1
$\begingroup$

Let A be a 3 × 3 real matrix with zero diagonal entries. If $1 + i$ is an eigenvalue of A, the determinant of $A$ equals-

I know the trace of the matrix is sum of eigen value but couldn't solve it.

$\endgroup$
  • $\begingroup$ What do you mean by "zero diagonal entries"? All zeros on the main diagonal? $\endgroup$ – imranfat May 23 at 9:39
  • 2
    $\begingroup$ Hint: If $1+i$ is an eigenvalue, then so is $1-i$. $\endgroup$ – Bungo May 23 at 9:39
4
$\begingroup$

If the entries are real, then $i + 1$ can only arise as the root of a polynomial with real coefficients. Such roots appear as conjugate pairs. So you know what 2 of the roots are. The last one is such as to make their sum equal to the trace.

This should be sufficient for you to be able to solve it.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Since $i+1$ is an eigenvalue, as already noticed by @PrimeMover and @Bungo, $1-i$ is also an eigenvalue. Then, we know that the trace, which is $0$, is equal to the sum of eigenvalues. Thus: $$ 2+ \lambda_3 = 0 $$ Therefore, the other eigenvalue is $-2$. Now recall that: $$ \det A= \prod_{i=1}^{3} \lambda_i = -2(1+i)(1-i)= -4 $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.