Soft question about Transpose of a matrix

Question

Does $AA^T=I$? Does the transpose of a matrix multiplied itself always give you the identity matrix? furthermore the review asks to show that: Assume that $U$ is a square matrix such that $U^TU=I$ show that the $Det(U )= \pm 1$

My thoughts are we can multiply both sides by the inverse of $U$

$U^{-1}U^TU=I \to U^T = U^{-1}$

Then take the determinant of both sides: $det(U^T)=det(U^{-1})$

giving: $Det(U)= det(U^{-1})$ well I am not even sure that I have done up to here is correct because I do not see any logical steps to follow what I have here

Revised:

$U^TU=I \to det(U^TU)=det(I) \to Det(U^T)det(U)= 1 \to det(U)det(U) = 1$ is this better can I proceed from here?

Answer 1

Hint - If $A$ is any square matrix, $det(A)=det(A^{T})$.

Answer 2

In general, no. The product of a matrix and its transpose generally do not yield the identity matrix. For example, let $$A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$ Then $A=A^{T}$ and $$AA^{T}=\begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \neq I$$ Moreover, your attempt at showing thet $\det(U)=\pm 1$ is correct. If $U^{T}U=I$, then $U$ is invertible with inverse $U^{T}=U^{-1}$. You can use the general fact about matrices that $\det(A)=\det(A^{T})$ to obtain: $$\det(U)=\det(U^{T})=\det{(U^{-1})}=\frac{1}{\det(U)} \implies \det(U)^2 = 1 \implies \det(U)=\pm1,$$ as required.