# Linear maps satisfying $T^2 = I_n$ and Jordan Form of $T(X) = AX - XA$

Let $$V$$ be any $$n$$-dimensional vector space and $$W = M_2(\mathbb{C})$$.

(a) Construct all linear maps $$T: V \to V$$ such that $$T^2 = I_n$$.

For $$T^2 = I_n$$, taking $$p(X) = X^2 - 1$$ we must have $$p(T) = 0$$. Since the minimal polynomial of $$T$$ divides $$p$$ and contains all eigenvalues, $$T$$ must be diagonalizable with eigenvalues $$1$$ and $$-1$$. Since the minimal polynomial is of degree $$\leq 2$$, the largest Jordan block of the Jordan Form of $$T$$ must have order $$2$$. Then the maps satisfying $$T^2 = I_n$$ are the maps with eigenvalues $$1$$ and $$-1$$ such that the Jordan Normal Form is $$J = (J^1,...,J^1)$$ where each Jordan block $$J^{i}$$ have order $$\leq 2$$.

(b) Find the Jordan Normal Form of $$T: W \to W$$ given by $$T(X) = AX - XA$$ with $$A = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right).$$

We have to find the characteristical polynomial of $$T$$. If

$$X = \left(\begin{array}{cc} x & y \\ z & w \end{array}\right),$$

then

$$T(X) = \left(\begin{array}{cc} z & w-x \\ 0 & -z \end{array}\right)$$

I have problem to find the matrix associate to $$T$$ so, I tried to use $$W \simeq \mathbb{R}^4$$ and define $$T: \mathbb{R}^4 \to \mathbb{R}^4$$ by

$$T(x,y,z,w) = (z, w-z,0,-z).$$

Thus, $$T(1,0,0,0) = (0,0,0,0) = 0e_1 + 0e_2 + 0e_3 + 0e_4,$$ $$T(0,1,0,0) = (0,0,0,0) = 0e_1 + 0e_2 + 0e_3 + 0e_4,$$ $$T(0,0,1,0) = (1,-1,0,-1) = 1e_1 -1e_2 + 0e_3 -1e_4,$$ $$T(0,0,0,1) = (0,1,0,0) = 0e_1 + 1e_2 + 0e_3 + 0e_4.$$

So, the matrix is $$T = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \end{array}\right).$$

Is that matrix correct? If yes, I can continue. If no, I would like some help.

• $p=(X-1)(X+1)$. Therefore T is diagonalizable and there are no Jordan block of size 2. – TomTom314 Sep 25 at 14:13
• By the way, $W\cong \Bbb C^4$, not $\Bbb R^4$. So you should have $T:\Bbb C^4\to \Bbb C^4$. – WE Tutorial School Sep 25 at 15:49

For (a), TomTom314 already mentioned that $$T$$ is diagonalizable with eigenvalues $$\pm1$$. In the case that all eigenvalues are the same, it is clear that $$T=\pm I$$. But for the rest, I think the best you can say about $$T$$ is $$T=XDX^{-1}$$ where $$X$$ is invertible and $$D$$ is a diagonal matrix of the form $$\operatorname{diag}(-1,-1,\ldots,-1,1,1,\ldots,1)$$ where $$-1$$ appears $$p$$ times and $$1$$ appears $$q$$ times ($$p,q$$ are positive integers such that $$p+q=n$$).

However, if this is what the problem means, we can find a 1-1 correspondence between all such $$T$$ and all projections on $$V$$. For a projection $$P:V\to V$$, set $$T_P=2P-I$$. Then $$T_P^2=I$$. For a map $$T$$ such that $$T^2=I$$, define $$P_T=\frac{1}{2}(T+I)$$ so $$P_T^2=P_T$$, making $$P_T$$ a projection.

For (b), the matrix of $$T$$ in your chosen basis is correct, but I'd like to point out that $$T$$ is nilpotent. So all eigenvalues are $$0$$. Note that $$T^2X=A(AX-XA)-(AX-XA)A=A^2X-2AXA-XA^2$$ and since $$A^2=O$$, we have $$T^2X=-2AXA.$$ That is, $$T^3X=A(-2AXA)-(-2AXA)A=-2A^2XA+2AXA^2=O.$$ Since $$T^2A^t=-2AA^tA\ne O$$, $$T$$ is nilpotent of depth $$3$$. That is $$T$$ has a Jordan block of size $$3$$, and since the space $$M_2(\Bbb C)$$ is $$4$$-dimensional, $$T$$ has another Jordan block of size $$1$$. That is, a Jordan normal form of $$T$$ is $$\left(\begin{array}{ccc|c}0&1&0&0\\0&0&1&0\\0&0&0&0\\\hline0&0&0&0\end{array}\right).$$

Hints

(1) As you've written, $$\require{cancel}T^2 = \operatorname{id}$$ implies that the minimal polynomial $$m(x)$$ of $$T$$ divides $$x^2 - 1 = (x - 1) (x + 1)$$. But as pointed out in the comments, (at least provided that the field $$\Bbb F$$ underlying $$V$$ does not have characteristic $$2$$) the linear factors of $$m(x)$$ do not repeat, so $$T$$ must be diagonalizable.

Thus, there is a basis of $$V$$ with respect to which the matrix representation $$[T]$$ of $$T$$ has the form $$\pmatrix{I_k \\ & -I_{n - k}} .$$ Put another way, if we fix any basis of $$V$$, the matrix representations $$[T]$$ of the linear transformations $$T$$ satisfying $$T^2 = \operatorname{id}$$ are precisely those similar to one of the above three matrices, that is, $$\pm I$$ and those of the form $$P^{-1} \pmatrix{I_k \\ & -I_{n - k}} P .$$

Remark The restriction on the characteristic is necessary: In characteristic $$2$$, $$x^2 - 1 = (x - 1)^2$$, so for example, that the nondiagonalizable matrix $$A = \pmatrix{1&1\\&1}$$ satisfies $$A^2 = I$$.

(2) In this case we can compute the Jordan Normal Form without working directly with the entries of $$X$$--though if you're not familiar with Lie algebras (which is surely the most important context wherein transformations of this form appear), computing $$T^k(X)$$ in terms of entries helps motivate this approach:

Notice that $$A^2 = 0$$---this makes easier the computation of powers of $$T$$: We have $$\begin{multline}T^2(X) = T(AX - XA) = A(AX - XA) - (AX - XA)A \\= \cancel{A^2 X} - 2 A X A + \cancel{X A^2} = -2 A X A,\end{multline}$$ and computing similarly gives $$T^3 = 0 .$$

Since $$T$$ is thus nilpotent, its only eigenvalue is $$0$$, and all that remains to find is the size of the Jordan blocks. But since $$T^3 = 0$$ and $$T^2 \neq 0$$, the Jordan normal form contains a $$3 \times 3$$ block, $$J_3(0)$$, so the remaining block is $$J_1(0) = \{0\}$$. Thus, the Jordan normal form of $$T$$ is $$J_3(0) \oplus J_1(0) = \pmatrix{0&1&\cdot\\\cdot&0&1\\ \cdot&\cdot&0\\&&&0} .$$

Remark Our computations of $$T^2, T^3$$ used only the definition of multiplication and the fact that $$A^2 = 0$$ (and not the specific form of $$A$$). But $$A$$ is the Jordan form of any nonzero $$2 \times 2$$ nilpotent matrix, so our conclusion applies to any such matrix.

You might find it instructive to repeat the exercise replacing $$A$$ with a general $$2 \times 2$$ matrix. The observation implicit in the previous remark shows that it's enough to consider just the $$A$$ in Jordan form, so those of the forms $$\lambda I, \quad \pmatrix{\lambda&1\\&\lambda}, \quad \operatorname{diag}(\lambda, \mu) .$$