Prove that v is an eigenvector of operator B

In summary, the conversation discusses finding an eigenvector and corresponding eigenvalue for a linear operator involving derivatives. The solution involves taking a double derivative of the given eigenvector and solving for the eigenvalue. The resulting eigenvalue is -16.
  • #1
Smazmbazm
45
0

Homework Statement



Let B be the linear operator

[itex](1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx}[/itex]

Show that

[itex]T_{4}(x) = 8x^{4} - 8x^{2} + 1[/itex]

is an eigenvector of B, and find the corresponding eigenvalue.

Attempt

Righto, I find these rather difficult so a step by step solution would be nice but I know that's a lot of typing and formatting so I'm grateful for any corrections.

Step 1: [itex]Bv_{4} = \lambda _{4} v_{4}[/itex]

Step 2: [itex]((1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx})(8x^{4} - 8x^{2} + 1) = \lambda _{4} (8x^{4} - 8x^{2} + 1)[/itex]

Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

[itex]\frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

\\ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

\\ -240x^4 + 192x^2 - 18 - 40x^4 + 18x^2 - 1 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

\\ -280x^4 + 216x^2 - 19 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}[/itex]

Is this all correct so far? If so, awesome. Now I don't really know how to go about solving for [itex]\lambda[/itex]

In general, is it enough to say that [itex]\lambda = \frac{-280x^4 + 216x^2 - 19}{8x^{4}- 8x^{2} + 1}[/itex]
 
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  • #2
Smazmbazm said:
Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

[itex] = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

You can do that, but you don't need the right hand side. It's probably easier to start by calculating
$$B v = \frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1)$$
and after simplifying you should get a multiple of ##8x^{4} - 8x^{2} + 1## from which you can read off the value of ##\lambda##.

$$ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) $$
No! You cannot move the x after the derivative operator:
$$x \frac{d}{dx} x^n \neq \frac{d}{dx} x^{n+1}$$
 
  • #3
Ok great, thanks for the response. By that logic, wouldn't it imply that you can't move the [itex](1-x^2)[/itex] to the right side either? So you just take the double derivative of the eigenvector then multiply by [itex](1-x^2)[/itex] minus the first derivative of the eigenvector multiplied by x then solve for lambda. If I follow that process, I get a lambda of -16 =] so I guess that's the way to do it
 

What is an eigenvector?

An eigenvector is a vector that, when multiplied by a linear operator, results in a scalar multiple of itself. In other words, the direction of the vector remains unchanged, but its magnitude may change.

What is an operator?

An operator is a mathematical function that takes in a vector as input and produces another vector as output. In the context of linear algebra, operators are often represented by matrices.

How do you prove that a vector v is an eigenvector of operator B?

To prove that v is an eigenvector of operator B, you must show that Bv is equal to a scalar multiple of v. This can be done by multiplying Bv and comparing it to the original vector v. If they are equal, then v is an eigenvector of B.

What is the significance of eigenvectors in linear algebra?

Eigenvectors are important in linear algebra because they help us understand how a linear operator transforms a vector. They also allow us to simplify difficult calculations by reducing them to scalar multiplication. Eigenvectors are also used in many applications, such as data compression and principal component analysis.

Can an operator have more than one eigenvector?

Yes, an operator can have multiple eigenvectors associated with it. In fact, most operators have an infinite number of eigenvectors. This is because any scalar multiple of an eigenvector is also an eigenvector of the same operator.

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