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.


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!

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

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