3
$\begingroup$

Let $\mathcal{G}$ be denote the set of all $3 \times 3$ real symmetric matrices and let $\mathcal{G}^+$ denote the set of all $3 \times 3$ positive semidefinite matrices (see definition).

Let $S: \mathcal{G} \to \mathcal{G} $ be a linear transformation satisfying the conditions: $S^2=S$ and $S(\mathcal{G}^+) \subseteq \mathcal{G}^+$.

Suppose $\dim(N(S))=1$, where $N(S)$ denotes the null-space of $S$. Can we find a non-zero positive semidefinite matrix in $N(S)$?

Thanks in advance.

$\endgroup$
2
$\begingroup$

The answer is yes, and moreover we may replace $3$ by any positive integer $n $.

Let $0 \ne R \in \mathcal N(S)$. If $R$ is either positive semidefinite or negative semidefinite, $R$ or $-R$ is a member of $\mathcal N(S) \cap \mathcal G^+$, so we assume it is indefinite. Then there is a nonzero vector $u$ such that $u^T R u = 0$.

Define the linear functional $\varphi$ on $\mathcal G$ by $S(A) - A =\varphi(A) R$. We can extend this linearly to a linear functional on $\mathbb R^{n \times n}$.
Thus there is some $n \times n$ matrix $B$ such that $\varphi(A) = \text{tr}(B A)$ for all $A$. Note that $\varphi(R) = -1$.

Now the question is, can this $S$ take $\mathcal G^+$ into itself? Consider the case $A = v v^T$ for some vector $v$. We have $S(v v^T) = v v^T + \varphi(v v^T) R = v v^T + (v^T B v) R$. If $u$ is any vector orthogonal to $v$, we need
$$0 \le u^T S(v v^T) u = (v^T B v) (u^T R u)$$ Since $R$ is indefinite and symmetric, it has orthogonal eigenvectors $u_+$ and $u_-$ corresponding to positive and negative eigenvalues $\lambda_+$, $\lambda_-$ respectively. Taking $u=u_+$ and $v = u_-$, we must have ${u_-}^T B u_- \ge 0$. Taking $u=u_-$ and $v = u_+$, ${u_+}^T B u_+ \le 0$. And if $v$ is any vector orthogonal to both $u_+$ and $u_-$, we have both $v^T B v \ge 0$ and $v^T B v \le 0$, so $v^T B v = 0$. But then using an orthonormal basis of eigenvectors of $R$, $$ -1 = \varphi(R) = \sum_i \lambda_i u_i^T B u_i = \lambda_+ u_+^T B u_+ + \lambda_- {u_-}^T B u_- \ge 0$$ contradiction!

$\endgroup$
  • $\begingroup$ Note that in the last step φ(R)≤0 ,how we are getting the contradiction?? $\endgroup$ – ashok Oct 11 '16 at 17:11

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.