Recent content by boo

OK I found the answer. In a complex inner product space every normal operator (commutes with it's adjoint) is guaranteed to have a full orthonormal set of basis vectors for that space. In a real inner product space every self adjoint operator does likewise.

Does the spectral theorem guarantee that algebraic and geometric multiplicity for all eigenvalues are always equal for normal operators?

Going through Axler's awful book on linear algebra. The complex spectral theorem (for operator T on vector space V) states that the following are equivalent: 1) T is normal 2) V has an orthonormal basis consisting of eigenvectors of T and 3) the matrix representation of T is diagonal with...

So going back to my original problem and using this technique I find two eigenvectors in the null space of ##\mathbf A~-~\lambda I## to be ##\begin{pmatrix}1\\0\\0\end{pmatrix}##and##\begin{pmatrix}0\\-1\\1\end{pmatrix}## but that the first eigenvector in the Jordan chain must have the form...

Yes, I checked out that link and none of the problems addressed this issue. They were all easily solved using the conventional technique of choosing one or another of the eigenvectors obtained in "step one" in order to find the generalized eigenvectors in step 2. This works "most" of the time...

Don't see any way to edit these posts which is frustrating when I see a typo. Anyway that last equation should be (A - \lambda I)*v3 = v2

Sorry, struggled with the Latek code which I don't use. The eigenvalues are 1 with algebraic multiplicity 3 and geometric multiplicity 2. The null space of (A - \lambda I) has 2 eigenvectors v1 = (1 0 0) and v2 = (0 -1 1) ( or any linear combination of these). Looking for the third...

Mentor note: The Tex shown below had to be modified a fair amount to conform to the MathJax on this site. Trying to calculate the modal matrix for the following ##A =\begin{pmatrix} 1 && 1/2 && 1/2 \\ 0 && 1/2 && -1/2 \\ 0 && 1/2 && 1 .5 \end{pmatrix}## there are two eigenvectors for this...

In case it's unclear above Vm is the voltage phasor magnitude and theta_v is the voltage phasor angle (phase).

Easiest way to solve this is to use phasors to calculate both current and voltage through resistor and inductor. (Of course the resistor and inductor voltages will be the same). Once you have their magnitude and phase use the following formulas for instantaneous power: P(resistor) =...

Would not recommend you torture yourself by expanding the terms using Eulers formula until the very end. Instead simply multiply both numerator and denominator by the complex conjugate of the given denominator. The denominator will then reduce to a very simple number. You can now expand the...

Seriously no need to invoke KCL. This is a simple voltage division problem. I'm sure that you know that if you have a voltage source in series with 2 resistors R and R2 that the voltage drop across R2 (your output voltage) is simply Vin*R2/(R + R2) In this case "R2" is the complex impedance of...

Sometimes it's hard answering these questions because I have no idea at what level an answer is expected. Essentially the question relates to plotting the magnitude of the transfer function for a simple passive low pass filter. Typically this is done on a logarithmic scale but that wasn't...

Whoa! Didn't realize that this was an ancient thread. Would not waded in. But agree that left going current is - Vo/4.

Let me also point out that the main reason you are getting stuck on this problem is because you are taking Vo as the voltage at that top node which it is not. The left going current is neither Vo/4 nor Vo/(4 + 6). The correct (left going) current is simply V/(4 + 6) where V, as above, is simply...