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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.

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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.

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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.

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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 $x$, $x^T(A+cI)x > 0$

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