Matrix proof of $A^2 = I_n$ $\implies A = I_n$

Question

I'm trying to find out whether $A^2 = I_n$ $\implies A = I_n$ is a true or false statement. Here's my thought process.

I'm going to be lazy and prove it for a $2 \times 2$ matrix, but I don't think the size should matter.

In an identity matrix, $$a_{11}= 1$$ $$a_{12}=0$$ $$a_{21}=0$$ $$a_{22}=1$$

So, I'm taking matrix $A$, and made an expression for each entry of $A^2$, defining it as $B$.

$$b_{11} = a^2_{11} + a_{12}a_{21}$$ $$b_{12}= a_{11}a_{12}+ a_{12}a_{22}$$ $$b_{21}= a_{21}a_{11}+a_{22}a_{21}$$ $$b_{22}= a^2_{22}+a_{21}a_{12}$$

By this logic, since $[a]_{ij} = [b]_{ij}$, we have the following:

$$a^2_{11} + a_{12}a_{21} = a^2_{22}+a_{21}a_{12}$$

$$a_{11}a_{12}+ a_{12}a_{22} = a_{21}a_{11}+a_{22}a_{21}$$

From here, however, I don't really know what to do to go any further. If my thought process is do-able for this proof, what should I do next?

Answer 1

$$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}^2=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

Answer 2

Just think about what you're doing: making a linear transformation which, applied twice, brings you back to where you started.

If you had a basis $x_1,x_2$, then you could just swap them: $T(x_1)=x_2$ and $T(x_2)=x_1$. Applying the transformation twice gets you the identity transformation.

So, such transformations exist, and don't have to be the identity.

Computing the matrix in this basis is straightforward: $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

Knowing this you should easily see how to do it for any $n>1$.

Alternatively, you could use a linear transformation that sends all the elements to their additive inverses. Compute the matrix for that. This version works for any $n> 0$.

Answer 3

Any reflection will do! Pick any axis , and reflect about it, reflecting again will give the identity.

To make the map linear, just make sure that you reflect about some codimension $1$ subspace.

In dimension $2$, this amounts to reflecting about a line $y=\lambda x$.

Take the line $y=\lambda x$, which is $\mathrm{span}(1,\lambda)$ and take a normal vector $\mathbf{N}=\frac{(-\lambda,1)}{\sqrt{(1+\lambda^2)}}$, where we divide justs to normalize the vector. Then a reflection is the same as taking your vector $v=(x,y)$, and figuring out its perpendicular distance to the line $y=\lambda x$. With vectors, this is the same as projecting onto the normal vector (draw a picture). Hence, the distance is $\bf{v} \cdot N$, where this is the dot product. We can then take $\mathbf{v}-2(\bf{v} \cdot N) \mathbf{N}$ as the suitable reflection.

In matrix notation, this is the transformation $$I-2NN^{T}=\begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix}-\frac{1}{1+\lambda^2}\begin{pmatrix} \lambda^2 & -\lambda\\-\lambda & 1\end{pmatrix}$$

You can see here for a proof that this algebraically squares to the identity.