Recent content by Inirit

<1,-2,2> would be the normal vector, but I don't understand where to go from here. Edit: Thank you for trying to help me, but it's obvious that I am too far lost with the entire concept. I was hoping that solving this problem would point me in the right direction for understanding, but seeing...

*sigh* I am having a hard time comprehending how this works, and now I feel more lost than ever. I just don't see where to go and how it all connects.

Yes it does take it to the origin, but I merely asked if that were a matrix that maps a vector in R^3 to the plane, not it's reflection (I know that's not the goal of the problem, but I am slowly trying to understand this). A vector only with a component that isn't within the space of the plane...

Yes that's what I meant, sorry. And the reflection would be (x,y,-z). That's what I was thinking, you just take the height of the vector relative to the plane and reverse it. But I am stuck on how to express it's height relative to the plane. But just to help me better understand what it...

I can't because I don't even understand the transformation. I don't get what kind of matrix would transform something to reflect it about the plane. Although... I've just had a thought. If you expressed the plane like this: x_1 = 2x_2 - 2x_3... would the matrix that transforms a vector onto...

Homework Statement Find a basis B of R^n such that the B matrix B of the given linear transformation T is diagonal. Reflection T about the plane x_1 - 2x_2 + 2x_3 = 0 in R^3. The Attempt at a Solution I just don't even know where to begin. I don't know how to interpret problem or how to...

t^2 - t is a linear combination of 1 - t and 1 - t^2 because 1(1 - t) - 1(1 - t^2) = t^2 - t, therefore it's merely within the space and doesn't form part of the basis. ...Right?

Really? So the independent combinations would be 1-t and 1-t^2, right? Therefore, the basis would be {1-t,1-t^2}, which would have a dimension of 2.

What's to stop you from differentiating a third time? Without the extra factor, it looks like the denominator would turn into a function of cosine, which wouldn't give you zero when you plug in zero.

Yes, that's what I meant, but I see that I worded it incorrectly before. To explain a little better, though, I meant that if you applied the rule to an equation over 0 (f(x)/0), it would still be over 0 no matter how many times you differentiated the top and bottom.

You'd use the rule if you got a 0 in the denominator after plugging in the limit of x, not if it were 0 to begin with. An equation with a 0 in the denominator would always have a 0 in the denominator no matter how many times you applied the rule.

Is it sin(5x-5x) or sin(5x)-5x? The first one seems like a trivial way to express an equation and would result in 0 anyway, making the original equation undefined.

Got another linear space question. I'm getting closer to understanding what's going on, but I'm not there yet. Homework Statement Find a basis for the space and determine its dimension. The space of all polynomials f(t) in P2 such that f(1) = 0. Homework Equations The Attempt at...

Awesome, thanks a lot for the help.

So... the neutral element of P2 has to equal 0 for all t? Therefore, a,b,c would have to all equal 0 to make that true, but that can't be the case because a has to equal 2, meaning it doesn't contain the neutral element and isn't a subspace of P2... right? If that's the case, then it makes a...