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seen Jul 23 '13 at 15:47

Jul
5
comment Confused about quadratic forms
You say the norm defined by q, but I'm not sure I understand that. Does that mean that if $f$ is the associated symmetric bilinear form then $f$ also defines an inner product in $R^n$ such that $f(u,v)=\langle u,v \rangle $ and $q(v)=\langle v,v \rangle = \lVert v \lVert^2$?
Jul
4
accepted Confused about quadratic forms
Jul
4
comment Confused about quadratic forms
@wild: Care to elaborate?
Jul
4
asked Confused about quadratic forms
Jun
23
accepted Finding $a_n$ using a given matrix
Jun
23
accepted If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
Jun
23
asked Finding $a_n$ using a given matrix
Jun
18
comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
Proof for lemma: let $u \in U^\perp$, then $<u, v>=0,\ \forall v \in U$. Since U is invariant, $Tv \in U$ as well, hence $<u, Tv>=<T^*u, v>=0$, therefore $T^*u \in U^\perp$. But how does this help?
Jun
17
comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
Ok, thanks for trying. So according to your proof, $T$ and $T^*$ can be represented by these diagonal matrices: $\text{diag}(a_{11},\ldots,a_{nn}),\ \text{diag}(\overline{a_{11}},\ldots,\overline{a_{nn}})$, and since diagonal matrices commute, so does the transformations?
Jun
17
comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
That's a very nice proof, thanks. Out of curiosity, can you think of one that uses Jordan forms? I'm trying to think how this relates to the material I just read.
Jun
17
comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
@Jonas: Yes to the last question. Never heard the term 'unitarily triangularized' though.
Jun
17
asked If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
Jun
4
comment Non diagonalizable matrix
I think I understand the example but could you also add a small intuitive explanation as to what's going on here?
Jun
4
accepted Non diagonalizable matrix
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
comment Positive semidefinite quadratic form
Thanks. But I think this: "The $v \in V$ such that $q(v)=0$ are then in the subspace" requires further explanation, such as the one provided by Plop.
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 :/