Reputation
1,116
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
9 24
Impact
~27k people reached

Jul
6
comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$
So if $A^2$ had a negative eigenvalue, say $-1$, the same argument would not work? Because then it's possible that $A$ characteristic polynomial has $i$ as one of its roots, which is not an eigenvalue of $A$.
Jul
6
comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$
@Chandru, not at all, thank you.
Jul
6
asked if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$
Jul
5
comment Confused about quadratic forms
I don't understand why this is true. By orthogonal diagonalization you can bring $q$ to the form $\sum \lambda_i x_i^2$, how did you conclude that $\lambda_i=1$ in this case? I know that there is a basis where $q(x)=\sum x_i^2$ but why is it necessarily orthogonal? Wikipedia says something similar too.
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?