Julia, 31 bytes

v->(m=1:max(v...)).*(@.v'÷m>0)

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This solution creates a boolean matrix in a different manner than the R¹ and Octave² answers: the vectors v and m are integer-divided. This results in a matrix filled with the integer numbers where v is greater or equal to m and zeroes otherwise. >0 turns it to a boolean matrix and multiplication by vector m gives the desired result.

EDIT Below is the port of the Octave answer (find 10 differences:)):

# Julia, 28 bytes

x->((y=1:max(x...)).<=x').*y

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APL (Dyalog Classic), 5 bytes

⍉∘↑⍳¨

Anonymous tacit function. Did you know that APL's Mix function automatically pads with zeros?

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Nekomata, 8 bytes

ṀR$ᵒ{≤0I

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Longer than other answers, because Nekomata doesn't have a padding builtin yet. I'll add one in some future version.

ṀR$ᵒ{≤0I
Ṁ           Maximum
 R          Range from 1 to that
  $         Swap
   ᵒ{       Make a 2D array by the following function
     ≤      If left <= right, return left
      0I    Otherwise, return 0

Nekomata, 8 bytes

R:∑0*ᵚ+Ť

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R:∑0*ᵚ+Ť
R           Range from 1 to input
 :          Duplicate
  ∑         Sum
   0*       Multiply by 0
     ᵚ+     Add this to each element of the other array
       Ť    Transpose

K (ngn/k), 17 bytes

{a*~x</:a:1+!|/x}

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J, 10 bytes

0|:>:@i."0

No surprises

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0|:>:@i."0
      i."0   NB. create 0..n-1 range for each atom
   >:@       NB. then increment
0|:          NB. transpose

Scala, 76 bytes

Golfed version. Try it online!

l=>l.flatMap(v=>(1 to l.max).map(i=>if(i<=v)i else 0)).grouped(l.max).toList

Ungolfed version. Try it online!

object Main {
  def f(l: List[Int]): List[List[Int]] = {
    val m = (1 to l.max).toList
    l.flatMap { value =>
      m.map { i =>
        if (i <= value) i else 0
      }
    }.grouped(m.size).toList
  }

  def main(args: Array[String]): Unit = {
    println(f(List(3, 5, 2, 1, 6)))
  }
}

K (ngn/k), 20 bytes

{+(1+!'x),'&'-x-|/x}

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TI-Basic, 48 bytes

Ans→A
List►matr(1 or Ans,[A]
[A]
For(I,2,max(ʟA
List►matr(I(I≤ʟA),[B]
augment(Ans,[B]
End
Ansᵀ

Takes input in Ans. Similar method to this answer.

Japt, 6 bytes

Would be 4 if Japt's transpose padded with 0 instead of null.

mõ zÔÕ

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Perl 5, 62 57 + 1 (-a) = 58 bytes

$,=$";say map$i*($_-->0),@F while++$i&&(sort{$b-$a}@F)[0]

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Uiua, 8 bytes

+1⍉⬚¯1∵⇡

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+1⍉⬚¯1∵⇡
      ∵⇡  # range of each
   ⬚¯1    # filling with -1
  ⍉       # transpose
+1        # increment

Java 10, 115 bytes

a->{int l=a.length,m=0;for(int j:a)m=j>m?j:m;var r=new int[m][l];for(;l-->0;)for(m=0;m<a[l];r[m][l]=++m);return r;}

Explanation:

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a->{                  // Method with integer-array parameter and integer-matrix return-type
  int l=a.length,     //  Length of the array
      m=0;            //  Largest integer in the array, 0 for now
  for(int j:a)        //  Loop over the array
    m=j>m?            //   If the current item is larger than `m`:
       j              //    Set `m` to this item as new max
      :               //   Else:
       m;             //    Leave `m` the same
  var r=new int[m][l];//  Result-matrix of size `m` by `l`, filled with zeroes by default
  for(;l-->0;)        //  Loop over the columns
    for(m=0;m<a[l];   //   Inner loop over the rows
      r[m][l]=++m);   //    Set the cell at position `m,l` to `m+1`
  return r;}          //  Return the result-matrix

JavaScript (Node.js), 55 bytes

x=>f=(t=1,s,e=x.map(y=>y<t?0:s|=t))=>s?[e,...f(++t)]:[]

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JavaScript (Node.js), 63 bytes

x=>[...Array(Math.max(...x,t=0))].map(_=>x.map(y=>y<t?0:t,++t))

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Add++, 37 bytes

D,g,@@*,bL_1+0XA$p$+bUp
L~*,€RAbM€gBc

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R, 40 38 bytes

function(l)outer(m<-1:max(l),l,"<=")*m

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Explanation:

outer applies its third argument (the function) to all combinations of elements of its first two arguments, generating a matrix of TRUE and FALSE where each column has TRUE where 1:max(l) is less than or equal to the corresponding element of l, for the example where l=c(3,5,2,1,6):

      [,1]  [,2]  [,3]  [,4] [,5]
[1,]  TRUE  TRUE  TRUE  TRUE TRUE
[2,]  TRUE  TRUE  TRUE FALSE TRUE
[3,]  TRUE  TRUE FALSE FALSE TRUE
[4,] FALSE  TRUE FALSE FALSE TRUE
[5,] FALSE  TRUE FALSE FALSE TRUE
[6,] FALSE FALSE FALSE FALSE TRUE

Then supposing the resultant matrix is A, then A*m -> A[i,j]=A[i,j]*i which coerces TRUE to 1 and FALSE to 0, producing the desired result.

Proton, 38 bytes

a=>[[i<j?-~i:0for j:a]for i:0..max(a)]

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MATL, 8 bytes

"@:]Xho!

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Explanation

"      % Implicit input, L. For each k in L
  @    %   Push k
  :    %   Range [1 2 ... k]
]      % End
Xh     % Collect all stack contents in a cell array
o      % Convert to double matrix. The content of each cell is
       % right-padded with zeros if needed
!      % Transpose. Implicitly display

C#, 136 bytes

Data

• Input Int32[] i An array of ints
• Output Int32[,] A bidimentional array.

Golfed

i=>{int m=System.Linq.Enumerable.Max(i),l=i.Length,x,y;var o=new int[m,l];for(y=0;y<m;y++)for(x=0;x<l;)o[y,x]=i[x++]>y?y+1:0;return o;};

Ungolfed

i => {
    int
        m = System.Linq.Enumerable.Max( i ),
        l = i.Length,
        x, y;
        
    var o = new int[ m, l ];
    
    for( y = 0; y < m; y++ )
        for( x = 0; x < l; )
            o[ y, x ] = i[ x++ ] > y ? y + 1 : 0;
            
    return o;
};

Ungolfed readable

// Take an array of Int32
i => {

    // Store the max value of the array, the length and declare some vars to save some bytes
    int
        m = System.Linq.Enumerable.Max( i ),
        l = i.Length,
        x, y;
        
    // Create the bidimensional array to output
    var o = new int[ m, l ];
    
    // Cycle line by line...
    for( y = 0; y < m; y++ )
    
        // ... and column by column...
        for( x = 0; x < l; )
        
            // And set the value of the line in the array if it's lower than the the value at the index of the input array
            o[ y, x ] = i[ x++ ] > y ? y + 1 : 0;
            
    // Return the bidimentional array.
    return o;
};

Full code

using System;
using System.Collections.Generic;

namespace TestBench {
    public class Program {
        // Methods
        static void Main( string[] args ) {
            Func<Int32[], Int32[,]> f = i => {
                int
                    m = System.Linq.Enumerable.Max( i ),
                    l = i.Length,
                    x, y;
                var o = new int[ m, l ];
                for( y = 0; y < m; y++ )
                    for( x = 0; x < l; )
                        o[ y, x ] = i[ x++ ] > y ? y + 1 : 0;
                return o;
            };

            List<Int32[]>
                testCases = new List<Int32[]>() {
                    new[] { 1, 2, 5, 6, 4 },
                    new[] { 3, 5, 2, 1, 6 },
                    new[] { 1, 2, 3, 4, 3, 2, 1 },
                };

            foreach( Int32[] testCase in testCases ) {
                Console.WriteLine( " INPUT: " );
                PrintArray( testCase );

                Console.WriteLine( "OUTPUT: " );
                PrintMatrix( f( testCase ) );
            }

            Console.ReadLine();
        }

        public static void PrintArray<TSource>( TSource[] array ) {
            PrintArray( array, o => o.ToString() );
        }
        public static void PrintArray<TSource>( TSource[] array, Func<TSource, String> valueFetcher ) {
            List<String>
                output = new List<String>();

            for( Int32 index = 0; index < array.Length; index++ ) {
                output.Add( valueFetcher( array[ index ] ) );
            }

            Console.WriteLine( $"[ {String.Join( ", ", output )} ]" );
        }

        public static void PrintMatrix<TSource>( TSource[,] array ) {
            PrintMatrix( array, o => o.ToString() );
        }
        public static void PrintMatrix<TSource>( TSource[,] array, Func<TSource, String> valueFetcher ) {
            List<String>
                output = new List<String>();

            for( Int32 xIndex = 0; xIndex < array.GetLength( 0 ); xIndex++ ) {
                List<String>
                    inner = new List<String>();

                for( Int32 yIndex = 0; yIndex < array.GetLength( 1 ); yIndex++ ) {
                    inner.Add( valueFetcher( array[ xIndex, yIndex ] ) );
                }

                output.Add( $"[ {String.Join( ", ", inner )} ]" );
            }

            Console.WriteLine( $"[\n   {String.Join( ",\n   ", output )}\n]" );
        }
    }
}

Releases

• v1.0 - 136 bytes - Initial solution.

Notes

• None

Perl 6, 39 bytes

{zip (1 X..$_).map:{|@_,|(0 xx.max-1)}}

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Expanded:

{                # bare block lambda with implicit parameter ｢$_｣

  zip

    (1 X.. $_)   # turn each input into a Range that starts with 1

    .map:        # map each of those Ranges using the following code

    {            # bare block lambda with implicit parameter ｢@_｣ 
                 # (｢@_｣ takes precedence over ｢$_｣ when it is seen)

      |@_,       # slip the input into a new list

      |(         # slip this into the list

        0        # a ｢0｣
        xx       # list repeated by

          .max   # the max of ｢$_｣ (implicit method call)
          - 1    # minus 1 (so that zip doesn't add an extra row)
      )
    }
}

Note that zip terminates once the shortest input list is exhausted.

Octave, 26 bytes

@(x)((y=1:max(x))'<=x).*y'

Anonymous function that inputs a row vector and outputs a matrix.

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Explanation

Consider input x = [3 5 2 1 6]. This is a row vector of size 1×5.

1:max(x) gives the row vector [1 2 3 4 5 6], which is assigned to variable y.

The transpose of that, i.e. the column vector [1; 2; 3; 4; 5; 6], is <=-compared (element-wise with broadcast) with the input [3 5 2 1 6]. The result is the 6×5 matrix

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

Finally, multiplying (element-wise with broadcast) by the column vector [1; 2; 3; 4; 5; 6], obtained as y transposed, gives the desired result:

[1 1 1 1 1;
 2 2 2 0 2;
 3 3 0 0 3;
 0 4 0 0 4;
 0 5 0 0 5;
 0 0 0 0 6]

Mathematica, 20 bytes

PadRight@Range@#&

Contains U+F3C7 (Mathematica's builtin Transpose function)

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Usage

PadRight@Range@#&[{3, 5, 2, 1, 6}]

{
 {1, 1, 1, 1, 1},
 {2, 2, 2, 0, 2},
 {3, 3, 0, 0, 3},
 {0, 4, 0, 0, 4},
 {0, 5, 0, 0, 5},
 {0, 0, 0, 0, 6}
}

Explanation

PadRight@Range@#&

         Range@#    (* Generate {1..n} for all elements of input *)
PadRight@           (* Right-pad 0s so that all lists are equal length *)
                   (* Transpose the result *)

Actually, 17 bytes

;M╗♂R⌠╜;0@α(+H⌡M┬

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Explanation:

;M╗♂R⌠╜;0@α(+H⌡M┬
;M╗                store the maximal element (M) of the input in register 0
   ♂R              range(1, n+1) for each n in input
     ⌠╜;0@α(+H⌡M   for each range:
      ╜;0@α          push a list containing M 0s
           (+        append to range
             H       take first M elements
                ┬  transpose

Jelly, 3 bytes

Rz0

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Explanation

Rz0  Input: array A
R    Range, vectorizes to each integer
 z0  Transpose and fill with 0

Pyke, 3 bytes

This uses the new feature of Pyke, hex encodings... The best part is that we tie Jelly! Raw bytes:

4D 53 AC

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The ASCII-Pyke equivalent would be 4 bytes:

MS.,

How?

4D 53 AC   - Full program.

4D         - Map.
   53      - Inclusive range.
      AC   - Transpose with zeroes.
           - Output implicitly.

-------------------------------------

MS.,   - Full program.

M      - Map.
 S     - Inclusive range.
  .,   - Transpose with zeroes.
       - Output implicitly.

Here is a pretty-print version with ASCII, and here is one with hex encodings.

C (gcc), 142 bytes

i,j,k;main(c,v)char**v;{for(;++i<c;k=k<*v[i]?*v[i]:k)printf("%c ",*v[i]);for(i=48;puts(""),i++<k;)for(j=1;j<c;)printf("%c ",i<=*v[j++]?i:48);}

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Pyth, 6 bytes

.tSMQZ

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Explanation

.tSMQZ   - Full program.

  SMQ    - Get the inclusive unary ranges for each.
.t       - Transpose, padding with copies of...
     Z   - ... Zero.
         - Implicit print.

A non-built-in transpose version would be:

mm*hd<dkQeS

This works as follows:

mm*hd<dkQeS   - Full program.

m        eS   - Map over [0, max(input)) with a variable d.
 m      Q     - Map over the input with a variable k.
   hd         - d + 1.
  *           - Multiplied by 1 if...
     <dk      - ... d is smaller than k, else 0.
              - Output implicitly.

05AB1E, 4 bytes

€L0ζ

Try it online! or as a nicely formatted test suite

€L   # For each: get the range 1..input
  0ζ # zip, padding with 0s

Python 3, 54 bytes

lambda x:[[-~i*(i<a)for a in x]for i in range(max(x))]

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Husk, 4 bytes

Returns a list of lists

T0mḣ

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Explanation

  m    Map over the input
   ḣ   Range from 1 to n
T0     Transpose, padding with 0s