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.

  • 2
    $\begingroup$ $\langle Ax,x\rangle \ge \lambda\langle x,x\rangle$ by the assumptions on $A$. $\endgroup$ – 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

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


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

now if

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

is the least eigenvalue, then (6) yields

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


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


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

(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. $OE\Delta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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