Recent content by psie

I get it now I think. We have a dot at ##(i,j)## if and only if ##p_j\geq i## and ##r_i\geq j##.

We know that a Jordan canonical form is simply the matrix representation of an operator (whose characteristic polynomial splits) with respect to a special basis called a Jordan canonical basis. This basis consists of a disjoint union of cycles/chains of generalized eigenvectors. Take all the...

I think I found a good explanation on Wikipedia for this.

I can accept that this is a tedious and messy exercise, but for instance, take case 3, we have $$AB=\begin{pmatrix}\cos\psi&-\sin\psi&0\\ -\sin\psi&-\cos\psi&0\\ 0&0&-1\end{pmatrix}.$$I can't even solve this case. I get that a row reduced form of the matrix ##AB-I## is...

The way I've approached the problem so far is that I'm looking for the set ##\{x\in\mathbb R^3: ABx=x\}##, since the rotation acts as the identity on the axis of rotation. The set is the null space of ##AB-I##, and $$AB=\begin{pmatrix}\cos\psi&-\sin\psi&0\\...

Ah ok, then I have an earlier printing of the book. It actually says (b) is true in my book. Thank you!

Conditioning doesn't refer to hair conditioners in this case, but what happens to the solution of a system of linear equations ##Ax=b## when we change say the vector ##b## slightly to ##b'##. If the relative change in ##b##, that is ##\frac{\|b-b'\|}{\|b\|}##, is close to the relative change in...

In this particular case. I'm confused how one would use the induction hypothesis.

Here ##A=B_1\oplus B_2\oplus \cdots\oplus B_k## means that ##A## is block diagonal with ##B_i## along the diagonal. A proof that I've seen goes as follows. The ##k=2## case boils down to realizing that if ##v\in\beta_1\subseteq\mathsf W_1##, then ##\mathsf T(v)## is a linear combination of...

I am not sure anymore. But I am pretty sure it is a typo in the exercise. 🙂 Thanks for the help though.

Well, they use the notation ##\mathsf C^\infty## in the exercise, which they defined previously (page 130) to be the subspace of ##\mathbb C^{\mathbb R}## with derivatives of all orders. Here's the exercise, word for word: I now suspect the equality in (b) is a typo and should simply be...

Yes, all functions are of the form ##\mathbb R\to\mathbb C##. But the problem is about a determinant of a matrix which depends on ##t\in\mathbb R##. If the columns of this matrix are linearly dependent, isn't it natural to expect that these columns sum to ##0## in a nontrivial way with...

I don't understand. Let ##y\in\mathsf N(\mathsf T)##, then ##\mathsf T(y)=0## is the zero function and consequently the determinant is zero for every ##t##. In other words, $$0=[\mathsf T(y)](t)=\det\begin{pmatrix}v(t)&v_1(t)&\cdots&v_n(t)\end{pmatrix}\quad\forall t\in\mathbb R.$$So for ##t=2##...

Kind of. The only thing that's missing is I think that we still need to show the ##\alpha_i## are constants, not functions of ##t##.

Is ##y(t)\in C^{\infty}(\mathbb C)##? I'm not sure it is differentiable at ##t=0##.