Orthogonal endomorphisms in finite dimension

Question

As far as I understand it, orthogonal endomorphisms in finite dimensions have an orthogonal matrix for representation only if the basis on which the matrix is created is orthonormal. Doesn't that pose a problem, as then there will exist some bases in which the orthonormal application is not orthonormal ?

Answer 1

Given a scalar product on a (finite dimensional) vector space $V$, an isometry $f:V\to V$ is a linear map preserving the distance induced by the scalar product (or, equivalently, that preserves the scalar product).

The fact that $f$ is an isometry is a base-free notion.

Now, if you fix a basis on $V$, then both the scalar product and $f$ have associated matrices. If $M$ is the matrix associated to the scalar product and $A$ is the matrix associated to $f$, then the fact that $f$ is an isometries, reads -- in COORDINATES relative to the fixed basis -- as

$$A^tMA=M$$

This holds true for ANY basis of $V$. Choosing a basis which is orthogonal w.r.t. the scalar product translates to the fact that $M$ is the identity matrix. As a consquence, the matrix of $f$ in any orthononmal basis is an orhtogonal matrix. But in general what holds true is just the above formula.

Answer 2

Whether an endomorphism is orthogonal is not basis-dependent. We say an endomorphism $T$ on a Hilbert space $V$ (e.g. a finite-dimensional inner product space) is orthogonal if $T^* T = I$, where $T^*$ is the adjoint of $T$, and $I$ is the identity map on $V$.

Now, orthogonal endomorphisms on finite-dimensional spaces do correspond to orthogonal matrices, but not as generally as some students tend to think. If $T$ is an orthogonal endomorphism and $B$ is an orthonormal basis, then the matrix $[T]_B$ of $T$, corresponding to $B$, is an orthogonal matrix. There is a converse to this too: if $B$ is orthonormal, and $[T]_B$ is orthogonal, then so is $T$.

But, the key fact here is that $B$ must be an orthonormal basis! If we take a non-orthonormal basis, we have no guarantee whatsoever that $[T]_B$ is orthogonal, even if $T$ is orthogonal. For example, take the orthogonal endomorphism: $$T : \Bbb{R}^2 \to \Bbb{R}^2 : (x, y) \mapsto (y, -x).$$ If you're interested, this is a clockwise rotation of the plane by $\pi/2$ radians. Under the standard basis $S$, we get $$[T]_S = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$ Note that the columns (and rows) form an orthonormal basis, hence the matrix is orthogonal. But, if we choose a non-orthonormal basis like $$B = \{(1, 0), (1, 1)\},$$ then $$[T]_B = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix},$$ which is not orthogonal.