# Is there a way to determine the matrix of $\Lambda^k(T)$ given the matrix of $T$?

Let $T$ be an endomorphism of a finite dimensional vector space $V$. Suppose that $(v_1,\ldots v_n)$ is an ordered basis of $V$. And let $[T]$ be the matrix of $T$ with respect to this basis.

Is there a way to compute the matrix of $\Lambda^k(T)$ with respect to the obvious basis given $[T]$? The 'obvious' basis is the $\binom{n}{k}$ $k$-wedge tuples from $\{v_1,\ldots,v_n\}$ with increasing indices.

There have been many questions on here about how to compute the characteristic polynomial and the coefficients $(-1)^k\text{tr}(\Lambda^k(T))$, but I haven't seen any interest in the matrix of $\Lambda^k(T)$ itself. I suspect it can be built from the minors of $[T]$, but I have no idea how to proceed.

• I recall reading about this calculation in books.google.com/books/about/… (the Advanced Calculus a Differential Forms Approach, by H.M. Edwards). Commented Jul 24, 2014 at 13:45

Let $T(v_i) = \sum_j a_{ji} \cdot v_j$, so that $(a_{ji})$ is the matrix of $T$ w.r.t. to $(v_1,\dotsc,v_n)$. The wedges $v_{i_1} \wedge \dotsc \wedge v_{i_k}$ with $i_1<\dotsc<i_k$ form a basis of $\Lambda^k(V)$. We have:

$$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} a_{j_1,i_1} \dotsc a_{j_k,i_j} \cdot v_{j_1} \wedge \dotsc \wedge v_{j_k}$$

If the indices $j_\ell$ are not pairwise distinct, the summand is zero. If not, we may choose a permutation to order the indices. The result is: $$\sum_{j_1<\dotsc<j_k} \sum_{\pi \in \Sigma_k} \mathrm{sgn}(\sigma) a_{j_{\pi(1)},i_1} \dotsc a_{j_{\pi(k)},i_j} \cdot v_{j_1} \wedge \dotsc \wedge v_{j_k}$$ Thus, the entry of the matrix of $\Lambda^k(T)$ in the row $(j_1<\dotsc<j_k)$ and the column $(i_1<\dotsc<i_k)$ is $$\sum_{\pi \in \Sigma_k} \mathrm{sgn}(\sigma) a_{j_{\pi(1)},i_1} \dotsc a_{j_{\pi(k)},i_j} .$$ This is nothing else than the determinant of the submatrix of $(a_{ji})$ corresponding to the rows $(j_1<\dotsc<j_k)$ and the columns $(i_1<\dotsc<i_k)$.

Conclusion: The entries of the matrix of $\Lambda^k(T)$ are the $k$-minors of the matrix of $T$.

I thought it might be worthwhile to add an explicit example. Consider $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ with $$[T] = \left[ \begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]$$ In particular, using the usual $e_1 = [ 1,0,0 ]^T=(1,0,0)$, $e_2 = (0,1,0)$, $e_3=(0,0,1)$ we have $$T(e_1) = (a,b,c), \ \ \ \ T(e_2) = (d,e,f),\ \ \ \ T(e_3) = (g,h,i).$$ Calculate the wedge products of these images \begin{align} T(e_1) \wedge T(e_2) &= (a,b,c) \wedge (d,e,f) \\ &= (ae-bd)e_1 \wedge e_2+(af-cd)e_1 \wedge e_3+(bf-ce)e_2 \wedge e_3 \end{align} Next, \begin{align} T(e_1) \wedge T(e_3) &= (a,b,c) \wedge (g,h,i) \\ &= (ah-gb)e_1 \wedge e_2+ (ai-cg)e_1 \wedge e_3+(bi-ch)e_2 \wedge e_3 \end{align} and \begin{align} T(e_2) \wedge T(e_3) &= (d,e,f) \wedge (g,h,i) \\ &= (dh-eg)e_1 \wedge e_2+(di-fg)e_1 \wedge e_3+(ei-fh)e_2 \wedge e_3 \end{align} The basis for $\Lambda^2 \mathbb{R}^3$ with increasing indices is $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$. Considering this the standard basis, we find in view of the calculations above and the definition $(\Lambda^2 T)(v,w) = T(v) \wedge T(w)$, $$[\Lambda^2 T] = \left[ \begin{array}{c|c|c} ae-bd & ah-gb & dh-eg \\ \hline af-cd & ai-cg & di-fg \\ \hline bf-ce & bi-ch & ei-fh \end{array}\right]$$ This is a matrix of minors of $[T]$. However, given our choice of basis, the minors are out of place relative to their usual assignment in the Laplace expansion. If we instead order the basis $\{ e_2 \wedge e_3, e_1 \wedge e_3, e_1 \wedge e_2 \}'$ then $$[\Lambda^2 T]' = \left[ \begin{array}{c|c|c} ei-fh & bi-ch & bf-ce \\ \hline di-fg & ai-cg & af-cd \\ \hline dh-eg & ah-gb & ae-bd \end{array}\right]$$ That's better. Now the minors are where they ought to be in my humble opinion. I suppose this is not terribly surprising, the strictly increasing basis is not natural for making the correspondence for vectors and one and two forms in $\mathbb{R}^3$.

• That was very illuminating. Thank you
– user123641
Commented Jul 25, 2014 at 4:18
• glad to help, I once made a sign-error in the 12-slot of the above. All the sudden I had students taking cell-phone pictures... Commented Jul 25, 2014 at 4:43