MHB Determining Value of a in Matrix A with $\lambda$ = 0

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Matrix Value
Yankel
Messages
390
Reaction score
0
Hello all,

Given the following matrix,

\[A=\begin{pmatrix} 2 & 6\\ 1 & a \end{pmatrix}\]

and given that

\[\lambda =0\]

is an eigenvalue of A, I am trying to determine that value of a.

What I did, is to create the characteristic polynomial

\[(\lambda -2)*(\lambda -a)+6=0\]

and given

\[\lambda =0\]

I got that a is -3.

Somehow I am not sure. Is there a way of finding the second eigenvalue before calculating a ?

Thank you !
 
Physics news on Phys.org
You could use that the determinant is the product of the eigenvalues.
(Also, the trace is the sum of the eigenvalues, but in this case you do not need that to determine $a$. You could use it to determine the second eigenvalue, though.)
 
Last edited:
Yankel said:
Hello all,

Given the following matrix,

\[A=\begin{pmatrix} 2 & 6\\ 1 & a \end{pmatrix}\]

and given that

\[\lambda =0\]

is an eigenvalue of A, I am trying to determine that value of a.

What I did, is to create the characteristic polynomial

\[(\lambda -2)*(\lambda -a)+6=0\]

and given

\[\lambda =0\]

I got that a is -3.

Somehow I am not sure. Is there a way of finding the second eigenvalue before calculating a ?

Thank you !
Check the signs in that calculation! I think that it should be $-6$ in the characteristic polynomial.
 
Opalg, you are correct, it is -6 and therefore a=3. Am I correct ?
 
Yankel said:
Somehow I am not sure. Is there a way of finding the second eigenvalue before calculating a ?

As Krylov pointed out, the trace is the sum of the eigenvalues.
So if one eigenvalue is 0, then the other is $\operatorname{Tr}A=2+a$.

And yes, a=3 is the correct solution.
 
Back
Top