1. Constructing a Matrix A

An $n x n$ matrix $A$ has the property that the sum of all its rows with odd numbers is equal to the sum of all its rows with even numbers. Find the determinant of the matrix $A$.

I am confident in my ability to compute the determinant, but I am not quite sure how to find $A$.

2. Re: Constructing a Matrix A

Let $A$ be any matrix satisfying that condition. You can perform certain elementary row operations on $A$ without changing its determinant. Suppose the rows of $A$ are $\begin{pmatrix}\vec{r_1} \\ \vec{r_2} \\ \vdots \\ \vec{r_n}\end{pmatrix}$. Let $E$ be a composition of elementary row operations so that $\vec{r_1} \leftarrow \sum_{i=1}^n(-1)^{i+1}\vec{r_i}$. The operation that adds a multiple of one row to another doesn't change the determinant. So, the determinant of $EA$ is the determinant of $A$.

3. Re: Constructing a Matrix A

Originally Posted by SlipEternal
Let $A$ be any matrix satisfying that condition. You can perform certain elementary row operations on $A$ without changing its determinant. Suppose the rows of $A$ are $\begin{pmatrix}\vec{r_1} \\ \vec{r_2} \\ \vdots \\ \vec{r_n}\end{pmatrix}$. Let $E$ be a composition of elementary row operations so that $\vec{r_1} \leftarrow \sum_{i=1}^n(-1)^{i+1}\vec{r_i}$. The operation that adds a multiple of one row to another doesn't change the determinant. So, the determinant of $EA$ is the determinant of $A$.
I am not sure I understand what $\vec{r_1} \leftarrow \sum_{i=1}^n(-1)^{i+1}\vec{r_i}$ is doing? But don't I need to construct A?

4. Re: Constructing a Matrix A

One obvious way to do that is to have every row the same! That way it is certainly true that "sum of all its rows with odd numbers is equal to the sum of all its rows with even numbers"

5. Re: Constructing a Matrix A

Originally Posted by vidomagru
I am not sure I understand what $\vec{r_1} \leftarrow \sum_{i=1}^n(-1)^{i+1}\vec{r_i}$ is doing?
It replaces $\vec{r}_1$ (i.e., row 1) with the sum of every row, but with alternating signs.

Originally Posted by vidomagru
But don't I need to construct A?
The matrix A is given to you. You are not told what it is exactly, but you are given the relevant property of A.

[rant]The other day I came across the following quote about Lorentz transformation in this document (PDF).

The transformation should be linear. Otherwise a specific system S, or a point in space or time would be distinct. (For a linear transformation, the inverse has the same form.)
Such considerations in physics, of the form "This function is nonnegative, so it must be x²", amaze me. They seem to be so contrary to how mathematics is done. However, they do help find a reasonable answer. In this problem, there is only one number that is distinguished in some sense with respect to matrices and determinants. You would not expect that the determinant of some arbitrary matrix is, say, 5, would you?[/rant]

Edit: Another way to solve the problem is to note that the property of A from post #1 implies that its rows are linearly dependent.

6. Re: Constructing a Matrix A

Originally Posted by emakarov
Such considerations in physics, of the form "This function is nonnegative, so it must be x²", amaze me. They seem to be so contrary to how mathematics is done. However, they do help find a reasonable answer. In this problem, there is only one number that is distinguished in some sense with respect to matrices and determinants. You would not expect that the determinant of some arbitrary matrix is, say, 5, would you?[/rant]
I have forgotten who all were involved, but there was a bit of angst between Physics and Mathematics in the mid 19th century. Many Mathematicians were disgruntled about the use Mathematics by Physicists because they thought the Math was too "beautiful" (I supplied the word here) for something "real." I think the funniest one was about non-commutative Mathematics. Who could possibly use that in a Physical theory? Then Heisenberg came along.... (I suspect that non-Riemannian geometry fell into that category as well.)

-Dan

7. Re: Constructing a Matrix A

Originally Posted by emakarov

Edit: Another way to solve the problem is to note that the property of A from post #1 implies that its rows are linearly dependent.
Why does "the property that the sum of all its rows with odd numbers is equal to the sum of all its rows with even numbers" imply linear dependence?

If every row is linearly dependent, then the determinant is zero, right?

8. Re: Constructing a Matrix A

Originally Posted by vidomagru
Why does "the property that the sum of all its rows with odd numbers is equal to the sum of all its rows with even numbers" imply linear dependence?

If every row is linearly dependent, then the determinant is zero, right?
$(r_1+ r_3+ r_5+ ...+ r_{2n-1}) - (r_2+ r_4+ r_6+ ...+ r_{2n})= 0$ is a linear combination of the rows equal to 0, with not all coefficients equal to 0. Therefore, the rows, thought of as vectors, are dependent.

9. Re: Constructing a Matrix A

Originally Posted by HallsofIvy
$(r_1+ r_3+ r_5+ ...+ r_{2n-1}) - (r_2+ r_4+ r_6+ ...+ r_{2n})= 0$ is a linear combination of the rows equal to 0, with not all coefficients equal to 0. Therefore, the rows, thought of as vectors, are dependent.
Well, the indices should only go up to $n$ since the matrix is $n\times n$.

So, let $a = \begin{cases} \dfrac{n}{2} & \text{if } n \text{ is even} \\ \dfrac{n-1}{2} & \text{otherwise}\end{cases}$

and let $b = \begin{cases} a & \text{if }n\text{ is even} \\ a+1 & \text{otherwise}\end{cases}$.

Then the correct sum would be $(r_1 + r_3 + \ldots + r_{2b-1}) - (r_2 + r_4 + \ldots + r_{2a}) = 0$