0
$\begingroup$

This is probably a matter of definition, but I would like to get more insight in the differences between orthogonal matrix, invertible matrix and orthonormality

Given an orthogonal matrix $U$:

$$ U \in \mathbb{R}^{n\times n} $$ is orthogonal, then $$ U^{-1} = U^\top $$

How can we show the difference between an orthogonal (which is automatically invertible) matrix and an invertible matrix (which is not automatically orthogonal)? I think I have trouble making the difference given the fact that the rows (respectively rows) of an invertible matrix have already to be orthogonal

Moreover, respect to orthonormality, Can we speak about orthonormal matrix or can we only speak about orthonormal vector basis which form a orthogonal matrix?

$\endgroup$
7
  • $\begingroup$ Compute $U^\top U$. $\endgroup$ Sep 21 '19 at 10:22
  • $\begingroup$ @RodrigodeAzevedo. $U^\top U = I$. but how can we highlight the difference between $U$ orthogonal and $A$ invertible but not orthogonal? $\endgroup$
    – ecjb
    Sep 21 '19 at 10:33
  • $\begingroup$ Then compute $A^\top A$. I am not sure I understand what you want to do. $\endgroup$ Sep 21 '19 at 10:34
  • $\begingroup$ @RodrigodeAzevedo Thank you for your comment. Ok so $U^\top U = I \implies U$ is orthogonal $A^\top A \ne I \implies A$ is not orthogonal. Correct? is there another way to exaplain it regarding the structure of the matrix? $\endgroup$
    – ecjb
    Sep 21 '19 at 10:48
  • $\begingroup$ Define "structure of matrix". An orthogonal matrix is square and has orthonormal columns. A non-orthogonal square matrix does not have orthonormal columns. $\endgroup$ Sep 21 '19 at 10:49
2
$\begingroup$

An orthogonal matrix is invertible by definition, because it must satisfy $A^TA=I$.

An invertible matrix need not be orthogonal: consider $$ A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$ which is not orthogonal, but is invertible, with $$ A^{-1}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\ne A^T $$

In an orthogonal matrix the columns are pairwise orthogonal and each is a norm $1$ vector, so they form an orthonormal basis.

$\endgroup$
0
$\begingroup$

Every column in the orthogonal matrix is orthogonize to other columns, but this fact is not exactly hold for any invertible matrix.

$\endgroup$
1
  • $\begingroup$ thank you for the answer. But I can we prove the difference between an orthogonal matrix and a simply inveritble (non-orthogonal) matrix? $\endgroup$
    – ecjb
    Sep 21 '19 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.