2
$\begingroup$

Let A be matrix. If $bi$ ($b$ is real) is eigenvalue then $-bi$ is also eigenvalue. Why can it alone exist? $bi$ is eigenvalue and $-bi$ is not eigenvalue.

My attempt at proving this: [eigenvalues are roots of characteristic equation. Any equation, complex roots exist in pair. But how to prove this?]

$Ax = bi x$ and say $$Ax \neq -bix \\ \implies Ax \neq -bix = - Ax \\ 2Ax \neq0 $$

It should lead to contradiction somehow. I could not see it.

Also $Ax \neq -bix, bix =Ax \neq -bix \implies 2bix \neq 0$ it should be possible right say of x = [1,2] then $bix = bi(1+2) $\neq$ 0 fine right?

Please help me in understanding this silly doubt

$\endgroup$
  • 1
    $\begingroup$ In the real case, the characteristic polynomial can only have conjugate roots. $\endgroup$ – Yves Daoust Nov 18 at 16:46
  • 2
    $\begingroup$ Terminology nit-pick: Complex doesn't really mean non-real. So we can have only one complex eigenvalue, assuming it is real. Many people (including me some times), when they say "complex" mean "complex and not real", but that's not quite right. $\endgroup$ – Arthur Nov 18 at 16:46
  • 2
    $\begingroup$ Just to emphasize what was said in the answer, if you're talking about matrices with complex entries, all bets are off. In particular, you can look at the diagonal matrices $\begin{bmatrix} i & 0 \\ 0 & 3\end{bmatrix}$ or $\begin{bmatrix} i & 0 \\ 0 & -3i\end{bmatrix}$. $\endgroup$ – Ted Shifrin Nov 18 at 16:47
  • $\begingroup$ @ThomasAndrews: Don't edit to make it wrong, please. The OP is talking about the pure imaginary $bi$. I have removed your edit. $\endgroup$ – Ted Shifrin Nov 18 at 16:49
4
$\begingroup$

If $A$ is a real matrix,

$$Ax=bx$$ implies by conjugation

$$Ax^*=b^*x^*$$

and $b^*$ is an Eigenvalue.


In the complex case, you can set the Eigenvalues that you want in a diagonal matrix (it can be simply $\begin{bmatrix}i\end{bmatrix}$ !).

$\endgroup$
  • $\begingroup$ this is easy to understand. Thank you.. But If A has complex entries then there can exist only 1 complex eigen value right? (please correct me if wrong) $\endgroup$ – Nascimento de Cos Nov 18 at 16:58
  • 1
    $\begingroup$ @NascimentodeCos: a diagonal matrix will have the Eigenvalues that you want. $\endgroup$ – Yves Daoust Nov 18 at 17:03
  • $\begingroup$ this is too good sir. Thank you. $\endgroup$ – Nascimento de Cos Nov 18 at 17:04
7
$\begingroup$

This is true if your matrix has real entries. As you said, eigenvalues are roots of the characteristic polynomial $f(x)$, which will have real coefficients. The conjugate root theorem then gives you what you need.

$\endgroup$
  • $\begingroup$ Sir kindly elaborate in lay man terms. If matrix has real entries, then its eigen values can have only 1 of 2 conjugate complex values. If matrix has complex entries, then it has to exist in pairs?(please correct me if i am wrong here) Any easy way to prove this? Else i will go thru conjugate root theorem $\endgroup$ – Nascimento de Cos Nov 18 at 16:50
  • 1
    $\begingroup$ Eigenvalues of your matrix are exactly the roots of the characteristic polynomial of the matrix. So we have reduced your problem to showing that if $f(x)$ is a real polynomial and $bi$ is a root of $f(x)$ (i.e. an eigenvalue of your matrix), then $-bi$ is also a root of $f(x)$ (i.e. another eigenvalue of your matrix). This is exactly what the conjugate root theorem proves. $\endgroup$ – Calder Nov 18 at 16:52
  • $\begingroup$ understood sir. Also went thru proof of conjugate root theorem. It was easy to understand. If a matrix has real values then if there is complex root then it has to exist in pairs(please correct me if wrong). If matrix has complex entries, then it becomes characteristic equation with coefficients as complex values. In that case, can only 1 complex eigen value exist? if so how does it look like? $\endgroup$ – Nascimento de Cos Nov 18 at 16:55
  • 1
    $\begingroup$ If the entries of the matrix are complex, the eigenvalues can be any complex numbers. (Take the diagonal matrix with diagonal entries equal to whatever eigenvalues you want.) $\endgroup$ – Calder Nov 18 at 17:01
-1
$\begingroup$

For real matrix we always have complex conjugate eigenvalues. Indeed recall that for a real matrix the trace is real and the trace is equal to the sum of eigenvalues.

Otherwise it is not necessarly the case.

$\endgroup$
  • 2
    $\begingroup$ Not really a proof, may have a $4 \times 4$ real matrix with real trace but all different imaginary parts which cancel out... $\endgroup$ – gt6989b Nov 18 at 16:45

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.