Conjugation with Pauli matrices Let $\{\sigma_j\}_{j=0}^3$ denote the Pauli basis of Hermitian matrices on $\mathbb C^2$ with $\sigma_0 := I$. Is it true that $$\frac{1}{4}\sum_{j=0}^3 \sigma_j A \sigma_j = \frac{\text{tr}(A)}{2}I$$ for any positive definite $2x2$ matrix $A$? If so, how would I go about showing this? I haven't been able to do so with the known properties of the Pauli matrices.
 A: Let
$$
P=\frac14\sum_{n=0}^3 \sigma_nA\sigma_n
=\frac14\left[ A - \sum_{n=1}^3 (-i\sigma_n)A(-i\sigma_n) \right].\tag{1}
$$
Using the isomorphism $I\mapsto 1,\ -i\sigma_1\mapsto i,\ -i\sigma_2\mapsto j,\ -i\sigma_3\mapsto j$ between the real vector space of all complex $2\times2$ matrices and the real algebra of quaterions, if we denote the quaternion representations of $P$ and $A$ by $p$ and $a$ respectively, we may rewrite $(1)$ as:
$$
p=\frac14(a - iai - jaj - kak).\tag{2}
$$
It is easy to verify that $wp=pw$ for $w=1,i,j,k$. For instance,
$$
ip = \frac14(ia + ai - kaj + jak)=\frac14(ai + ia + jak - kaj)=pi.
$$
Hence $P$ commutes with all Pauli matrices and in turn, also with all complex $2\times2$ matrices. Thus $P=cI$ for some scalar $c$. Now, by the tracial property,
$$
2c=\operatorname{tr}(P)=\frac14\sum_{n=0}^3\operatorname{tr}(A\sigma_n^2)=\frac14\sum_{n=0}^3\operatorname{tr}(A)=\operatorname{tr}(A).
$$
Therefore $c=\frac{\operatorname{tr}(A)}2$ and $P=\frac{\operatorname{tr}(A)}2I$.
A: To show
$$\frac{1}{4}\sum_{j=0}^3 \sigma_j A \sigma_j = \frac{\text{tr}(A)}{2}I$$
for any $2 \times 2$ matrix $A$, it's enough to show this when $A$ is $\sigma_0, \sigma_1, \sigma_2$ or $\sigma_3$, since these four matrices form a basis of the $2 \times 2$ matrices, and each side of your identity is linear in $A$.  This may not be the most elegant approach, but it has this advantage: it reduces the problem to a calculation you can do using standard Pauli matrix identities without any brilliant insights!
For example when $A = \sigma_1$ you need to prove
$$\frac{1}{4}(\sigma_0 \sigma_1 \sigma_0 + \sigma_1 \sigma_1 \sigma_1 + \sigma_2 \sigma_1 \sigma_2 + \sigma_3 \sigma_1 \sigma_3)= 0$$
Using the rule for multiplying Pauli matrices with $j, k = 1,2,3$:
$$\sigma_j \sigma_k = i \epsilon_{jk\ell} \sigma_\ell + \delta_{jk} I $$
together with $\sigma_0 = I$, the formula we need to prove becomes
$$\frac{1}{4}(I \sigma_1 I + I \sigma_1 - i \sigma_3 \sigma_2 + i \sigma_2 \sigma_3)= 0$$
and using the rule again it becomes
$$\frac{1}{4}(\sigma_1 + \sigma_1 - \sigma_1 - \sigma_1)= 0$$
which is true.  The calculation works the same sort of way when $A$ is $\sigma_2$ or $\sigma_3$.  The case $A = \sigma_0$ is even easier, and this is the only Pauli matrix with a nonzero trace.  In this case we need to show
$$\frac{1}{4}(\sigma_0 \sigma_0 \sigma_0 + \sigma_1 \sigma_0 \sigma_1 + \sigma_2 \sigma_0 \sigma_2 + \sigma_3 \sigma_0 \sigma_3)= I$$
but since $\sigma_0 = I$ this amounts to
$$\frac{1}{4}(I + \sigma_1^2 + \sigma_2^2 + \sigma_3^2)= I$$
and since every Pauli matrix squares to $I$ this is true.
