# Find Minimal Polynomial and Jordan Normal Form of $T(A)=-A^t$

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

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 $$J_A=J_6(-1), J_3(1)$$