1. ## Special orthogonal group

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

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

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

Tonio

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

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

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

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 $I$ is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors $X$ for which $X^tX=0$...! 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 $I$ is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors $X$ for which $X^tX=0$...! Thus, the claim is false

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

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

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

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