Proving Idempotency for general matrix

Question

Suppose that $a_1,...,a_n,b_1,...,b_n ∈ F $ are such that $\sum a_ib_i = 1_F$.

Let $J : F^n → F^n $ be the linear transformation whose standard matrix has $ij^{th}$ entry $a_ib_j$.

Prove that $J^2 = J$.

So I think I've figured out that the index in the matrix $F^2$ given by

$u_{ij} = a_ib_j = \sum_{c=1}^{n}\sum_{d=1}^{n} (a_ib_c)(b_ja_d)$

given that

$ \sum (a_ib_i) = 1 $

I think I got the above right, not 100% sure, but I still don't know where to go from this point onwards.

Could someone help me out with that.

Thanks.

Answer 1

It might help you to know that, if you call $$a=\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix},\ \,b=\begin{pmatrix}b_1\\b_2\\\vdots\\b_n\end{pmatrix}$$

Then $J=a\cdot b^t$

So $J^2=\left(a\cdot b^t\right)\cdot\left(a\cdot b^t\right)=a\cdot\left(b^t\cdot a\right)\cdot b^t=a\cdot b^t=J$

If you like the calcs, by the way,

$$u_{ij}=\sum_{k=1}^n a_ib_ka_kb_j=a_ib_j\sum_{k=1}^na_kb_k=a_ib_j$$

Answer 2

Let $X=[x_{ij}]$ and $Y=[y_{ij}]$ be matrices, the first being $m\times n$ and the second being $n\times p$; then, for $1\le i\le m$ and $1\le j\le p$, the coefficient in place $(i,j)$ of $XY$ is $$ \sum_{k=1}^n x_{ik}y_{kj} $$ In your case, $x_{ik}=a_ib_k$ and $y_{kj}=a_kb_j$, so you have $$ \sum_{k=1}^n a_ib_ka_kb_j $$

There is no double summation for the entries of the matrix product.