Finding parameters a,b in a matrix with given eigenvalues

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.

• Do you know $|A-\lambda I|=0$? – Shubham Johri May 21 at 19:12
• I do not. After i get the A−λI matrix should i solve such that det(A−λI) = 0? – user569685 May 21 at 19:15
• 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 $0$ – Shubham Johri May 21 at 19:20
• Alright il try to do that. Thank you :) – user569685 May 21 at 19:23
• 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 $b$. – 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 $$\lambda=-1$$,$$\begin{vmatrix}8&-4&0\\12&-6&b\\3&-2&1\end{vmatrix}=0\implies b=0$$
• Why did you equate $|A|=0$? – Shubham Johri May 21 at 20:17