I have a linear operator $T:V \to V$ (where $V$ is a finite-dimensional vector space) such that $T^9=T^8$ and $T$ is normal, I need to prove that $T$ is self-adjoint and also that $T^2=T$.

Would appreciate any help given.

Thanks a million!

  • $\begingroup$ Is your vector space finite dimensional? $\endgroup$ – Cameron Williams Jul 1 '13 at 22:05
  • $\begingroup$ Yes it is finite dimensional $\endgroup$ – NBP Jul 1 '13 at 22:06
  • $\begingroup$ Does this help? (I am not good at extending Matrix Algebra to Linear Algebra). (Contradiction) \begin{align} T^2&\neq T\\\implies T.T^2&\neq T.T\\ &\vdots\\ \implies T^7.T^2&\neq T^7.T\\ \implies T^9&\neq T^8\\ \end{align} $\endgroup$ – Inquest Jul 1 '13 at 22:33
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    $\begingroup$ Google the second line of the first paragraph (the third link gives a mostly coherent argument). $\endgroup$ – David Mitra Jul 1 '13 at 22:34
  • $\begingroup$ @DavidMitra How did you come up with that? Do you usually google questions? $\endgroup$ – Git Gud Jul 1 '13 at 22:35

Any normal matrix is diagonalizable and thus we can write $T = U\Lambda U^*$ for some unitary matrix $U$ - the matrix of eigenvectors - and diagonal matrix $\Lambda$ - the matrix of eigenvalues.

If $T^9 = T^8$, we have the following:

$$T^9v = T^8 v.$$

Suppose $v$ is an eigenvector of $T$, then $Tv = \lambda v$ and

$$T^9v = \lambda^9 v = \lambda^8 v = T^8 v$$

Thus we have that

$$\lambda^8(\lambda-1)v = 0.$$

Therefore $\lambda = 0,1$ and the eigenvalues are real.

We wish to show that $T^2 = T$. Using the fact that $T = U\Lambda U^*$, we have that

$$T^2 = TT = (U\Lambda U^*)(U\Lambda U^*) = U\Lambda^2 U^*.$$

This last equality follows from that fact that $U$ is unitary. However since $\lambda = 0,1$, we have that $\Lambda^2 = \Lambda$ and so $T^2 = T$.

To show $T = T^*$, we can again use the fact that the eigenvalues are real and that $T$ is normal:

$$T^* = (U\Lambda U^*)^* = (U^*)^*\Lambda^*U^* = U\Lambda^*U^*.$$

However since $\Lambda$ is a matrix of real numbers and is diagonal, $\Lambda^* = \Lambda$ and we are done.

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    $\begingroup$ The operation $T$ certainly does not need to be invertible; the diagonal elements of $\Lambda$ need not all be non-zero. (E.g., the zero mapping is normal and satisfies $0^9 = 0^8$.) $\endgroup$ – Mees de Vries Jul 1 '13 at 22:45
  • $\begingroup$ Oh gosh you're absolutely right. I'm not sure how I made that mistake. I'll fix it up. $\endgroup$ – Cameron Williams Jul 1 '13 at 22:46
  • $\begingroup$ @MeesdeVries it's fixed now. Thanks for that! That was a very embarrassing mistake. $\endgroup$ – Cameron Williams Jul 1 '13 at 22:55

Hint. Call the underlying field $\mathbb{F}$. As $T$ is normal and its characteristic polynomial can be split into linear factors $\underline{\text{over }\, \mathbb{F}}$ (why?), $T$ is unitarily diagonalisable over $\mathbb{F}$. Now the rest is obvious.


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