Question

Show that all the diagonal elements of the skew-symmetric matrix are zero

Answer 1

Hint: A matrix $ A = \left[ {{a_{ij}}} \right] $ can be said to be skew—matrix if $ A' = - A $ or $ A = - A' $ , so from this $ {a_{ij}} = - {a_{ji}} $ for all the values which are possible of $ i\& j $ . In a matrix when the rows and columns are being changed then we call it to transpose the matrix. By using all this we can prove it.

Complete step-by-step answer:
For showing the diagonals to be zero we will assume the square matrix $ A = \left[ {{a_{ij}}} \right] $ , where $ i $ will be the number which is in a row and $ j $ will be the number which is in a column.
As we know from the hind that for a skew-symmetric matrix the condition will be $ A' = - A $ . And here, $ A' $ will be the transpose of the matrix.
So it can be written as $ {a_{ij}} = - {a_{ji}} $
Since the elements are in diagonal, so we will substitute $ i = j $
And on putting it, we get
 $ \Rightarrow 2{a_{ii}} = 0 $ , for all the value of $ i's $
Hence, on solving it we will get the equation as
 $ \Rightarrow {a_{ii}} = 0 $
So in a square matrix, the diagonal elements will be
 $ A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) $
And from this, the elements $ {a_{11}},{a_{22}},{a_{33}} $ will be the diagonal elements.
Since, $ {a_{ii}} = 0 $
Therefore, the elements $ {a_{11}} = {a_{22}} = {a_{33}} = 0 $
Hence, it is proved that the diagonal will be zero for the skew-symmetric matrix.
So, the correct answer is “ $ {a_{11}} = {a_{22}} = {a_{33}} = 0 $”.

Note: When a question comes like this, we just need to do one thing that is by using some concepts of it we can easily solve it. One thing we should know about the diagonal of the skew-symmetric matrix is, it can be imaginary too. Since we know a symmetric matrix is a square matrix and will be equal to its transpose. These are the important points we should know and remember.