Question

Find a matrix such that

1) components of each of its column vectors add up to 1 and
2) components of each of its tow vectors add up to 0

Answer 1

*row

Answer 2

Assume matrix [a, b; c, d]. Then we wish
\[(1)~ a+b=0\]
\[(2)~ c+d=0\]
\[(3)~ a+c=1\]
\[(4)~ b+d=1\]
Subbing (1) and (2) into (4), we get that:
\[-a-c=1\]
which contradicts (3)
Therefore there are no solutions for 2x2 matrices at least...

Answer 3

Oh... in the general case we are trying to find matrices such that (1,1,....,1) is an eigenvector with eigenvalue 1 whose transpose has (1,1,...,1) in its nullspace. This is a contradiction since the transpose has the same eigenvalues. Therefore no such matrices exist, at least for square matrices.

Answer 4

[1, -1; 1, -1; -1, 1] is an example of a matrix satisfying your properties :D

Answer 5

Lol, i hope you don't mind my train of thoughts :P Anyways I believe my argument for the square matrix thing is wrong since I was confusing eigenvectors with eigenvalues

Answer 6

2x2 case looks nice, thank you :)

This example
[1, -1; 1, -1; -1, 1]
The components of second column don't add up to 1 right ?

Answer 7

Oops, true. Consider the matrix:
\[ [~\vec{v_1}|\vec{v_2}|\vec{v_3}|....|\vec{v_n}~] \]
Now,
\[\vec{v_1}+....+\vec{v_n}= \vec{0}\]
and
\[\vec{v_i} \cdot \vec{1}=\vec{1}, ~i=1,2,..,n\]
Which implies,
\[\vec{1}=\vec{v_1} \cdot \vec{1}=-(\vec{v_2}+...\vec{v_n}) \cdot \vec{1} = -(n-1) \cdot \vec{1}\]
which is a contradiction since n is greater than 0. Therefore there is no solution to your problem.

Does my logic seem right? :)

Answer 8

Beautiful!
Basically you've added up all the elements in the matrix in two ways :

1) if the components of each row add up to 0, then the sum of all the elements of the matrix will be 0.

2) if the components of each column add up to 1, then the sum of all the elements of the matrix will equal the number of columns in the matrix.


0 does not equal the number of columns in the matrix

Answer 9

I basically did what @Bobo-i-bo did, but in a probably more confusing looking way. Oh well I had fun, so take it or leave it I guess. I attempted to make it a little clearer by putting the dimension of the matrix and vectors as subscripts, \(A_{mn}\) just means it's an m by n matrix. You can completely remove the m and n subscripts and just imagine the matrices.

\(\vec 1_n\) is a column vector of all 1s of dimension n and \(\vec 0_n\) is a column vector of all 0s of dimension n.

OK, now to the good stuff:

This represents adding all the components of the column vectors together to get 1.
\[\vec 1_m^\top A_{mn} = \vec 1_n^\top\]

This represents adding all the components of the row vectors together to get 0.
\[A_{mn}\vec 1_n = \vec 0_m\]

Now here's the contradiction, you can complete each of these left hand sides to look like the other one,

\[\vec 1_m^\top A_{mn}\vec 1_n = \vec 1_n^\top\vec 1_n = n\]\[\vec 1_m^\top A_{mn}\vec 1_n =\vec 1_m^\top \vec 0_m =0\]

Since \(\vec 1_m^\top A_{mn}\vec 1_n \) can't be both n and 0 at the same time, there's no solution!

Answer 10

Looks more beautiful !