When is simultaneous diagonalisation of two 2x2 symmetric traceless matrices possible?

Question

Consider A and B as 2 by 2 traceless symmetric matrices. Both of them are real. When is it possible to simultaneously diagonalize them?

Answer 1

The underlying principle here is that two square matrices $A$ and $B$ may be simultaneously diagonalized if and only if they commute one with the other, $AB = BA$. In what follows, we will work out this condition in terms of the matrix entries for two symmetric, traceless, $2 \times 2$ matrices $A$ and $B$.

Observe that, with

$D = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \tag 1$

and

$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} , \tag 2$

we have

$DP = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = J \tag 3$

and

$PD = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -J; \tag 4$

it is also to see that

$D^2 = P^2 = I. \tag 5$

A symmetric $2 \times 2$ matrix $A$ with vanishing trace evidently may be written as

$A = \begin{bmatrix} a & b \\ b & -a \end{bmatrix} = aD + bP; \tag 6$

if

$B = cD + dP \tag 7$

is another such matrix, then

$AB = (aD + bP)(cD + dP) = acD^2 + bdP^2 + adDP + bcPD$ $= (ac + bd)I + adJ - bcJ = (ac + bd)I + (ad - bc)J, \tag 8$

and similarly

$BA = (ac + bd)I - (ad - bc)J; \tag 9$

thus

$AB = BA \Longleftrightarrow (ad - bc)J = - (ad - bc)J \Longleftrightarrow ad = bc. \tag{10}$

We may gain more detailed information by examining cases:

In the event that

$ad = bc \ne 0, \tag{11}$

so that

$a, b, c, d \ne 0, \tag{12}$

we may write

$\dfrac{a}{c} = \dfrac{b}{d} = \rho \in \Bbb F, \tag{13}$

where $\Bbb F$ is the field from which the entries of $A$ and $B$ are taken. (I assume we are dealing with $2 \times 2$ matices over $\Bbb F$, and that

$\text{char} \; \Bbb F \ne 2.)\tag {14}$

From (13),

$a = \rho c, \; b = \rho d, \tag{15}$

which affirms that $A$ and $B$ are multiples one of the other; indeed, (15) immediately yields

$A = \rho B; \tag{16}$

going the other way, this trivially implies that $A$ and $B$ commute, $AB = BA$.

If on the other hand

$ad = bc = 0, \tag{17}$

but

$a \ne 0, \tag{18}$

we have

$d = 0, \tag{19}$

and then

$c = 0 \Longrightarrow B = 0, \tag{20}$

so $A$ and $B$ trivially commute; with

$c \ne 0 \tag{21}$

it is easy to see that

$A = aD = \rho cD = \rho B, \; \rho = \dfrac{a}{c}; \tag{22}$

also,

$d \ne 0 \Longrightarrow a = 0; \tag{23}$

if $b = 0$, then $A = 0$ and there is nothing to prove; if $b \ne 0$, then $c = 0$ and thus

$A = bP, \; B = dP, \tag{24}$

and in this case again we easily verify that $AB = BA$.

Answer 2

In general, two diagonalisable matrices are simultaneously diagonalisable if and only if they commute. In case they are commuting real symmetric matrices, they can even be simultaneously orthogonally diagonalised. Proofs of these facts can be found in many linear algebra textbooks.

In your case, you may solve the problem directly without using these facts. Note that all traceless diagonal $2\times2$ matrices are scalar multiples of $\operatorname{diag}(1,-1)$. Therefore $A$ and $B$ are simultaneously diagonalisable if and only if one of them is a scalar multiple of the other.