Let $A$ be symmetric positive definite matrix and $E$ is symmetric with $||E||_{2} < ||A^{-1}||^{-1}_{2}$ then prove that $A+E$ is symmetric positive definite.

-- \ Observation; Since $A$ is invertible and $A+E = A(I+A^{-1}E)$ since $||A^{-1}E||_{2} \leq ||A^{-1}||_{2}||E||_{2}< 1$ by assumption then $A+E$ is invertible. But I don't see the connection to show that eigenvalues of $A+E$ are positive.

Any hint would be appreciated.

  • 1
    $\begingroup$ You could replace $E$ by $tE$ and your argument shows that $A+tE$ is invertible for all $0\leq t\leq1$. As $t$ goes from $0$ to $1$, I think you could show that eigenvalues must all remain positive by continuity. $\endgroup$ – alex.jordan Dec 8 '11 at 18:46

The eigenvalues for $A^{-1}$ are the reciprocals of those for $A.$ So the 2-norm gives the largest reciprocal. Taking the reciprocal of that gives you the smallest eigenvalue of $A,$ still positive, call it $\lambda.$ This is what we call a "coercive" estimate, for a column vector $x$ and its transpose $x^T,$ we have $$ x^T A x \geq \lambda \; x^T x = \lambda \; x \cdot x.$$ The statement on $E$ is that, while it might have negative eigenvalues, in any case all are smaller then $\lambda$ in absolute value, and strictly larger than $- \lambda.$ So $$ x^T E x \geq - \lambda \; x^T x = - \lambda \; x \cdot x, $$ with equality only if $x=0.$ Put together, $$ x^T (A + E) x \geq \lambda \; x \cdot x - \lambda \; x \cdot x = \; 0 ,$$ with equality only when $x=0.$


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.