9203224f   
   From: tjoberts137@sbcglobal.net   
      
   On 9/22/24 3:22 PM, D. Goncz wrote:   
   > Let's consider square matrices   
   > The identity matrix is a diagonal from upper left to lower right of all   
   > ones   
   > The transpose matrix is, if I remember Wikipedia correctly gosh I'm sorry,   
   > a single diagonal from upper right to lower left of all ones   
      
   No. The transpose of the (square) identity matrix is the identity   
   matrix. This is easy to see algebraically: the identity matrix is:   
   	I_ij = d_ij   
   where d is the Kronecker delta, which is symmetric in its indices:   
   	d_ij = {1 if i=j, 0 otherwise} = d_ji   
      
   So the transpose of the identity is:   
   	I^T_ij = d_ji = d_ij = I_ij   
      
   Indeed the transpose of any diagonal matrix is itself. Proof left to the   
   reader.   
      
   > Clearly the transpose of the transpose is identity making transpose the   
   > second square root of the identity matrix   
      
   That's not how "square root" works -- transpose is irrelevant. You must   
   MULTIPLY the square root by itself to get the original matrix.   
      
   That said, the (square) "backwards diagonal" matrix with 1's from top   
   right to bottom left and 0's everywhere else, when multiplied by itself,   
   yields the identity matrix. So it is indeed a square root of the   
   identity matrix. Thee are others....   
      
   > [.. too many fundamental errors to bother reading the rest]   
      
   Tom Roberts   
      
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