Why is the eigenspace of D^2 two-dimensional? One "vector" would be a constant function, say y=5, and all the scalar multiples of it. What is the other "vector"? Is it y=2x+1 and all scalar multiples of it? Or is it y=3x+1? or maybe y=x+152? Clearly these lines are not scalar multiples of each other.
You said the only functions whose derivatives produce a multiple of the original function are constant functions... but what about exponential functions??? Even the example you write on the board meets this requirement! for e^2x, both first derivative (2e^2x) and second derivative (4e^2x) are multiples of e^2x - am I wrong? Please clarify, thanks.
You answered your question in your first thought: "only functions whose derivatives produce a multiple of the original function..." Since any derivative of e^2x is a constant multiplied by the original function, then that is a linear operation. In general, a linear operation follows the notion that some operator 'A' acting on some function 'f' gives some constant 'a' multiplied by that same function. Or in equation form, A f = a f. Here, 'a' is an eigenvalue of the operator 'A' when it acts on 'f', and 'f' is the eigenfunction. To address your "constant" function worries - he's just mentioning a separate class. So, you can have the case discussed above OR a constant function. Those are the two primary cases. A constant function is not dependent on any variable. The derivative of such a function is zero. Therefore, if A = d/dx and if f = constant, then A f = 0. But this can be viewed as A f = 0 f, where 0 is an eigenvalue of this equation. It's kind of a "null" case (I'm sure some people can contest this point), but that's what he was referring to. Any other polynomial equation cannot be acted on by a derivative operator and still be linear.
You are right. Every function e^ax is an eigenfunction of the derivative operator, including e^0x (constant function). However, I think here he is only considering polynomial functions, so the only eigenfunction in that space is the constant function. See 9:50
@@loremazza5191 don't think so because sinpx and sin-px are linearly dependent. I think its just that sine and cosine are not linearly dependent, so they are the two dimensions
@michaelbobman7964 according to that logic (that sine and cosine represent different components of a vector), then any function y of x apart from the constant function (where y is fixed) has a 2D eigenspace, is that right?
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Just want to say that 7 years later this is still super helpful.
This is the first TH-cam Video without any dislikes
He reminds me of David Wallace from the office
Amazing
Why is the eigenspace of D^2 two-dimensional? One "vector" would be a constant function, say y=5, and all the scalar multiples of it.
What is the other "vector"? Is it y=2x+1 and all scalar multiples of it? Or is it y=3x+1? or maybe y=x+152? Clearly these lines are not scalar multiples of each other.
Great point! Zero is a "repeated" or "multiple" eigenvalue which means that there's a corresponding two-dimensional eigenspace.
i like your short but very informative videos
thank you sir
Very interesting video! Thanks again! =)
You said the only functions whose derivatives produce a multiple of the original function are constant functions... but what about exponential functions??? Even the example you write on the board meets this requirement! for e^2x, both first derivative (2e^2x) and second derivative (4e^2x) are multiples of e^2x - am I wrong? Please clarify, thanks.
You answered your question in your first thought: "only functions whose derivatives produce a multiple of the original function..." Since any derivative of e^2x is a constant multiplied by the original function, then that is a linear operation. In general, a linear operation follows the notion that some operator 'A' acting on some function 'f' gives some constant 'a' multiplied by that same function. Or in equation form, A f = a f. Here, 'a' is an eigenvalue of the operator 'A' when it acts on 'f', and 'f' is the eigenfunction.
To address your "constant" function worries - he's just mentioning a separate class. So, you can have the case discussed above OR a constant function. Those are the two primary cases. A constant function is not dependent on any variable. The derivative of such a function is zero. Therefore, if A = d/dx and if f = constant, then A f = 0. But this can be viewed as A f = 0 f, where 0 is an eigenvalue of this equation. It's kind of a "null" case (I'm sure some people can contest this point), but that's what he was referring to. Any other polynomial equation cannot be acted on by a derivative operator and still be linear.
You are right. Every function e^ax is an eigenfunction of the derivative operator, including e^0x (constant function).
However, I think here he is only considering polynomial functions, so the only eigenfunction in that space is the constant function. See 9:50
Sir, what is the analogous case of 1D Eigen-space in geometric vectors (i.e. geometric operation similar to D operator in function vectors)?
Can anyone show what he says at the end?
I thought it was because both D''sin(px) and D''sin(-px) have an eigenvalue of -p^2
@@loremazza5191 That's what I think it is.
@@loremazza5191 don't think so because sinpx and sin-px are linearly dependent. I think its just that sine and cosine are not linearly dependent, so they are the two dimensions
@michaelbobman7964 according to that logic (that sine and cosine represent different components of a vector), then any function y of x apart from the constant function (where y is fixed) has a 2D eigenspace, is that right?