Why if add a multiple of $I$ to a symmetric matrix then we have positive semidefinite matrix

Question

I'm dealing with the following problem:

If we have a symmetric real matrix $S$ then there is a multiple of $I$ ,the identity matrix, such that $S+cI$ is positive semidefinite. Could you please help me to prove this? I think if use the matrix norm we can prove this fact but I could not.

Answer 1

If the eigenvalues of $S$ are $a_1,\ldots,a_n$, then the eigenvalues of $S+\lambda\operatorname{Id}$ are $a_1+\lambda,\ldots,a_n+\lambda$. So, take $\lambda$ so large that $a_1+\lambda,\ldots,a_n+\lambda\geqslant0$.

Answer 2

Let $S$ be a $n\times n$ matrix and let $T=S+\alpha I_n$. Note that every eigenvalue $\lambda$ of $S$ satisfies $S\vec{v}=\lambda\cdot\vec{v}$ for some nonzero $\vec{v}\in\mathbb{R}^n$. It follows that $$ T\vec{v}=S\vec{v}+\alpha\cdot\vec{v}=\lambda\cdot\vec{v}+\alpha\cdot\vec{v}=(\lambda+\alpha)\cdot\vec{v} $$ This argument proves that $\lambda$ is an eigenvalue of $S$ if and only if $\lambda+\alpha$ is an eigenvalue of $T$.

Now, the spectral theorem implies that the eigenvalues of $S$ are all real. Let $\alpha$ be the smallest eigenvalue of $S$. Then the eigenvalues of $T=S+\left\lvert\alpha\right\rvert\cdot I_n$ are all nonnegative. Hence $T$ is positive semidefinite.

Answer 3

What you are interested is not $S+I$ but rather to find a $c \ge 0$ such that $S+cI$ is positive semidefinite.

Since $S$ is symmetric, we can find an orthogonal matrix such that $S=UDU^T$, $$S+cI=U(D+cI)U^T$$

Hence we can use $c$ to update the eigenvalues, choose it such that each number is nonnegaitve.

Answer 4

We don't actually need anything about eigenvalues to prove this:

Since $x^T A x = \lVert x \rVert^2 (x/\lVert x \rVert)^T A (x/\lVert x \rVert) $, $A$ is positive semidefinite if and only if $ x^T A x $ is nonnegative for every $ x \in B = \{x : \lVert x \rVert = 1 \}$. $x^T S x$ is a continuous function and $B$ is a compact set, so $\min_{x \in B} x^T S x = \alpha$ exists. But then $$ \min_{x \in B} x^T (S+\lvert\alpha\rvert I)x = \alpha+\lvert \alpha \rvert \geq 0, $$ and by the above this is sufficient for $S+\lvert \alpha \rvert I$ to be positive-semidefinite.