1
$\begingroup$

Let T be a linear transformation such that $T(A)=-A^t$, in the space $M_{n \times n}^{\Bbb R}$, and $-A^t$ is donated as $-A$ Transpose, with the standard inner product.

  1. Find $T$'s Characteristic Polynomial and Minimal Polynomial.
  2. Find The Jordan Normal Form of T, when $trace(T)=-3$

I was able to prove that T is Unitarian since $T^2(A)=A \Rightarrow T^2(A)=I$

$T$ therefore satisfies the equation $A^2-I=0 \Rightarrow (t-1)(t+1)=0$

Therefore the eigenvalues of T are $\lambda \in \left \{-1,1 \right\}$ and T is Normal (Unitarian and all the eigenvalues equal $|(1)|$.

However this is the best I could do here. I know the characteristic polynomial is supposed to be $$P_A=(t-1)^i(t+1)^{n-i}$$

But what is the minimal polynomial? How do I determine Jordan Normal Form?

Thanks,

Alan

$\endgroup$
0
$\begingroup$

I believe I can solve the question with the following:

$$T(A)=-A^t \Rightarrow T^2(A)=(-A^t)(-A^t)=A \Rightarrow T^2=I$$

Therefore we can say the minimal polynomial is:

$$(t^2-1)=(t-1)(t+1)$$

Therefore the eigenvalues are $t=1, t=-1$.

In addition: $$TT^*(A)=(-A^t)(A)=T*T^2=T*I=T$$

$$T^*T(A)=A(-A^t)=T^2*T=I*T=T$$

So $T$ is Normal, and thus Unitarian (eigenvalues are $t=|1|$).

The dimension of the symmetric and anti symmetric matrices is known:

$$S_N=\frac{n(n+1)}{2}$$ $$A_N=\frac{n(n-1)}{2}$$

Therefore the characteristic polynomial is $$P_A=(t-1)^{\frac{n(n-1)}{2}} * (t+1)^{\frac{n(n+1)}{2}}$$

From the fact that trace=$-3$ and that the trace is the sum of eigenvalues, we solve the equation $$1*\frac{n(n-1)}{2} + (-1)*\frac{n(n-1)}{2} \Rightarrow n=3$$

Therefore, $$P_A=(t-1)^3(t+1)^6$$

And the Jordan Normal Form is $$J_A=J_6(-1), J_3(1)$$

$\endgroup$

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.