=input
⩥ïⓜ         // create a range [0,ï) and map over it
    ãĉ⇝      // get a copy of range ã and map over it
       +($≔a // output 1 or 0 depending on whether item in first map == item in current map
             // implicit output

Answer 64

Scala, 53 51 bytes

(n:Int)=>Seq.tabulate(n,n){(i,j)=>if(i==j)1 else 0}

Answer 65

Samau, 5 bytes

,;`=o

It's a function that takes a number and return a 2D array.

,;`=o
,      range from 0 to n-1
 ;     duplicate
    o  take the outer product by
  `=   equality

Answer 66

Mouse, 75 bytes

N:1I:(I.N.1+<^1J:(J.I.<^0!32!'1J.+J:)1!32!'(J.N.<^0!32!'1J.+J:)10!'1I.+I:)$

This is a full program that reads from STDIN and prints to STDOUT. Each line will have a single trailing space and there will be one trailing newline at the end of the output.

Explanation:

~ Read N from STDIN
? N:

~ Initialize a row index
1 I:

~ Loop while we've printed fewer than N+1 rows
( I. N. 1 + < ^

  ~ Initialize a column index
  1 J:

  ~ Print I-1 space-separated zeros
  ( J. I. < ^
    0 !
    32 !'
    1 J. + J:
  )

  ~ Print 1 and a space
  1 !
  32 !'

  ~ Print the remaining zeros
  ( J. N. < ^
    0 !
    32 !'
    1 J. + J:
  )

  ~ Newline
  10 !'

  ~ Increment the row index
  1 I. + I:
)
$

Answer 67

Java, 200 Bytes

This is the shortest working java code i could come up with. Nothing to special ;)

class a{public static void main(String[]a){int c=Integer.parseInt(a[0]);int[][]b=new int[c][c];for(int d=0;d<c;d++){for(int e=0;e<c;e++){b[d][d]=1;System.out.print(b[d][e]);}System.out.println();}}}

Partly posting this hoping someone could explain why TNT 's answer is valid (it has no class declaration and no main(String[] args))? I'm new to code golf and i thought that answers had to be a working program?

Answer 68

Python 3, 22 bytes

from numpy import*;eye

It's a bit boring, but it's short! Because of discussion on meta, I think this follows the spirit of the rules a bit more closely.

Answer 69

Ruby 42 bytes

->n{([1]+[0]*(n-1)).permutation.to_a.uniq}

creates every permutation of the array [1,0,0,0...] and then takes the unique ones. this will end up with the ones we want, in the correct order.

Answer 70

C++, 111 bytes

I love abusing for loops.

#include <stdio.h>
void f(int n){for(int j,i=0;i++<n;putchar(10))for(j=0;j++<n;putchar(32))putchar(i-j?48:49);}

Answer 71

Oracle SQL 11.2, 102 bytes

SELECT SUBSTR(RPAD(0,:1,0),:1-LEVEL+2)||1||SUBSTR(RPAD(0,:1,0),LEVEL+1)FROM DUAL CONNECT BY LEVEL<=:1;

Answer 72

Python 3, 116 107 characters

def M(n):
 x=[]
 for i in range(n):
  r=[]
  for j in range(n):
   r+=[1 if i==j else 0]
  x+=[r]
 return x

Output

iM(1) 
[[1]]
iM(2)
[[1, 0], [0, 1]]
iM(3)

Answer 73

Jellyfish, 12 bytes

P`N(Rri
   -

Try it online! This entry is non-competing.

Explanation

• i is input, and r computes the range from 0 to input-1.
• The operator ( is a left hook. On inputs - and R, it applies R (rotate) to the range and its negated version. This gives the i×i matrix whose rows are the ranges ri, rotated k steps to the right for each row k.
• `N computes the logical negation of each entry: 0 goes to 1 and everything else to 0. Since the 0s are on the diagonal, this gives the identity matrix.
• P prints the matrix.

Answer 74

picTuring - 16 States (320 bytes)

I thought this challenge might be a good chance to show off my new Turing Machine interpreter.

0 * 0 * r
0 _ 1 * r
1 * 2 1 l
2 * 2 * l
2 _ d * l
d _ f * r
d 0 d 1 l
d 1 p 0 r
p * p * r
p _ q * r
q * q * r
q _ 2 0 l
f * f _ r
f _ g * r
g 1 g _ r
g 0 r 1 r
r * r * l
r _ c * r
c _ n * d
c 0 o * d
c 1 i * d
o * e 0 u
i * j 0 r
j * h 1 u
h * k * r
k _ halt * *
k * c * *
e * c * r
n * r * l

The input must be in binary (1 / 0), with the number terminating at the head (the white circle).

How to Run it:

In case you didn't realize that the link in the header leads to my interpreter, you can find it here -> http://fred-choi.com/projects/picTuring/index.html

Here is the compressed test case (since my interpreter does not support permalinks yet):

AwAlH0QRhBaGLTgJjCSMg AwAgVCoQTgUKB9EBGcI6ogJhSANrDtmgTkgCYmyyVIBmaclolqBrIADpOtdxP17ckAR0awxESbzFIcoArAYQGSOKpABzcbG2pta3T2i44sEzComkAYx12kAOzTlYd0AHsXb3AEtvsF4QAKY8AK7U-hAAVsawsRAAFrgRyRAA1uKZSIkAhngALmhgsJkQdhAlsKHlOs6WEHhAA

To use it, just hover over save, paste the compressed code in the text box, and click load.

To edit the tape, make sure you're in "Type" mode (Edit -> Paint), then double click the tape, and a red box should show up, indicating that you're now editing the tape.

Once you're finished, hit Edit -> Ok to save the tape, then Controls -> Run. Controls -> Reset will restore the tape to when you hit Edit -> Ok.

Again, this is interpreter is still indev, I don't have a manual yet, so feel free to leave comments if I left anything out.

Answer 75

PHP, 106 bytes

<?php $i=$argv[1];for($j=0;$j<$i;$j++){for($h=0;$h<$i;$h++){$l="0 ";if($h==$j)$l="1 ";echo$l;}echo"\n";}?>

Based on a double for loop, it uses the command-line argument for the input.

php 70365.php 3
1 0 0 
0 1 0 
0 0 1

Expanded code:

<?php
    $i = $argv[1];
    for($j = 0; $j < $i; $j++)
    {
        for($h = 0; $h < $i; $h++)
        {
            $l="0 ";
            if($h == $j)
                $l="1 ";
            echo $l;
        }
        echo "\n";
    }
?>

Answer 76

V, 16, 11 bytes (non-competing?)

Ài0 ÀÄòjl

Try it online!

Answer 77

Common Lisp, 96 94 92 bytes

(lambda(n)(set'a(make-array(list n n):initial-element 0))(dotimes(i n)(setf(aref a i i)1))a)

Try it online!

Answer 78

PHP, 55 bytes

while($p**.5<$z=+$argn)echo$p%-~$z?0:1,"
 "[++$p%$z>0];

requires PHP 5.6 or later for the power operator. Note that the second line begins with a space!

Run as pipe with -nR or try it online.

Answer 79

Charcoal, 7 bytes

ＥθＥθ⁼κμ

Try it online!

This returns an array, which in Charcoal is not readable in its default format. A version using Wolfram Lists is included below for readability.

Explanation

Ｅθ       Map for each number from 0 to first input
   Ｅθ    Map for each number from 0 to first input
      ⁼κμ Return whether both loop indices (yes, indices) are the same. The variables here are: i = first loop variable, k = first loop index, l = second loop variable, m = second loop index.

Readable version
Even accepts input as unary :P

Answer 80

PHP 88 Bytes

Try it online

Code

<?php for($i=0;$i++<$argv;){$a=array_fill(1,$argv,0);$a[$i]=1;echo implode(",",$a)."
";}

Explanation

<?php 
  for($i=0;$i++<$argv;){          // loop up to the argument
      $a=array_fill(1,$argv,0);   // create an array fill with 0 with length = $argv value
      $a[$i]=1;                   // use the current iteration to place the "1"
      echo implode(",",$a)."      // implode to output the matrix
     ";}

Output

  $argv=5

  1,0,0,0,0
  0,1,0,0,0
  0,0,1,0,0
  0,0,0,1,0
  0,0,0,0,1

Answer 81

Forth (gforth), 48 bytes

: f dup 0 do dup 0 do 0 i j = - . loop cr loop ;

Try it online!

Explanation

uses nested loops to output the matrix. For each element it outputs 1 if the indexes are equal, and 0 otherwise.

Code Explanation

:f           \ start a new word definition
  dup        \ duplicate input
  0 do       \ outer counted loop from 0 to input - 1
    dup      \ duplicate input again 
    0 do     \ inner counted loop from 0 to input - 1
      0      \ put a 0 on the stack (used to convert -1 to 1)
      i j =  \ -1 if i equals j, 0 otherwise
      - .    \ subtract from 0 to convert -1 to 1, and output
    loop     \ end inner loop
    cr       \ output a newline
  loop       \ end outer loop
;            \ end word definition

Answer 82

D, 114 bytes

void f(int n){import std.stdio;int N=n*n;while(N--){if(N%(n+1))write("0 ");else write("1 ");if(N%n==0)writeln();}}

Try it online!

Answer 83

Seriously, 24 bytes

,;╗r`╙k╜#"{:0%sb}"%f#`MR

Try it online!

Explanation:

,;╗r`╙k╜#"{:0%sb}"%f#`MR
,;╗                       get input (n), save a copy in register 0
   r                      push [0..n-1]
    `                `M   map the function:
     ╙k                     push [2**a]
       ╜#"{:0%sb}"%         push "{:0nb}"
                   f#       format and explode string
                            (push binary representation, padded with zeroes to n digits)
                       R  reverse list

Answer 84

Sage, 15 bytes

identity_matrix

Exactly what it says on the tin

Try it online

Thanks to Lynn for pointing out that I was being a doofus.