I've just encountered a matrix which seems to display nothing special to me: $$B=\begin{pmatrix}1&4&2\\0 &-3 &-2\\ 0 &4 &3 \end{pmatrix}$$ But further observation reveals something stunning: $$B^n=\cases{{I}&n is even\\{B}&n is odd}$$ So it leads me to wonder if there is indeed some special properties of this matrix $B$, or more probably, $B$ belongs to a whole special class of matrices whose name I don't know? Could you drop me a hint? Thanks in advance.

EDIT I think I was being a bit stupid... It suddenly hit me that any matrix $B$ such that $B^2=I$ will have this property.

• Asking questions is never stupid! The matrix describes a reflection leaving a fixed a plane and a line which you can compute. Such reflections are indeed important examples of involutions. – Michael Le Barbier Grünewald Apr 6 '15 at 20:12
• Why didn't I pick the blue pill? – jxh Apr 6 '15 at 21:03

If you want a name it is called an involution. That is any matrix whose square is identity, or one which is its own inverse. All such matrices have the property you mentioned. Matrices corresponding to reflection are such kind.

• Yeah I just realized that any matrix whose square is identity possesses such property. But the name involution is indeed new to me. – Vim Apr 6 '15 at 13:58
• Actually the name involution is not specific to linear algebra: it is used in general group theory for any element of order 2: here we apply to group of non-singular matrices. – P Vanchinathan Apr 6 '15 at 14:04
• "[An involution] is any matrix whose square is identity" except for the identity matrix itself! As @PVanchinathan says, involutions are group elements of order exactly two; the identity has order 1 so is not an involution. – David Richerby Apr 6 '15 at 16:20

the eigenvalues of $B$ are given by $$0=\det\pmatrix{1-\lambda&4&2\\0&-3-\lambda&-2\\0&4&3-\lambda} =(1-\lambda)(\lambda^2 -1)=-(\lambda -1)^2(\lambda+1)$$ eigenvectors corresponding to the eigenvalue $1$ are $u = (1,0,0)^T, v = (0,1, -2)^T$ and an eigenvector corresponding to the eigenvalue $-1$ is $w=(1,-1,1).$ therefore the spectral decomposition of $B$ is $$B = \frac 1{u^Tu}uu^T+ \frac 1{v^Tv}vv^T- \frac 1{w^Tw}ww^T.$$ this implies that $$B^n = \begin{cases} I & \text{ if } n \text{ is even }\\ B & \text{ if } n \text{ is odd }\end{cases}$$

• You may replace det with \det to make the symbol upright. – CiaPan Apr 7 '15 at 8:46

$B$ is a square root of unity. See http://en.wikipedia.org/wiki/Square_root_of_a_matrix for more about square roots of matrices.

A reasonably short computation shows that your matrix has eigenvalues $-1,1,1$. Your computation shows it has minimal polynomial $t^2-1=0$, which forces the Jordan blocks in Jordan Normal Form to be diagonal. Hence you can diagonalise it to $$R J R^T,$$ where $R$ is an orthogonal matrix and $J=\operatorname{diag}{(1,1,-1)}$. It's then clear what happens when you take powers of it.

It is special in the sense that it is a representation of a generating element for $C_2$, the cyclic 2 group. Which is one of the most basic groups. This means you can find a block diagonalization such that the square of the blocks are the identity matrix. For instance this could be -1 or $\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, since their squares are both some identity matrix. It is also possible that you can find a diagonalization in the field you are already working in. If Chappers above is correct then it is possible to do in this case, but in general this is not the case for groups and a block-diagonal will have to do.

Another example is the matrix $\left(\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right)$. This matrix is a generating element for the $C_3$ group as - you guessed it - it's powers of 3 is the identity.

However, it has complex eigenvalues which are the roots of unity for 3, so if working in ${\mathbb R}$ a simple diagonalization would not work and we have to accept a block-diagonalization.