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With T: R^m -->R^n be linear transformation T(x) = B*x and if psi sub I is an elementary alternating k-tensor on R^n, then T*psisub I has the form:

$$ T^**\psi_I $$ = sigma sub [J] cJ*psi[J] where psiJ are the elementary alternating k-tensors on R^m and

where $I = (i_1,\ldots, i_k)$ and we can let $I_\sigma = (i_{\sigma(1)},\ldots, i_{\sigma(k)})$.

I'm trying to determine the coefficients of C sub J.

My proof:

T*f(x) = f(T(x)) = f(Bx) = ABx so the matrix T*f is AB

And if f = sigma sub [I] dsubI * psisub I is an alternating k-tensor on R^n, how can T*f be expressed in terms of the elementary k-tensors on R^m?

Thanks

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1  
Why don't you write in TeX? –  Riccardo.Alestra Mar 20 '12 at 9:15
    
I'm trying to learn –  mary Mar 20 '12 at 9:20
2  
What is $\phi$? What is $\psi$? –  martini Mar 20 '12 at 9:31
    
The tensors phi sub I are elementary k-tensors on V corresponding to a basis a1,...an for V. –  mary Mar 20 '12 at 9:52
1  
Mary, after visiting this Latex site, you should be able to create equations without having to write "psi sub I" etc. –  The Chaz 2.0 Mar 20 '12 at 13:02

1 Answer 1

if I understand you correctly, the $\phi_I$ are given by the property that $\phi_I(a_{j_1}, \ldots, a_{j_k}) = 1$ iff $(j_1, \ldots, j_k) = (i_1, \ldots, i_k)$ and $ = 0$ otherwise.

Supposing that, we have for $1 \le j_1, \ldots, j_k \le n$ \begin{align*} \phi_{I_\sigma}^\sigma(a_{j_1}, \ldots, a_{j_k}) &= \phi_{I_\sigma}(a_{\sigma(j_1)}, \ldots, a_{j_{\sigma(j_k)}})\\ &= \begin{cases}
1 & (\sigma(j_1), \ldots, \sigma(j_k)) = (\sigma(i_1), \ldots, \sigma(i_k))\\ 0 & \text{\otherwise} \end{cases}\\ &= \begin{cases} 1 & (j_1, \ldots, j_k) = (i_1,\ldots, i_k)\\ 0 & \text{\otherwise} \end{cases}\\ &= \phi_I(a_{j_1}, \ldots, a_{j_k}) \end{align*} so $\phi_{I_\sigma}^\sigma = \phi_I$. Regarding $\psi$, I still don't understand what you want to show,

AB,

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