# If $A$ is Symmetric Positive Definite(SPD) matrix, is $A+E$ SPD?

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.

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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. –  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 $x=0.$