We would like to show that the basis $E_{ij}$ of matrix units of $M_{d_{\lambda }}\left(ℂ\right)$ is contained in $W^{\lambda }\left(A\right),$ whence that $M_{d_{\lambda }}\left(ℂ\right)=W^{\lambda }\left(A\right)\text{.}$ We need to produce scalars $t_{\lambda }\in ℂ$ such that $$E_{ij}=\sum _g{t}_{\lambda }{W}^{\lambda }\left(g^{*}\right)W^{\lambda }\left(g\right)\text{.}$$

By analogy with the way the matrix units arise in the case of the regular representation (do either of you want me to type that up as well?), wherein the dual $g^{*}$ are constructed using the ordinary trace tr, one may hope that $$E_{ij}=\sum _g{d}_{\lambda }{W}_{ji}^{\lambda }\left(g^{*}\right)W_{ij}^{\lambda }\left(g\right)$$ will hold in this more general situation. In fact, $$\begin{array}{ccc}\sum _g{d}_{\lambda }{W}_{ji}^{\lambda }\left(g^{*}\right)W_{ij}^{\lambda }\left(g\right)& =& \sum _g{d}_{\lambda }{W}_{ji}^{\lambda }\left(g^{*}\right)I_{d_{\lambda }}{W}_{ij}^{\lambda }\left(g\right)\\ & =& \sum _g{d}_{\lambda }{W}_{ji}^{\lambda }\left(g^{*}\right)\sum _{m=1}^{d_{\lambda }}{E}_{mm}{W}^{\lambda }\left(g\right)\\ & =& \sum _g\sum _{m=1}^{d_{\lambda }}{d}_{\lambda }{E}_{mj}{W}^{\lambda }\left(g^{*}\right)E_{im}{W}^{\lambda }\left(g\right)\\ & =& \sum _{m=1}^{d_{\lambda }}{d}_{\lambda }{E}_{mj}\frac{{\delta }_{im}}{d_{\lambda }}{I}_{d_{\lambda }}\\ & =& \sum _{m=1}^{d_{\lambda }}{E}_{mj}{\delta }_{im}=E_{ij}\end{array}$$

Thus we have shown that all of the matrix units of $M_{d_{\lambda }}\left(ℂ\right)$ are in $W^{\lambda }\left(A\right),$ that is $M_{d_{\lambda }}\left(ℂ\right)=W^{\lambda }\left(A\right),$ and thus $$A\simeq V\left(A\right)\simeq ⨁W^{\lambda }\left(A\right)\simeq ⨁M_{d_{\lambda }}\left(ℂ\right)\text{.}$$

$\square$ $1\to 3$

1: Let $a\in \sqrt{A},$ and $c\in A\text{.}$ Then for every $b\in A,$ $$\overrightarrow{t}\left(b\hspace{0.17em}ac\right)=\overrightarrow{t}\left(ba\hspace{0.17em}c\right)=\overrightarrow{t}\left(c\hspace{0.17em}ba\right)=\overrightarrow{t}\left(cb\hspace{0.17em}a\right)=0\text{.}$$

The second equality holds since $\overrightarrow{t}$ is a trace; the last since $a$ was chosen to be in the radical.) Thus $\sqrt{A}$ is invariant under right multiplication from $A\text{.}$ Similar arguments show that $\sqrt{A}$ is invariant under left multiplication, and closed under addition.

2: Let $J⊲A$ be an ideal of nilpotent elements, and let $a\in J\text{.}$ To show that $a\in \sqrt{A},$ consider $ba,$ for any $b\in A\text{.}$ Since $J$ is an ideal, we have $ba\in J,$ whence $ba$ is nilpotent. Thus $\varphi \left(ba\right)$ is a nilpotent matrix in $M_d\left(ℂ\right),$ and so its trace $\text{tr}\left(\varphi \left(ba\right)\right)=0\text{.}$ Thus $\overrightarrow{t}\left(ba\right)=0,$ and so $a\in \sqrt{A},$ as desired.

3: Given an element $a\in \sqrt{A},$ show that $a$ is nilpotent. Note that for every $k>0,$ we have $\overrightarrow{t}\left(a^k\right)=\overrightarrow{t}\left(a^{k-1}a\right)=0\text{.}$ Thus we have that the trace $\text{tr}\left(\varphi {\left(a\right)}^k\right)=\text{tr}\left(\varphi \left(a^k\right)\right)=0,$ for all $k>0\text{.}$ We want to show that this implies that the matrix $\varphi \left(a\right)$ is nilpotent.

I can’t think without symmetric functions. – A.Ram

We know that $\varphi \left(a\right)$ is similar to an upper triangular matrix, and what the $k\text{-th}$ powers of such a matrix are: $$\begin{array}{cccc}& \varphi \left(a\right)& \approx & \left(\begin{array}{ccc}{\lambda }_1& & *\\ & \ddots \\ 0& & {\lambda }_d\end{array}\right)\\ \text{And thus}& \varphi {\left(a\right)}^k& \approx & \left(\begin{array}{ccc}{\lambda }_1^k& & *\\ & \ddots \\ 0& & {\lambda }_d^k\end{array}\right)\end{array}$$

Thus for every $k>0$ we know that ${\lambda }_1^k+\dots +{\lambda }_d^k=0\text{.}$
Define $$P_k({\lambda }_1,\dots ,{\lambda }_d)={\lambda }_1^k+\dots +{\lambda }_d^k,$$ and $$e_k({\lambda }_1,\cdots ,{\lambda }_d)=\sum _{1\le i_1<\cdots <i_k\le d}{\lambda }_{i_1}\cdots {\lambda }_{i_k}$$

One has the following Newton identity (c.f. Macdonald Symmetric functions and Hall polynomials):

For our current purposes, the point is that $P_r\left(\lambda \right)=0$ for all $r>0,$ and therefore all of the $e_k\left(\lambda \right)$ are zero.

Now consider the characteristic polynomial of $\varphi \left(a\right)\text{.}$

Thus the characteristic polynomial of the matrix $\varphi \left(a\right)$ is ${(-t)}^d,$ and so we have that $\varphi {\left(a\right)}^d=0\text{.}$

Once $\varphi \left(a\right)$ is nilpotent, we know that a is nilpotent as well.

$\square$