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Here is the question:

Let $A$ $=$ $\begin{pmatrix} 1 & 2\\2 & 3 \end{pmatrix}$ and let $I$ be the $2$ x $2$ identity matrix. Find the value of $\alpha$ such that $A + \alpha I$ is positive semi-definite, but not positive definite

I understand that I am looking for values of $\alpha$ such that it will give an eigenvalue of 0 and one positive.

So, as I also understand it, I'm looking for $|A + \alpha I - \lambda I|$

We can solve $A + \alpha I$ to be $\begin{pmatrix} 1 + \alpha & 2\\2 & 3+\alpha \end{pmatrix}$

Thus, we can find the determinate of: $$|\begin{pmatrix} 1 + \alpha & 2\\2 & 3+\alpha \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\0 & \lambda \end{pmatrix}| = det(\begin{pmatrix} 1 + \alpha - \lambda & 2\\2 & 3+\alpha - \lambda \end{pmatrix}$$

Solving forward, I get to the equation: $7 + 4\alpha - 4\lambda - 2\alpha\lambda+\lambda^2$

However, I'm not sure how to solve this polynomial or provide mathematical reasoning for good choices of $\alpha$ other than guessing and checking. I think I'm at the 90% mark for this problem, but I'm just a little caught up on the last piece. Or perhaps I have other thought this problem and it could be significantly more easier than I am making it. Either way, this seems like a good juncture to ask for some assistance with this problem. Would anyone be able to guide me in the right direction (especially if I've lost my way)? Thank you for your help in advanced!

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    $\begingroup$ The $A$ in the question and the $A$ in $A+\alpha I$ seems inconsistent. $\endgroup$ – Siong Thye Goh Sep 10 '18 at 3:46
  • $\begingroup$ ah, so it is. Good catch! The $A + \alpha I$ was actually correct, but the original $A$ was not $\endgroup$ – Billy Thorton Sep 11 '18 at 1:45
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Consider $\det(A+\alpha I)=(1+\alpha)(3+\alpha)-4=\alpha^2+4\alpha-1$. For positive semi-definite matrix, the determinant of every upper-left submatrices are $\geq 0$. So we need, $\alpha^2+4\alpha-1 \geq 0$ and $1+\alpha \geq 0$. Now solve for $\alpha$ and rule out the case when both are strict inequalities.

Added solution: So we want

  1. Either $\alpha^2+4\alpha-1 > 0 \wedge 1+\alpha=0$.
  2. Or $\alpha^2+4\alpha-1 = 0 \wedge 1+\alpha>0$.
  3. OR $\alpha^2+4\alpha-1 = 0 \wedge 1+\alpha=0$.

Clearly we can rule out cases 1 and 3. So we only need to resolve case 2. The quadratic gives $\alpha=\frac{-4\pm\sqrt{20}}{2}=-2\pm \sqrt{5}$. The only value which will also satisfy the other constraint $1+\alpha >0$ is $\color{red}{\alpha=-2+\sqrt{5}}$.

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  • $\begingroup$ Note: Adjusted original matrix above. How do you rule out the case when they are both strict inequalities? Would it be easier to just make one of them not strict to begin with? $\endgroup$ – Billy Thorton Sep 11 '18 at 1:53
  • $\begingroup$ In other words, do I solve $\alpha^2 + 4\alpha - 1 \geq 0 $ and $1 + \alpha \geq 0$? In doing so, we know $\alpha \geq -1$, but plugging $-1$ into the original equation, you get a false inequality ($-4 \geq 0)$. Is there another way to approach the problem? Can you guess any $\alpha \geq -1 $ such that it fits the original first inequality? $\endgroup$ – Billy Thorton Sep 11 '18 at 2:16
  • $\begingroup$ @BillyThorton I have added more explanation to my solution. $\endgroup$ – Anurag A Sep 11 '18 at 7:52
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You want to find $\alpha$ such that $A + \alpha I$ has one zero eigenvalue and one positive eigenvalue.

If $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$, then $\lambda_1 + \alpha$ and $\lambda_2 + \alpha$ are the eigenvalues of $A + \alpha I$. (Why?) Thus you should choose $\alpha = - \min\{\lambda_1, \lambda_2\}$.

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