0

  1. In linear algebra, an $n$-dimensional orthogonal matrix refers to $Q \in M(R, n)$ such that $Q^tQ = QQ^t = I$. If $T \in L(R^n)$ is an isometry, then we have that $T^*T = I$, where $T^*$ is adjoint of $T$. This implies that a matrix representation of an isometry w.r.t. any orthonormal basis is an orthogonal matrix. What's the relationship between these two?

  2. We know that $T^*T = I$. Is there a nice interpretation/intuition why it is true? I remember seeing $T^*$ is actually not an operator acting on $R^n$, but an operator acting on dual space of $R^n$. How do these pieces fit together to give a picture: $T^*T = I$?

Thank you

James C
  • 1,338
  • $T^$ is the adjoint* of $T$, not the "self-adjoint" of $T$. – Ben Grossmann Jun 25 '21 at 00:14
  • For point two, consider the fact that you need $\langle Tx, Tx \rangle = \langle x, x \rangle$ and recall how adjoint works with respect to the inner product – Osama Ghani Jun 25 '21 at 00:18
  • As you note: if $T$ is an operator on a vector space $V$ without any further structure (for instance an inner product), then $T^$ is a map over $V^$, the dual space of $V$. In this setting, the composition $T^T$ is not defined. Without something like an inner product, the expression $T^T$ can't be interpreted. – Ben Grossmann Jun 25 '21 at 00:29
  • @Paul Also, your statement "this implies that a matrix representation of an isometry w.r.t. any basis is an orthogonal matrix" is incorrect. This is only true for orthonormal bases. – Ben Grossmann Jun 25 '21 at 00:35
  • @BenGrossmann Thanks for adjusting my word, "adjoint." I'll fix it. And right. I am implicitly assuming the existence of an inner product. – James C Jun 25 '21 at 00:38
  • @OsamaGhani Yes, that's the standard proof why $T^*T = I$. I want a more intuitive picture, if there is one. – James C Jun 25 '21 at 00:38
  • @BenGrossmann Yes, also, matrix representation of $T$ needs to be w.r.t. an orthonormal basis. Thanks for the correction. (Meanwhile, I am wondering if the result is false if I have a general basis instead of orthonormal one. I'll work on a counter example.) – James C Jun 25 '21 at 00:42
  • @Paul For a counterexample, consider the fact that the matrices $$ \pmatrix{1&0\0&-1}, \quad \pmatrix{1&1\0&-1} $$ are similar, but the second of these fails to be orthogonal – Ben Grossmann Jun 25 '21 at 00:57
  • @Paul You might find my description of $T^*T$ on this post to be helpful. Notably, we will have $|Tx| = |\sqrt{T^*T}x|$ for all $x \in \Bbb R^n$. – Ben Grossmann Jun 25 '21 at 01:00
  • I found a counterexample. I set a 90 degree rotation matrix $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$ and set basis vector $v_1 = (1,0), v_2 = (1,1)$. Explicit calculation gives me a counterexample. Just wondering why does your example show the result that I wanted? (i.e. for an arbitrary basis, matrix representation of an isometry is not necessarily orthogonal.) Is it because $P^{-1} A P$ is a matrix representation of switching from one coordinate to the other? – James C Jun 25 '21 at 01:03
  • @Paul Yes, that's exactly why. – Ben Grossmann Jun 25 '21 at 01:04
  • I see. Thank you very much. I'll see if your post helps resolve my question why $T^*T = I$. – James C Jun 25 '21 at 01:05

0 Answers0