If $A$ is a $10\times10$ matrix with entries from the set $\{0, 1, 2, 3\}$ and if $AA^T$ is of the form: $$\begin{pmatrix} 0 & * & * & \cdots & * \\ * & 0 & * & \cdots & * \\ * & * & 0 & \cdots & * \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ * & * & * & \cdots & 0 \end{pmatrix}$$

Then the number of such matrices $A$ is:
A) $(4^3)^{10}$ B)$(4^2)^{10}$ C)$4^{10}$ D) $1$

Since all the diagonal elements of $AA^T$ is zero I could realize that it is a skew-symmetric matrix. But, I'm no able to understand how I can use this result for finding the possibilities of the original matrix. I would like hints rather than answers.

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    $\begingroup$ Say that $A=(a_{ij})$. Then the $i$-th diagonal entry of $AA^t$ is $(AA^{t})_{ii}=\sum_{j=1}^{n}a_{ij}^2=0$. If the sum of two nonnegative numbers is zero, then the summands must be zero. Consequently, $a_{ij}^2=0$ for all $i,j$. So every entry of the matrix you started with was zero. $\endgroup$ – Adam Azzam Aug 26 '13 at 2:23
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    $\begingroup$ He asked for hints :) $\endgroup$ – Avraham Aug 26 '13 at 2:36

Since you mentioned symmetry, try simplifying the expression $(AA^T)^T$. What do you get, and what does this say about $AA^T$?

To find the possibilities of the original matrix, try writing the $(i,i)$-th diagonal element of $AA^T$ in terms of $A$, then equate the result with $0$.


I do not know if all such matrices are skew-symmetric, but if the original matrix has to be skew-symmetric, what do you know about the relationship between $a_{ij}$ and $a_{ji}$? Which members of the set {0, 1, 2, 3} exhibit that property?.


Even if the original is not skew-symmetric, try writing out $AA^T$ for a 3x3 matrix—what are you multiplying for each element of the diagonals? Try writing what the first diagonal entry would be in matrix multiplication for $AA^T$ in a 10x10. The second. What does this mean for each entry on the diagonal? Which members of the set {0, 1, 2, 3} can provide the needed main diagonal of 0's in the answer?

  • $\begingroup$ But, it is not given that the original matrix is skew-symmetric. $\endgroup$ – Rajath Radhakrishnan Aug 26 '13 at 1:54
  • $\begingroup$ You are correct. However, look at the edited portion of the above answer for a further hint. $\endgroup$ – Avraham Aug 26 '13 at 2:17
  • $\begingroup$ When I take 10*10 with general elements a,b,c.....etc and multiply it with its transpose I get the diagonal elements as the sum of the squares of the elements of the particular row. But, only zero can satisfy this condition. Thus, the original matrix is a 10*10 zero matrix. So,there is only one possibility. Is my thinking correct? $\endgroup$ – Rajath Radhakrishnan Aug 26 '13 at 12:25
  • $\begingroup$ Exactly! The diagonals are the sum of the squares of all of the elements in each row, which must be nonnegative, so every element must be 0. $\endgroup$ – Avraham Aug 26 '13 at 12:35

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