Show that if $T: V \rightarrow V$ is linear, then $T'' \circ \Lambda = \Lambda \circ T$ So I've got this subquestion on my fortnightly Linear Algebra problem set which I've been staring at for the past 3 hours;

The double dual space of $V$, denoted by $V''$, is defined to be the dual space of $V$, in other words, $V'' = (V')'$. Define,
      $$\Lambda : V \rightarrow V'', \ \ (\Lambda v)(\phi) = \phi (v) $$ for $v \in V$, and $\phi \in V'$

The first question was "Show that $\Lambda$ is linear" which was pretty trivial, but the next one:

Show that if $T: V \rightarrow V$ is linear, then $T'' \circ \Lambda = \Lambda \circ T$

I am completely stuck on. I've figured out that for some fixed $v \in V$, fixed $\phi \in V'$ where we define $T(\bar{v}) = v$, that
$$(\Lambda \circ T \bar{v})(\phi) = \phi(v)$$
And
$$(T'' \circ \Lambda v)(\phi) = T''(\phi(v))$$
But I'm not sure how to show that they're equal, since they're essentially at different stages of evaluation? I've tried going for a contradiction assuming $T'' \circ \Lambda \neq \Lambda \circ T$, and getting a break in linearity of one of the functions or compositions, which didn't work, or simply that evaluating $(T'' \circ \Lambda)(v) - (\Lambda \circ T)(v)$ to equal $0$, which I couldn't do. I get the feeling I've just muddled notation somewhere.
Any hints or corrections in my interpretation would be really appreciated!
 A: Note that $T'' \circ \Lambda$ and $\Lambda \circ T$ are maps from $V$ to $V''$.  We want to show, then, that for any $x \in V$, the outputs $[T'' \circ \Lambda](x)$ and $[\Lambda \circ T](x)$ are the same elements of $V''$.
In other words, we want to show that for any $\phi \in V'$, $[[\Lambda \circ T](x)](\phi)$ is the same element of $\Bbb R$ (or $\Bbb F$ for an arbitrary field) as $[[T'' \circ \Lambda](x)](\phi)$.
With that in mind, we proceed.  For the first, we note that
$$
[[\Lambda \circ T](x)](\phi) = (\Lambda[T(x)])(\phi) = \phi([T(x)]) = [\phi\circ T](x)
$$
For the other, we're going to use the definition of the dual map.  In particular: if $T:V \to V$, then for $\phi \in V'$ and $\alpha \in V''$, we have
$$
T'(\phi) = \phi \circ T, \qquad T''(\alpha) = \alpha \circ T'
$$
With that in mind,
$$
[[T'' \circ \Lambda](x)](\phi) = [T'' (\Lambda(x))](\phi) =
[(\Lambda(x)) \circ T'](\phi) = [\Lambda(x)](T'(\phi)) =\\
[\Lambda(x)](\phi \circ T) = [\phi \circ T](x)
$$
So, we see that the two functions do the same thing to their inputs.  In particular, if we think of each as a function $f:V \to V''$, then we find that 
$[f(x)](\phi) = \phi(T(x))$.
So, they are indeed the same linear map.
