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

Clearly the transpose of the transpose is identity making transpose the
second square root of the identity matrix

Let us consider now

The negative identity matrix with a diagonal from upper left to lower right
of all negative ones

Multiplying by this twice and gives identity so it is yet another square
root of the identity matrix

Consider now the negative transpose matrix with a diagonal from upper right
to lower left of all negative one s. Multiplying by this twice gives the
original matrix. So we see that the negative transpose matrix is yet
another square root of the identity matrix.

I wonder if these four square roots of unity

Can be extended to interpret the negative identity matrix as something
which itself might have a square root. My guess is the imaginary identity
matrix with a diagonal from upper left to lower right of all i, where I is
the square root of negative one, would suffice

Many thanks to John Baez to introducing me to the matrix multiplication
operation around a dozen years ago thanks again

[[Mod. note -- Wikipedia has lots of information about matrix square roots:
https://en.wikipedia.org/wiki/Square_root_of_a_matrix
-- jt]]

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