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For two singular matrices $A=\left[\begin{array}{ll}1&0\\0&0 \end{array}\right]$ and $B=\left[\begin{array}{ll}0&0\\0&1 \end{array}\right]$, the sum of them is not singular.


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Yes, but it's identically zero. $\wedge^2 V^{\ast}$ and $\wedge^3 V^{\ast}$ are, in general, nonisomorphic irreducible representations of $GL(V)$, so there are no nonzero natural maps between them. You can think of $\wedge^k V^{\ast}$ as being the $k$-forms on $V$ with "constant coefficients"; all of these have exterior derivative $0$. The next most ...


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There are two relationships along these lines. The one you are referring to, I think, is the linear isomorphism between a Clifford algebra of $(V,Q)$ and the exterior algebra for $V$ mentioned in the wiki article. This lets you apply some intuition from exterior algebra inside the Clifford algebra. This linear isomorphism lets you identify a linear copy of ...


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This only answers question 2: There is no hope to do anything independent of a choice of bases here. If you have two linear maps $f,g:V\to W$, which have the same rank, then there are linear isomorphisms $S:V\to V$ and $T:W\to W$ such that $g=T\circ f\circ S$. Equivalently, you can choose bases such that the two maps correspond to the same matrix. To see ...


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Here is what they actually do. It is a worthwhile exercise to confirm that everything is well-defined, that changing basis in the underlying vector space does not alter anything.


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Suppose $U$ is a vector space over a field of characteristic different from $2$. If you have any linear map $\alpha$ on $U$ such that $\alpha^{2} = I$, then either $\alpha = \pm I$, or $x^{2} - 1$ is the minimal polynomial of $\alpha$. In the latter case the minimal polynomial has two distinct roots $1$ and $-1$, thus $U$ decomposes as the direct sum of ...



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