Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in

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In summary: Similarly, applying T to te^t gives 1(e^t)+ 1(te^t)+ 0(t^2e^t) so the second column of the matrix is\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}. And applying T to e^t gives 0(e^t)+ 1(te^t)+ 0(t^2e^t) so the first column of the matrix is\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}. The matrix representation of T, with respect to the given basis is \begin{bmatrix}0 & 1 & 2 \\ 1 &
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renolovexoxo
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Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in the infinite continuous functions. Let T:V-->V be give by T(f)=f''-2f'+f. Decide whether T is diagonalizable.

We learned a theorem that this will be diagonalizable if and only if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity.

What I am having trouble with is translating this into a way to find geometric and algebraic multiplicity. I'm not entirely sure what to do when I'm not given a matrix, since that's how we did it in class.
 
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Try to find the matrix associated to this linear map.

Choose a basis of V, and try to construct the matrix.
 
  • #3


So would I choose f1, f2, f3 as a basis for V and then construct a matrix
[f1 f2 f3]? Something like that?
 
  • #4


You need to construct a matrix with respect to the basis [itex](f_1,f_2,f_3)[/itex]. Do you know how to do that?
 
  • #5


I know I should know how to do that, but I get tripped up in functions, especially involving e. After looking at it, I can't figure it out.
 
  • #6


You are given the basis vectors for a space, V, and a linear transformation from V to itself. To find the matrix representation of the linear transformation, apply the transformation to each basis vector, in turn, and write the result as a linear combination of the basis vectors. The coefficients give a column of the matrix.

Here, the given basis is [itex]\{e^t, te^t, t^2e^t\}[/itex] and the linear transformation is T(f)= f''- 2f'+ f. Applying that to, say, [itex]t^2e^t[/itex], gives [itex]2e^t= 2(e^t)+ 0(te^t)+ 0(t^2e^t)[/itex] so that the third column of the matrix is
[tex]\begin{bmatrix}2 \\ 0 \\ 0\end{bmatrix}[/tex]
 

What is the span of the functions f1, f2, and f3?

The span of a set of functions is the set of all possible linear combinations of those functions. In this case, the span of f1, f2, and f3 is the set of all possible combinations of e^t, te^t, and t^2e^t, where t is any real number.

What is the purpose of defining V as the span of f1, f2, and f3?

Defining V as the span of f1, f2, and f3 allows us to represent any function that is a linear combination of these three functions. This can be useful in applications such as solving differential equations or approximating more complex functions.

How do the functions f1, f2, and f3 relate to each other?

These functions are related through multiplication by t and t^2. Specifically, f1(t) = f2'(t) and f2(t) = f3'(t). This means that the derivative of f2 is equal to f1, and the derivative of f3 is equal to f2. This relationship can also be seen in the coefficients of t and t^2 in each function.

Is V a vector space?

Yes, V is a vector space. It satisfies all the properties of a vector space, including closure under addition and scalar multiplication, and contains the zero vector (the function f(t) = 0).

Can V be spanned by any other set of functions?

Yes, V can be spanned by an infinite number of sets of functions. As long as the set of functions is linearly independent (meaning no function in the set can be written as a linear combination of the other functions), it can span V. In this case, f1, f2, and f3 are a convenient and commonly used set of functions to span V.

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