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Dec
20
awarded  Constituent
Dec
20
answered What are some interesting sole exceptions or counterexamples?
Dec
10
awarded  Caucus
Nov
24
comment Confusing Analysis proof
Because you represent a bilinear form $f \colon V \otimes V \rightarrow \mathbb{F}$ and not a linear map $L \colon V \rightarrow V$. You can read about it in any reference that discusses bilinear forms - for example, people.virginia.edu/~mah7cd/Math5651/Lecture15.pdf.
Nov
23
revised Confusing Analysis proof
Addressed an additional question
Nov
23
comment Confusing Analysis proof
I'll add an explanation in an hour or so, but the point is that a bilinear form doesn't have a well-defined notion of a determinant as a number, only the ratio makes sense, and you can compute it any way you want.
Nov
23
comment How to show existence of an orthogonal map?
Hint: Complete $\frac{x}{||x||}$ to an orthonormal basis $(e_1, \ldots, e_n)$, complete $\frac{y}{||y||}$ to an orthonormal basis $(f_1, \ldots, f_n)$ and define $T$ by describing how it acts on $e_i$.
Nov
23
answered Confusing Analysis proof
Nov
17
revised Adaptions I have to make to go from integer coefficients to coefficients in $R$
edited tags
Sep
30
awarded  Explainer
Sep
6
comment Can there exist a non diagonal matrix whose inverse is diagonal matrix?
Yes, it is like in a field with respect to addition or multiplication or like in a vector space with respect to addition.
Sep
2
answered Kernel of linear transformation composition
Sep
2
comment Can there exist a non diagonal matrix whose inverse is diagonal matrix?
The inverse matrix of a diagonal matrix $\mathrm{diag}(\lambda_1, \ldots, \lambda_n)$ is the diagonal matrix $\mathrm{diag}(\frac{1}{\lambda_1}, \ldots, \frac{1}{\lambda_n})$ and the inverse is unique so no.
Sep
1
answered If $\mathcal{B}^*$ is a basis for $V^*$, then $V^*$ is finite dimensional.
Sep
1
revised Integration against divergence free vector fields
added 401 characters in body
Sep
1
answered Integration against divergence free vector fields
Aug
30
comment Integration against divergence free vector fields
Is this true when $n = 1$?
Aug
30
revised proof that a continuous additive homomorphism $\mathbb{R}^n\to\mathbb{R}^m$ is $\mathbb{R}$-linear
added a hypothesis that appears in the title but not in the question itself
Jul
29
answered What's going on here with zero divisors and invertibility?
Jul
21
answered diagonalization of a matrix over finite fields