Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Without using the fact that symmetric matrices can be diagonalized: Let $A$ be a real symmetric matrix. Show that there exists a real number $c$ such that $A+cI$ is positive.

That is, if $A=(a_{ij})$, one has to show that there exists real $c$ that makes $\sum_i a_{ii}x_i^2 + 2\sum_{i<j}a_{ij}x_ix_j + c\sum_i x_i^2 > 0$ for any vector $X=(x_1,...,x_n)^T$.

This is an exercise in Lang's Linear Algebra.

Thank you for your suggestions and comments.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

If $c$ is sufficiently big, you can complete the squares with the mixed terms and rewrite the left side as a sum of squares.

share|improve this answer

Whether $x^TAx$ is positive doesn't depend on the normalization of $x$, so you only have to consider unit vectors. The unit sphere is compact, so the sum of the first two sums is bounded. The third sum is $1$, so you just have to choose $c$ greater than minus the lower bound of the first two sums.

share|improve this answer
    
This proof doesn't even require $A$ symmetric, right? –  Thomas Andrews Oct 20 '11 at 18:33
    
@Thomas: It depends on your definition of "positive" whether it applies to non-symmetric matrices. If it does, then a non-symmetric matrix $A$ is positive iff the symmetric matrix $A+A^T$ is positive (since $x^TAx=(x^TAx)^T=x^TA^Tx$), so the claim doesn't depend on the symmetry of $A$, anyway. –  joriki Oct 20 '11 at 18:40
    
I just forgot that symmetry was part of the (standard) definition of positive. –  Thomas Andrews Oct 20 '11 at 18:56

You can use the fact $$ \lambda_{min}x^Tx \leq x^TAx\leq \lambda_{max}x^Tx $$ (which is not difficult to prove) and suppose that the matrix is already negative definite hence all the eigenvalues are negative. This means $x^TAx<0$ for all non-zero $x$. This allows us to write $$ \lambda_{min}\|x\|^2 \leq x^TAx\leq \lambda_{max}\|x\|^2 < 0 $$

But, consider the following: $$ x^T(A+cI)x = x^TAx +cx^Tx \geq (\lambda_{min}+c)\|x\|^2 $$ If we select $c>|\lambda_{min}|$ we obtain a positive definite matrix since for every non-zero $x$, $x^T(A+cI)x > 0$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.