# Five linearly independent 3X3 matrices

1. Aug 3, 2009

### Dafe

1. The problem statement, all variables and given/known data
In the space of 2 by 2 matrices, find a basis for the subspace of matrices whose row sums and column sums are all equal. (Extra credit: Find five linearly independent 3 by 3 matrices with this property)

3. The attempt at a solution

The first one is ok. The matrix is symmetric and toeplitz :

$$\left[ \begin{array}{cc} a_{ii} & a_{ij} \\ a_{ij} & a_{ii} \end{array} \right]$$

A basis would then be:

$$\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

and

$$\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$$

Now for that yummy extra credit (which I never get):

$$\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{11} & a_{23} \\ a_{13} & a_{23} & a_{11} \end{array} \right]$$

I think I can find four linearly independent 3 by 3 matrices:

$$\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

$$\left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right]$$

$$\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$

$$\left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]$$

I do not see where I could fit a fifth one..
Any suggestions?

Thanks!

2. Aug 3, 2009

### aostraff

I'm not too sure about what you mean by all row-sum and all column-sum are equal. Does this mean every row has the same sum and every column has the same sum? If so, I don't think your last 3 matrices work.

But if you meant the sum of row i = sum of column i, then I think I can help you on that one. You have 9 degrees of freedom here (basis will have 9 matrices), so you have quite a few options to choose. Since you have the idea of obtaining symmetric matrices to fulfill the row-sum = column-sum criterion. Remember that:

$$\left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right]$$

is a symmetric matrix as well.

If you are interested in testing linear independence, you can think of them as $$\mathbb{R}^{9}$$ vectors.

3. Aug 3, 2009

### D H

Staff Emeritus

The second matrix does not satisfy the given conditions. Some of the rows and columns in that matrix sum to one, others sum to zero. (The same goes for the unquoted third and fourth matrices.)

Denoting

\begin{aligned} a_{i*} &= \sum_j a_{ij} \\ a_{*j} &= \sum_i a_{ij} \end{aligned}

You need to find matrices with $a_{i*} = a_{*j} = c$, where $c$ is some constant. $c=1$ is a reasonable choice, and so is the use of binary matrices as you did for the first part.

4. Aug 3, 2009

### Dafe

$$A = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

$$B = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right]$$

$$C = \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right]$$

$$D = \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right]$$

$$E = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

As far as I can see, those five matrices are linearly independent and the sum of the rows equals the sum of the columns.

As I on the right track here?

Thanks

5. Aug 3, 2009

### D H

Staff Emeritus
There you go. There are of course six permutation matrices, not five. The sixth is not linearly independent of the other five. Proving linear independence is a bit easier if you use this sixth permutation matrix in lieu of the identity matrix.

6. Aug 3, 2009

### Dafe

I understand, and here is the sixth permutation matrix just for good measure:

I really appreciate your help, thank you!

$$\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]$$

7. Aug 3, 2009

### D H

Staff Emeritus
You're welcome.