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

[Yankel]

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 !

 

[S.G. Janssens]

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.)

 

[Opalg]

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.

 

[Yankel]

Opalg, you are correct, it is -6 and therefore a=3. Am I correct ?

 

[I like Serena]

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.