Mixing
Show that K>0 is an eigenvalue by finding an Eigenvecter corresponding to k
I'm having trouble answering these questions:
Consider the map $D : P_inftyto P_infty$ defined by: $D(f)=(x+2)f'$
a) For any non-negative integer k, show that k is an eigenvalue
of D by finding a nonzero eigenvector corresponding to k.
I've tried something along the lines of $D(x^k)=(x+2)(kx^{k-1})=kx^k+2kx^{k-1}$, as well as trying $D(a_1+a_2x+a_3x^2+...+a_nx^{n-1})=(x+2)(a_2+a_3x+...+a_n(n-1)x^{n-2})$ but I don't know what to do from here (or if either of these are the right approach at all).
b) Find the kernel of D and state its dimension.
With this I think that that $Ker(D)={a_1,a_2,a_3,...,a_n:a_1in RBbb text{and} a_2,a_3,...,a_n=0}$ but I'm not sure how to find out the dimension of this.
Any help would be greatly appreciated.
linear-algebra eigenvalues-eigenvectors linear-transformations
asked Nov 21 '18 at 17:18
O.D
102
add a comment |
I'm having trouble answering these questions:
Consider the map $D : P_inftyto P_infty$ defined by: $D(f)=(x+2)f'$
a) For any non-negative integer k, show that k is an eigenvalue
of D by finding a nonzero eigenvector corresponding to k.
I've tried something along the lines of $D(x^k)=(x+2)(kx^{k-1})=kx^k+2kx^{k-1}$, as well as trying $D(a_1+a_2x+a_3x^2+...+a_nx^{n-1})=(x+2)(a_2+a_3x+...+a_n(n-1)x^{n-2})$ but I don't know what to do from here (or if either of these are the right approach at all).
b) Find the kernel of D and state its dimension.
With this I think that that $Ker(D)={a_1,a_2,a_3,...,a_n:a_1in RBbb text{and} a_2,a_3,...,a_n=0}$ but I'm not sure how to find out the dimension of this.
Any help would be greatly appreciated.
linear-algebra eigenvalues-eigenvectors linear-transformations
asked Nov 21 '18 at 17:18
O.D
102
add a comment |
I'm having trouble answering these questions:
Consider the map $D : P_inftyto P_infty$ defined by: $D(f)=(x+2)f'$
a) For any non-negative integer k, show that k is an eigenvalue
of D by finding a nonzero eigenvector corresponding to k.
I've tried something along the lines of $D(x^k)=(x+2)(kx^{k-1})=kx^k+2kx^{k-1}$, as well as trying $D(a_1+a_2x+a_3x^2+...+a_nx^{n-1})=(x+2)(a_2+a_3x+...+a_n(n-1)x^{n-2})$ but I don't know what to do from here (or if either of these are the right approach at all).
b) Find the kernel of D and state its dimension.
With this I think that that $Ker(D)={a_1,a_2,a_3,...,a_n:a_1in RBbb text{and} a_2,a_3,...,a_n=0}$ but I'm not sure how to find out the dimension of this.
Any help would be greatly appreciated.
linear-algebra eigenvalues-eigenvectors linear-transformations
asked Nov 21 '18 at 17:18
O.D
102
I'm having trouble answering these questions:
Consider the map $D : P_inftyto P_infty$ defined by: $D(f)=(x+2)f'$
a) For any non-negative integer k, show that k is an eigenvalue
of D by finding a nonzero eigenvector corresponding to k.
I've tried something along the lines of $D(x^k)=(x+2)(kx^{k-1})=kx^k+2kx^{k-1}$, as well as trying $D(a_1+a_2x+a_3x^2+...+a_nx^{n-1})=(x+2)(a_2+a_3x+...+a_n(n-1)x^{n-2})$ but I don't know what to do from here (or if either of these are the right approach at all).
b) Find the kernel of D and state its dimension.
With this I think that that $Ker(D)={a_1,a_2,a_3,...,a_n:a_1in RBbb text{and} a_2,a_3,...,a_n=0}$ but I'm not sure how to find out the dimension of this.
Any help would be greatly appreciated.
linear-algebra eigenvalues-eigenvectors linear-transformations
linear-algebra eigenvalues-eigenvectors linear-transformations
asked Nov 21 '18 at 17:18
O.D
102
asked Nov 21 '18 at 17:18
O.D
102
asked Nov 21 '18 at 17:18
O.D
102
asked Nov 21 '18 at 17:18
O.D
102
asked Nov 21 '18 at 17:18
O.D
102
102
add a comment |
add a comment |
1 Answer
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For part $(a)$, you want a function $f in P_infty$ that satisfies
$$ D(f) = kf quad iff quad (x+2)f'= kf, $$
which is a simple separable differential equation.
If $x=-2$, we see from the above that we must have $f(-2) = 0$, and otherwise if $xnot=-2$, we have
$$ int frac{mathrm df}{f} = kint frac{mathrm dx}{x+2} quad iff quad log f = k log (x+2) + C quad iff quad f = (x+2)^k, $$
where in the last step we set $C=0$ due to the condition $f(-2)=0$.
For part $(b)$, you want all functions $f in P_infty$ such that
$$ D(f) = (x+2)f' = 0. $$
Since this should hold for all $x$, we see that $f' = 0$, i.e. $f = text{const.}$ The dimension of this kernel is evidently equal to $1$.
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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active
oldest
votes
For part $(a)$, you want a function $f in P_infty$ that satisfies
$$ D(f) = kf quad iff quad (x+2)f'= kf, $$
which is a simple separable differential equation.
If $x=-2$, we see from the above that we must have $f(-2) = 0$, and otherwise if $xnot=-2$, we have
$$ int frac{mathrm df}{f} = kint frac{mathrm dx}{x+2} quad iff quad log f = k log (x+2) + C quad iff quad f = (x+2)^k, $$
where in the last step we set $C=0$ due to the condition $f(-2)=0$.
For part $(b)$, you want all functions $f in P_infty$ such that
$$ D(f) = (x+2)f' = 0. $$
Since this should hold for all $x$, we see that $f' = 0$, i.e. $f = text{const.}$ The dimension of this kernel is evidently equal to $1$.
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
add a comment |
For part $(a)$, you want a function $f in P_infty$ that satisfies
$$ D(f) = kf quad iff quad (x+2)f'= kf, $$
which is a simple separable differential equation.
If $x=-2$, we see from the above that we must have $f(-2) = 0$, and otherwise if $xnot=-2$, we have
$$ int frac{mathrm df}{f} = kint frac{mathrm dx}{x+2} quad iff quad log f = k log (x+2) + C quad iff quad f = (x+2)^k, $$
where in the last step we set $C=0$ due to the condition $f(-2)=0$.
For part $(b)$, you want all functions $f in P_infty$ such that
$$ D(f) = (x+2)f' = 0. $$
Since this should hold for all $x$, we see that $f' = 0$, i.e. $f = text{const.}$ The dimension of this kernel is evidently equal to $1$.
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
add a comment |
For part $(a)$, you want a function $f in P_infty$ that satisfies
$$ D(f) = kf quad iff quad (x+2)f'= kf, $$
which is a simple separable differential equation.
If $x=-2$, we see from the above that we must have $f(-2) = 0$, and otherwise if $xnot=-2$, we have
$$ int frac{mathrm df}{f} = kint frac{mathrm dx}{x+2} quad iff quad log f = k log (x+2) + C quad iff quad f = (x+2)^k, $$
where in the last step we set $C=0$ due to the condition $f(-2)=0$.
For part $(b)$, you want all functions $f in P_infty$ such that
$$ D(f) = (x+2)f' = 0. $$
Since this should hold for all $x$, we see that $f' = 0$, i.e. $f = text{const.}$ The dimension of this kernel is evidently equal to $1$.
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
For part $(a)$, you want a function $f in P_infty$ that satisfies
$$ D(f) = kf quad iff quad (x+2)f'= kf, $$
which is a simple separable differential equation.
If $x=-2$, we see from the above that we must have $f(-2) = 0$, and otherwise if $xnot=-2$, we have
$$ int frac{mathrm df}{f} = kint frac{mathrm dx}{x+2} quad iff quad log f = k log (x+2) + C quad iff quad f = (x+2)^k, $$
where in the last step we set $C=0$ due to the condition $f(-2)=0$.
For part $(b)$, you want all functions $f in P_infty$ such that
$$ D(f) = (x+2)f' = 0. $$
Since this should hold for all $x$, we see that $f' = 0$, i.e. $f = text{const.}$ The dimension of this kernel is evidently equal to $1$.
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
edited Nov 21 '18 at 17:32
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
answered Nov 21 '18 at 17:25
MisterRiemann
5,8291624
5,8291624
add a comment |
add a comment |
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