1. ## Special orthogonal group

Let $\displaystyle P$ be a matrix in $\displaystyle SO_3(\mathbb{C})$.
Prove that if $\displaystyle X$ ia an eigenvector with eigenvalue $\displaystyle 1$ and if $\displaystyle P \ne I$, then $\displaystyle X^t X\ne 0$.

2. Originally Posted by KaKa
Let $\displaystyle P$ be a matrix in $\displaystyle SO_3(\mathbb{C})$.
Prove that if $\displaystyle X$ ia an eigenvector with eigenvalue $\displaystyle 1$ and if $\displaystyle P \ne I$, then $\displaystyle X^t X\ne 0$.

Perhaps I'm misunderstanding something here, but isn't $\displaystyle X^tX$ just the inner (scalar) product of the vector $\displaystyle X$ with itself, also known as $\displaystyle ||X||^2$ ?
If so then it's obvious it can't be zero, and this much is true for ANY eigenvector of ANY operator/matrix, since $\displaystyle ||X||=0\Longleftrightarrow X=0$ , and zero cannot be an eigenvector...

Tonio

3. Originally Posted by tonio
Perhaps I'm misunderstanding something here, but isn't $\displaystyle X^tX$ just the inner (scalar) product of the vector $\displaystyle X$ with itself, also known as $\displaystyle ||X||^2$ ?
If so then it's obvious it can't be zero, and this much is true for ANY eigenvector of ANY operator/matrix, since $\displaystyle ||X||=0\Longleftrightarrow X=0$ , and zero cannot be an eigenvector...

Tonio
Base field is $\displaystyle \mathbb{C}$!!!
For example, if $\displaystyle P=I$, then any non zero column vector $\displaystyle X$ can be eigenvector.
If $\displaystyle X=\begin{bmatrix}1\\i\\0\end{bmatrix}$,then $\displaystyle X^tX=1-1+0=0$

4. Originally Posted by KaKa
Base field is $\displaystyle \mathbb{C}$!!!
For example, if $\displaystyle P=I$, then any non zero column vector $\displaystyle X$ can be eigenvector.
If $\displaystyle X=\begin{bmatrix}1\\i\\0\end{bmatrix}$,then $\displaystyle X^tX=1-1+0=0$

Yeah, I totally missed the complex thing: I'm used to call a complex matrix unitary...but then you have a counterexample: the unity matrix $\displaystyle I$ is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors $\displaystyle X$ for which $\displaystyle X^tX=0$...! Thus, the claim is false

Tonio

5. Originally Posted by tonio
Yeah, I totally missed the complex thing: I'm used to call a complex matrix unitary...but then you have a counterexample: the unity matrix $\displaystyle I$ is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors $\displaystyle X$ for which $\displaystyle X^tX=0$...! Thus, the claim is false

Tonio
Originally Posted by KaKa
Let $\displaystyle P$ be a matrix in $\displaystyle SO_3(\mathbb{C})$.
Prove that if $\displaystyle X$ ia an eigenvector with eigenvalue $\displaystyle 1$ and if $\displaystyle \bold{P \ne I}$, then $\displaystyle X^t X\ne 0$.

6. Originally Posted by KaKa
Let $\displaystyle P$ be a matrix in $\displaystyle SO_3(\mathbb{C})$.
Prove that if $\displaystyle X$ ia an eigenvector with eigenvalue $\displaystyle 1$ and if $\displaystyle P \ne I$, then $\displaystyle X^t X\ne 0$.
If $\displaystyle P\in SO_3(\mathbb{C})$ then the product of the three eigenvalues of P is $\displaystyle \det (P) = 1$. If one of the eigenvalues is 1 then the other two must be $\displaystyle \lambda$ and $\displaystyle \lambda^{-1}$, where $\displaystyle \lambda\in\mathbb{C}$ and $\displaystyle \lambda \ne 1$ because $\displaystyle P\ne I$.

Let $\displaystyle X,\;Y,\;Z$ be eigenvectors for the three eigenvalues. Then $\displaystyle PX=X$ and so $\displaystyle X = P^{-1}X = P^{\textsc t}X$ (because P is orthogonal). Thus $\displaystyle X^{\textsc t}Y = (P^{\textsc t}X)^{\textsc t}Y = X^{\textsc t}PY = \lambda X^{\textsc t}Y$. Since $\displaystyle \lambda\ne1$, it follows that $\displaystyle X^{\textsc t}Y = 0$. Similarly $\displaystyle X^{\textsc t}Z = 0$.

If in addition $\displaystyle X^{\textsc t}X = 0$ then $\displaystyle X^{\textsc t}W = 0$ for all $\displaystyle W\in\mathbb{C}^3$ (because $\displaystyle \{X,Y,Z\}$ is a basis for $\displaystyle \mathbb{C}^3$). In particular, $\displaystyle X^{\textsc t}\overline X = 0$, where $\displaystyle \overline X$ is the vector whose coordinates are the complex conjugates of those of X. But that implies that $\displaystyle X=0$: contradiction. Therefore $\displaystyle X^{\textsc t}X \ne 0$.