Im preparing for a linear algebra exam and im trying to solve the next exercise:

We are given a matrix A:

7 −4 0
a −7 b
3 −2 0

which eigenvalues are −1 and 1.Find the parameters a, b∈R.

I honestly have no idea how to solve this but it seems like it should be easy. I know how to find eigenvalues given a matrix. It is done by finding the characteristic polynomial.

And the eigenvectors can be found by solving the system:

7x-4y   = lambx
ax-7y+b = lamby
3x-2y   = lambz

If i am correct,i can assign a value to x and solve for y and z. It feels like im missing something obvious.

  • $\begingroup$ Do you know $|A-\lambda I|=0$? $\endgroup$ – Shubham Johri May 21 at 19:12
  • $\begingroup$ I do not. After i get the A−λI matrix should i solve such that det(A−λI) = 0? $\endgroup$ – user569685 May 21 at 19:15
  • $\begingroup$ Well, for the eigenpair $(\lambda,x)$, you have $Ax=\lambda x\implies(A-\lambda I)x=0$. Since $x\ne0$ (eigenvectors are non-zero), this represents a homogeneous system of equations with non-trivial solutions, which means $|A-\lambda I|=0$. Just find $|A-\lambda I|$ for $\lambda=\pm1$ and equate it to $0$ $\endgroup$ – Shubham Johri May 21 at 19:20
  • $\begingroup$ Alright il try to do that. Thank you :) $\endgroup$ – user569685 May 21 at 19:23
  • $\begingroup$ Start by figuring out what the third eigenvalue must be. This matrix is traceless, so neither $1$ nor $-1$ can be a repeated eigenvalue. Once you know all three eigenvalues, you can use the fact that their product is equal to the determinant to get a simple equation for $b$. $\endgroup$ – amd May 21 at 19:41

Let $\lambda$ be an eigenvalue of $A$. Then$$|A-\lambda I|=\begin{vmatrix}7-\lambda&-4&0\\a&-7-\lambda&b\\3&-2&-\lambda\end{vmatrix}=0$$Putting $\lambda=1$,$$\begin{vmatrix}6&-4&0\\a&-8&b\\3&-2&-1\end{vmatrix}=0\implies a=12$$Putting $\lambda=-1$,$$\begin{vmatrix}8&-4&0\\12&-6&b\\3&-2&1\end{vmatrix}=0\implies b=0$$

  • $\begingroup$ Worked out :) thank you $\endgroup$ – user569685 May 21 at 20:08
  • $\begingroup$ Note: I think b=0 since |A−λI|= 0 = 4b. Plugged it into calculators,seems to work out. $\endgroup$ – user569685 May 21 at 20:16
  • $\begingroup$ Why did you equate $|A|=0$? $\endgroup$ – Shubham Johri May 21 at 20:17
  • $\begingroup$ Typo. Meant |A−λI| $\endgroup$ – user569685 May 21 at 20:18
  • $\begingroup$ Yes, I made a calculation error. Thanks for noting. $\endgroup$ – Shubham Johri May 21 at 20:19

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