Recent content by Oliviacarone

Homework Statement Find the distance from point P (1,7,3) to the line (x,y,z) = (-2,1,4) + s(1,-3,4), s is a free variable Homework Equations projnQP = ( QP⋅n/(lengthQP)(lengthn) )(n) The Attempt at a Solution I'm not quite sure about how to find the normal (n) here, but if I make s=0, I'm...

Hmm should I just try s=0 and t=0? Not sure what else I would use.

Thank you!

Homework Statement Find the cosine of the angle determined by the intersecting lines m1: (x,y,z) = (-2,1,4)+s(1,-3,4) m2: (x,y,z) = (-2,1,4)+t(-2,3,2) s and t are free variables Homework Equations a⋅b = length(a)length(b) cosθ The Attempt at a Solution I just did this equation using no...

I totally understand now. Thank you so much. Now, if the Eigen vectors HAD been the same, how would I prove this It says: Prove that if λ and μ correspond to the same Eigen vector x then λ+μ is an Eigen value of A+B and λμ is an eigenvalue of AB. So that equation would prove this?

I totally understand now. Thank you so much. Now, if the Eigen vectors HAD been the same, would I just prove this by saying (A+B)v = (λ+μ)v ?

But these have Eigen values of 2, 1 for A and 3,1 for B. A+B has Eigen values of 3 and 4, which can be found by adding Eigen values of A and B. BUT A and B also gives other Eigen values (2+3 = 5, for example) ... is this why it works??

Okay so I can just answer this question with A = [1 0 0 2] and B = [10 0 0 20] and write out all the Eigen values and that there's no Eigen value of 12 or 21 for A+B? To me, the question was saying that A and B only had one Eigen value each that's why I was confused

hmmm no, only 11 and 22

No, you're right (at least for A+B, I haven't checked for AB e Okay yeah so then those don't work? A= 2 and 1, B= 3 and 4. A+B = 1 and 3, and 2+1 =3

because if I just answer it with that A and B, the Eigen values are 2 and 1 for A, and 1 and 3 for B. For AB, the Eigen values are 3 and 4 so Eigen value A + Eigen value B can = Eigen value A+B

So my answer is the A and B i gave + that the Eigen vectors are different so it doesn't matter if the (A+B) Eigen value is the same as the Eigen value A + Eigen value B?

Okay, so I get that, but they could still have (λ+μ) = Eigen value of (A+B) no ?

Okay... So I found A = [2 1 0 1] and B = [1 0 1 3] which fits your equations, but these both have 2 Eigen values each, which isn't what I'm looking for... really not sure what you're trying to say

I'm sorry I honestly don't understand what you mean by saying I only used v. When I wrote (A+B)w= Aw+Bw, aren't I using w??