Csdrhrt

I would like to get all matrices $N in M_2(mathbb R)$ such that $N^2 = -I_2$.

To start with, I know that
$N_0=begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}$ works, and we can prove that every matrices $N$ that are similar to $N_0$ work.

$i.e.$ Let $N in M_2(mathbb R)$ if $exists P in mathrm{GL}_2(mathbb R)$ such that $N = PN_0P^{-1}$, then $N^2 = -I_2$.

My question is, is the converse true?

Are all matrices $N in M_2(mathbb R)$ such that $N^2 = -I_2$ similar to $N_0$?

For $A, B in M_n(mathbb{C})$, write $A sim B$ if $A$ and $B$ are similar in $mathbb{C}$.

Assume that $N in M_2(mathbb{R})$ satisfies $N^2 + I_2 = 0$. Then the minimal polynomial of $N$ is $X^2+1$ and this factors into distinct linear factors in $mathbb{C}$. So $N$ is diagonalizable in $mathbb{C}$ with eigenvalues $i$ and $-i$, i.e., $N sim operatorname{diag}(i, -i) $ in $mathbb{C}$. By the same reason, $N_0 sim operatorname{diag}(i, -i)$ and hence $N sim N_0$. Then the desired claim follows from the following proposition.

Proposition. Let $A, B in M_n(mathbb{R})$. Suppose that $A sim B$. Then $A$ and $B$ are similar in $mathbb{R}$.

Proof. Choose $P in mathrm{GL}_n(mathbb{C})$ such that $A = PBP^{-1}$. Write $P = Q+iR$ for $Q, R in M_n(mathbb{R})$ and define $P_z = Q + zR$ for $z in mathbb{C}$.

1. Since $AQ = QB$ and $AR = RB$, we have $AP_z = P_z B$ for all $z in mathbb{C}$,




2. $det(P_i) = det(P) neq 0$, hence $det(P_z)$ is a non-zero polynomial in $mathbb{R}[z]$.

So we can find $x in mathbb{R}$ such that $P_x$ is invertible. Then $P_x in operatorname{GL}_2(mathbb{R})$ and $A = P_x BP_x^{-1}$, hence $A$ and $B$ are similar in $mathbb{R}$ as required. $Box$

It's not very hard to find all matrices $N in M_2(Bbb R)$ such that

$N^2 = -I; tag 1$

for let

$N = begin{bmatrix} n_{11} & n_{12} \ n_{21} & n_{22}end{bmatrix}; tag 2$

then

$begin{bmatrix} n_{11}^2 + n_{12}n_{21} & n_{11}n_{12} + n_{12} n_{22} \ n_{21}n_{11} + n_{22} n_{21} & n_{21}n_{12} + n_{22}^2 end{bmatrix} = begin{bmatrix} n_{11} & n_{12} \ n_{21} & n_{22}end{bmatrix}begin{bmatrix} n_{11} & n_{12} \ n_{21} & n_{22}end{bmatrix} = N^2 = -I; tag 3$

therefore,

$n_{11}^2 + n_{12}n_{21} = n_{21}n_{12} + n_{22}^2 = -1, tag 4$

$(n_{11} + n_{22})n_{12} = (n_{11} + n_{22})n_{21} = 0; tag 5$

we see from this equation that

$text{Tr}(N) = n_{11} + n_{22} tag 6$

is a determinative, classifying factor; if

$text{Tr}(N) ne 0, tag 7$

then from (5),

$n_{21} = n_{12} = 0, tag 8$

whence from (4),

$n_{11}^2 = n_{22}^2 = -1; tag 9$

clearly there are no real solutions in this case (7); thus we take

$text{Tr}(N) = 0, tag 8$

and see that

$-n_{22} = n_{11} = alpha in Bbb R; tag 9$

now from (4),

$n_{12}n_{21} ne 0, tag{10}$

implying

$n_{12} ne 0 ne n_{21}; tag{11}$

thus we may write

$n_{21} = -dfrac{1 + alpha^2}{n_{12}}; tag{12}$

if we set

$n_{12} = beta ne 0, tag{13}$

then we may write

$n_{21} = -dfrac{1 + alpha^2}{beta}; tag{14}$

taken together, (9)-(14) provide a parameterized family of $2 times 2$ matrices

$N(alpha, beta) = begin{bmatrix} alpha & beta \ -dfrac{1 + alpha^2}{beta} & -alpha end{bmatrix} tag{15}$

such that

$N^2(alpha, beta) = -I. tag{16}$

It is easy to see that the set of admissible parameters is characterized by

$alpha in Bbb R, ; 0 ne beta in Bbb R. tag{17}$

Careful scrutiny of the above argument reveals that we have in fact demonstrated that every matrix in $M_2(Bbb R)$ satisfying (1) is of the form (15); therefore our parametric representation (15) is complete.

Since any $2 times 2$ real matrix satisfting (1) is of the form (15), and we see that

$text{Tr}(N(alpha, beta)) = alpha + (-alpha) = 0, tag{18}$

and

$det(N(alpha, beta)) = -alpha^2 + beta dfrac{1 + alpha^2}{beta} = 1, tag{19}$

and thus all have characteristic polynomial

$chi_N(x) = x^2 + 1, tag{20}$

we may affirm the the eigenvalues of any $N(alpha, beta)$ are $pm i$, and the eigenvectors satisfy

$begin{bmatrix} alpha & beta \ -dfrac{1 + alpha^2}{beta} & -alpha end{bmatrix} begin{pmatrix} v_1 \ v_2 end{pmatrix} = pm ibegin{pmatrix} v_1 \ v_2 end{pmatrix}, ; v_1, v_2 in Bbb C; tag{21}$

then with eigenvalue $i$,

$alpha v_1 + beta v_2 = i v_1, tag{22}$

which since $beta ne 0$ yields

$v_2 = dfrac{i - alpha}{beta} v_1; tag{23}$

since eigenvectors are only determined up to a scaling factor, we can in fact take $v_1 = 1$ and then

$ begin{pmatrix} v_1 \ v_2 end{pmatrix} =begin{pmatrix} 1 \dfrac{i - alpha}{beta} end{pmatrix}; tag{24}$

since $N(alpha, beta)$ is a real matrix it follows that the eigenvector associated with $-i$ is

$overline{ begin{pmatrix} v_1 \ v_2 end{pmatrix}} = begin{pmatrix} v_1 \ bar v_2 end{pmatrix} =begin{pmatrix} 1 \dfrac{- i - alpha}{beta} end{pmatrix}; tag{25}$

the diagonalizing matrix is thus

$V = begin{bmatrix} 1 & 1 \ dfrac{i - alpha}{beta} & dfrac{-i - alpha}{beta} end{bmatrix}, tag{26}$

with

$det(V) = -dfrac{2i}{beta}, tag{27}$

we have

$V^{-1} = dfrac{beta i}{2}begin{bmatrix} dfrac{ -i - alpha}{beta} & -1 \ -dfrac{i - alpha}{beta} & 1 end{bmatrix} = begin{bmatrix} dfrac{ 1 - ialpha}{2} & -dfrac{beta i}{2} \ -dfrac{-1 - ialpha}{2} & dfrac{beta i}{2} end{bmatrix}; tag{28}$

therefore

$V^{-1} N(alpha, beta)V = begin{bmatrix} i & 0 \ 0 & -i end{bmatrix}; tag{29}$

finally, $N$ from (1) itself may be so represented via application of the above formulas; therefore any $N(alpha, beta)$ is similar to $N$, since each is similar to the diagonal matrix $text{diag}(i, -i)$.

We have thus reached an affirmative answer to our OP Euler Pythagoras' closing question, even if our path has been a tad on the round-about side. We did however discover the formula (15) for solutions to (1), which I think is pretty cool.

Something more general holds. If $f(t) = t^n + a_{n-1}t^{n-1} + dots + a_1t + a_0$ is a monic polynomial then the matrix

$$ C_f = begin{pmatrix}
0 & 0 & 0 & dots & 0 & -a_0 \
1 & 0 & 0 & dots & 0 & -a_1 \
0 & 1 & 0 & dots & 0 & -a_2 \
0 & 0 & 1 & dots & 0 & -a_3 \
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & dots & 1 & -a_{n-1}
end{pmatrix}
$$

is called the companion matrix of $f$. A simple calculation shows that $C_f$ satisfies $f$, meaning that
$$ f(C_f) = C_f^n + a_{n-1}C_f^{n-1} + dots + a_1 C_f + a_0I = 0 $$

Moreover, if $C_f$ satisfies any other polynomial equation $g(C_f) = 0$ then $f$ divides $g$. We say that $f$ is the minimal polynomial of $C_f$.

There's a result in algebra called rational normal form (or "canonical form") that says that every matrix is similar to a block diagonal matrix of companion matrices. Specifically,

If $A$ is a matrix that satisfies a polynomial $f$ (for example its characteristic polynomial) then $A$ is similar to a block diagonal matrix whose blocks are companion matrices. I.e. $$A sim operatorname{diag}(C_{g_1},dots,C_{g_r}).$$
Moreover:

1. 
   $g_i$ divides $f$ for $i = 1, dots, r$
   


2. We can make it so that $g_i$ divides $g_{i + 1}$ for $i = 1,dots,r-1$
   


3. If 2. holds, you can check that $g_r$ is the minimal polynomial of $A$ in the sense described above.

One idea here is that if $h$ is any polynomial then
$$h(operatorname{diag}(C_{g_1},dots,C_{g_r})) = operatorname{diag}(h(C_{g_1}),dots,h(C_{g_r})). $$
This is true generally for block-diagonal matrices.

The point is that for the polynomial $f(t) = t^2 + 1$ the only real factors of $f$ are $1$ and $t^2 + 1$. On the other hand, $C_1$ is the empty $0times 0$ matrix which just leaves us with $C_{t^2 + 1}$ is exactly what you called $N_0$.