daniel.jackson
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 Jun 4 comment Non diagonalizable matrix @Aaron: by Cayley-Hamilton, $Q$ is the characteristic polynomial of $A$, but I'm not sure I followed the rest of your suggestion. Jun 4 comment Non diagonalizable matrix @Mark: that makes sense, but where do I go from there? Jun 4 asked Non diagonalizable matrix May 28 accepted Positive semidefinite quadratic form May 28 revised Positive semidefinite quadratic form must have 6 characters in edit... just wanted to fix the latex :/ May 28 suggested approved edit on Positive semidefinite quadratic form May 28 asked Positive semidefinite quadratic form May 2 comment Fun Linear Algebra Problems @Moron: very nice, got it :). May 1 comment Fun Linear Algebra Problems @Moron: ok, unfortunately the only time I saw $F_2$ was to assume the exact opposite, so I have no idea how that helps. Also, I assume it's obvious but an inhabitant can be a member of several clubs? (otherwise it seems trivial) May 1 comment Fun Linear Algebra Problems I'm having some trouble solving this with my linear algebra knowledge, any hints? (also, just so I'm sure, what does $\mathbb{F}_{2}$ mean?) Apr 20 comment Spectral decomposition of a normal matrix @Arturo: I'm stuck after finding the eigenspaces. I got $$V_{\lambda_1}=sp{(\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2})}, V_{\lambda_2}=sp{(\frac{-1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3})}, V_{\lambda_3}=sp{(\frac{-1}{\sqrt 6}, \frac{-2}{\sqrt 6}, \frac{1}{\sqrt 6})}$$ And I know that if $v=v_1+v_2+v_3, v_i \in V_{\lambda_i}$ then $P_i = v_i$. But how do I actually find $[P_i]$? I'm confused. Apr 19 revised Spectral decomposition of a normal matrix fixed error Apr 19 asked Spectral decomposition of a normal matrix Apr 13 comment Norm of a symmetric matrix? Ross, how is your last sentence justified? Also I'm not entirely sure I got it right, what did you mean by "...input vector is along the eigenvector associated with the largest eigenvalue."? Which eigenvector? if $v=\alpha_1 v_1 +...+\alpha_n v_n$ is the input vector as you put it, then it's a linear combination of eigenvalues where $||v||=1$. What are you arguing about $||Av||$? Mar 31 accepted Multiplication of block matrices Mar 31 asked Multiplication of block matrices Mar 26 comment Adjoint of a linear transformation in an infinite dimension inner product space I tried understanding the links you provided but I'm still in doubt: is it possible to find an infinite inner product space and a linear transformation $T$, such that no linear transformation $S$ exists such that $(Tv, u)=(v, Su)$? Mar 26 comment Adjoint of a linear transformation in an infinite dimension inner product space Thanks. I'm indeed unfamiliar with Hilbert spaces. I was trying to think of $R[x]$ with the inner product $\int_{a}^{b} f(x)g(x)dx$. Mar 26 asked Adjoint of a linear transformation in an infinite dimension inner product space Mar 25 comment Conjugate-transpose of a linear transformation Yes that's what I meant.. Or transformation matrix. Thanks.