# Sum of Symmetric Positive Definite Matrix and Scalar of Identity

If $$A$$ is an $$n\times n$$ symmetric positive definite matrix with the smallest eigenvalue $$\lambda$$, then for any $$\mu>-\lambda$$, $$A+\mu I$$ is positive definite.

I am trying to show this, but I am stuck on one part. Here is what I have so far: \begin{align*} \langle x,\left(A+\mu I\right)x\rangle&=\langle x,Ax+\mu x\rangle\\ &=\langle x,Ax\rangle+\langle x,\mu x\rangle\\ &>0+\mu\langle x,x\rangle\\ &>-\lambda\langle x,x\rangle. \end{align*}

I'm stuck on showing that $$\mu\langle x,x\rangle$$ is positive because I only know that $$\mu>-\lambda$$. Any help would be appreciated.

• $\langle Ax,x\rangle \ge \lambda\langle x,x\rangle$ by the assumptions on $A$. – DisintegratingByParts Dec 10 '18 at 18:22

If $$A$$ is a positive definite symmetric matrix with smallest eigenvalue $$\lambda$$, then for all vectors $$x$$ we have

$$\langle x, Ax \rangle \ge \lambda \langle x, x \rangle; \tag 1$$

the easiest way I know to see this is to diagonalize $$A$$ by a suitable orthogonal matrix $$O$$, which will preserve the inner product:

$$\langle Oy, Ox \rangle = \langle y, O^TOx \rangle = \langle y, Ix \rangle = \langle y, x \rangle, \tag 2$$

where we have used the fact that

$$O^TO = OO^T = I \tag 3$$

in (2); then we have

$$OAO^T = \Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \tag 3$$

where $$\lambda_1, \lambda_2, \ldots, \lambda_n$$ are the eigenvalues of $$A$$. It is well-known that $$A$$ is also possessed of an orthonormal eigenbasis of vectors $$e_i$$ such that

$$A e_i = \lambda_i e_i, \; 1 \le i \le n; \tag 4$$

we may then write

$$x = \displaystyle \sum_1^n x_i e_i, \tag 5$$

and

$$\langle x, Ax \rangle = \left \langle \displaystyle \sum_1^n x_ie_i, \sum_1^n x_i Ae_i \right \rangle = \left \langle \displaystyle \sum_1^n x_ie_i, \sum_1^n x_i \lambda_i e_i \right \rangle = \displaystyle \sum_{i, j = 1}^n x_ix_j \langle e_i, \lambda_j e_j \rangle$$ $$= \displaystyle \sum_{i, j = 1}^n x_ix_j \lambda_j \langle e_i,e_j \rangle = \sum_{i, j = 1}^n x_ix_j \lambda_j \delta_{ij} = \sum_1^n \lambda_i x_i^2; \tag 6$$

now if

$$\lambda = \min \{\lambda_1, \lambda_2, \ldots, \lambda_n \} > 0 \tag 7$$

is the least eigenvalue, then (6) yields

$$\langle x, Ax \rangle = \displaystyle \sum_1^n \lambda_i x_i^2 \ge \sum_1^n \lambda x_i^2 = \lambda \sum_1^n x_i^2 = \lambda \langle x, x \rangle; \tag 8$$

therefore,

$$\langle x, (A + \mu I)x \rangle = \langle x, Ax \rangle + \mu \langle x, x \rangle \ge \lambda \langle x, x \rangle + \mu \langle x, x \rangle = (\lambda + \mu) \langle x, x \rangle; \tag 9$$

since

$$\mu > - \lambda \Longleftrightarrow \mu + \lambda > 0, \tag{10}$$

(9) becomes

$$\langle x, (A + \mu I)x \rangle \ge (\mu + \lambda ) \langle x, x \rangle > 0, \tag{11}$$

which shows that $$A + \mu I$$ is positive definite, the desired result. $$OE\Delta$$.