Re: Imaginary matrices

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