Special orthogonal group

**KaKa** · December 1st 2009, 01:28 AM

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 $X^t X\ne 0$ .

**tonio** · December 1st 2009, 01:56 AM

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

**KaKa** · December 1st 2009, 03:22 AM

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 $X^tX=1-1+0=0$

**tonio** · December 1st 2009, 05:12 AM

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

**Defunkt** · December 1st 2009, 06:03 AM

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 $X^t X\ne 0$ .

**Opalg** · December 3rd 2009, 12:15 AM

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 $X^{\textsc t}X \ne 0$ .

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