How can I show that $T \in L(V)$ can be written in a unique way as $T=S_1+iS_2$? Suppose $V$ is a finite dim inner product space. I am trying to show that any operator $T \in L(V)$ can be written in a unique way as $T=S_1+iS_2$ for some $S_1$ and $S_2$ that are self-adjoint. Here, $i=\sqrt{-1}$.
I have been trying to use the spectral theorem, stated:
Suppose $V$ is a complex inner-product space and $T \in L(V)$. Then $V$ has an orthonormal basis consisting of eigenvectors of $T$ if and only if $T$ is normal.
I am running into trouble using this theorem. Will anyone be kind enough to tell me if I am on the right track using this theorem? Thank you!
 A: Not to put too fine a point on it, but I believe invoking the spectral theorem here is, to borrow an idiom, like killing flies with your daddy's new .45!  (As in "Daddy's got a new .45", from the song Santeria, by Sublime.)
Here's more of a .22 caliber way to shoot this one down:
Note that any linear map $T \in L(V)$ may be uniquely decomposed as the sum of a self-adjoint operator $S_1$ and a skew-adjoint operator $K$.  This is easy to see, for if we set
$S_1 = \dfrac{1}{2}(T + T^\dagger), \tag{1}$
and
$K = \dfrac{1}{2}(T - T^\dagger), \tag{2}$
then we have
$S_1^\dagger = \dfrac{1}{2}(T + T^\dagger)^\dagger = \dfrac{1}{2}(T^\dagger + (T^\dagger)^\dagger) = \dfrac{1}{2}(T + T^\dagger) = S_1, \tag{3}$
and 
$K^\dagger = \dfrac{1}{2}(T - T^\dagger)^\dagger = \dfrac{1}{2}(T^\dagger - (T^\dagger)^\dagger) = -\dfrac{1}{2}(T - T^\dagger) = -K, \tag{4}$ 
and clearly
$T = S_1 + K, \tag{5}$
a fact the readers may easily verify for themselves.  Furthermore, such a decomposition is unique, for if there were $R = R^\dagger$ and $Z = -Z^\dagger$
with $T = R + Z = S_1 + K$, then $R - S_1 = K - Z$; but $R - S_1$ is self-adjoint whilst $K - Z$ is skew; the only way this can happen is if both sides vanish.  (Use $Y = Y^\dagger = -Y^\dagger \Rightarrow Y^\dagger = -Y^\dagger \Rightarrow 2Y^\dagger = 0 \Rightarrow Y^\dagger = 0 \Rightarrow Y = 0$!)  Thus $S_1 = R$ and $K = Z$, so the decomposition is unique.  Now set
$S_2 = -iK, \tag{6}$
then
$S_2^\dagger = iK^\dagger = -iK = S_2, \tag{7}$
i.e., $S_2$ is self-adjoint.  Furthermore $iS_2 = i(-iK) = K$, so the decomposition
$T = S_1 + iS_2 = S_1 + K \tag{8}$
is also unique.  QED!!!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: I assume $L(V)$ is the set of operators on $V$. Then the spectral theorem won't help, since it only applies to normal operators. You want to prove a statement for all operators. 
For motivation, let's consider the case of complex numbers. If $z$ is a complex number, then $z=a+bi$ for some real numbers $a$ and $b$. We can represent $a$ and $b$ in terms of $z$.
$$a=\frac{z+\bar z }{2}$$
$$b=\frac{z-\bar z}{2i}$$
These are standard formulas that you should have learned when you were taught about complex variables. 
In the realm of operators, the adjoint takes the place of conjugation. This leads us to guess 
$$S_1=\frac{T+T^*}{2}.$$
I leave it to you to find $S_2$ and verify that the $S_i$ satisfy the requirements of the problem.
A: Its a little trick people sometimes use in different contexts .Clearly $T=\frac{T+T^*}{2}+i\frac{T-T^*}{2i}$ .Now take  $S_1=\frac{T+T^*}{2}$ and $S_2=\frac{T-T^*}{2i}$. Its very  easy to check that $S_1$ and $S_2$ are self-adjoint.
