What can we say about the eigenvalues if ##L^2=I##?

[Karl Karlsson]

Homework Statement

Let V be a vectorspace with scalars in $\mathbb{R}$. Let L: V$\rightarrow$V be an operator.

If $L(L( \vec x)) = \vec x$ for all $\vec x \in$ V , what can we say about the eigenvalues of L? Give examples of V and L where different possibilities occur.

Relevant Equations

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This was a problem that came up in my linear algebra course so I assume the operation L is linear. Or maybe that could be derived from given information. I don't know how though. I don't quite understand how L could be represented by anything except a scalar multiplication if L: V$\rightarrow$V is to be satisfied.

Edit: Also isn't V just the same as $\mathbb{R}$ according to the definition of vector spaces if $\vec v$ is in V then $c\cdot\vec v$ must be in V also.

I can only come up with the specific examples of L and V:
If $L(\vec x) = \vec x$ then L has the eigenvalue 1

Thanks in advance!

 

[PeroK]

There is nothing special about having an operator with $L^2 = I$, as $L^2 = I$ satifies this in any vector space. There's no reason to conclude that $V = \mathbb R$.

Note that, $L^2 = I$ does not imply that $L$ has eigenvalue $1$. The question may want you to find an example of this.

 

[Karl Karlsson]

There is nothing special about having an operator with $L^2 = I$, as $L^2 = I$ satifies this in any vector space. There's no reason to conclude that $V = \mathbb R$.

Note that, $L^2 = I$ does not imply that $L$ has eigenvalue $1$. The question may want you to find an example of this.

Somebody else changed the title of this post to $L^2 = 1$ . No where in the problem is it stated that L is a linear operation but I assumed that it had to be. But okay if it is linear then $L^2=I$ but how could $x^2-1$ be a minimalpolynomial if the Function goes from the scalars to the scalars. L could be -1 is that what they were after?

 

[PeroK]

You misunderstand. This condition:

If $L(L( \vec x)) = \vec x$ for all $\vec x \in V$

Is equivalent to $L^2 = I$, where $I$ is the identity operator. Physicists sometimes write $1$ instead of $I$, hence the title.

 

[PeroK]

L could be -1 is that what they were after?

Yes, that's an example. Generally there will be other operators that satisfy the condition $L^2 = I$.

 

[Karl Karlsson]

Yes, that's an example. Generally there will be other operators that satisfy the condition $L^2 = I$.

When you mean "generally" you don't mean in this particular case where both the input and output of the function is V, the scalars ?

 

[PeroK]

When you mean "generally" you don't mean in this particular case where both the input and output of the function is V, the scalars ?

$V$ is any real vector space. That's what "scalars in $\mathbb R$" means. As opposed to a compolex vector space (with scalars in $\mathbb C$) or any other field of scalars.

 

[Karl Karlsson]

$V$ is any real vector space. That's what "scalars in $\mathbb R$" means. As opposed to a compolex vector space (with scalars in $\mathbb C$) or any other field of scalars.

Is there any reason why they would write V instead of $\mathbb R$ ? Also L must be represented by a 1x1 matrix that is either 1 or -1, right? Do we know from given information that L is linear? x and y are vectors in V
$L(L(cx+dy)) =cx+dy=c*L(L(x))+d*L(L(y))$ but this only shows that L(L(x)) is linear not that L is linear

 

[PeroK]

Is there any reason why they would write V instead of $\mathbb R$ ?

I have no idea why you would assume that a general vector space $V$ is only the real numbers. $V$ could be $\mathbb R^n$ or a space of polynomials or matrices or functions, or vectors of any description.

 

[Karl Karlsson]

I have no idea why you would assume that a general vector space $V$ is only the real numbers. $V$ could be $\mathbb R^n$ or a space of polynomials or matrices or functions, or vectors of any description.

Oh, maybe I misinterpreted the problem where it says "Let V be a vectorspace with scalars in $\mathbb R$". I see now, then it makes sense. L could be represented as the linear transformation for a reflection. But still does it have to be linear?

 

[PeroK]

Oh, maybe I misinterpreted the problem where it says "Let V be a vectorspace with scalars in $\mathbb R$". I see now, then it makes sense. L could be represented as the linear transformation for a reflection. But still does it have to be linear?

A reflection is a linear transformation.

If you allow $L$ to be non-linear that will give you even more examples. I would assume they meant Linear Operator, but it doesn't really matter if all they want are some examples.

 

[Delta2]

I think you can prove that if L is linear and satisfies L(L(x))=x then the eigenvalues are +1 and -1.

 

[Mark44]

Somebody else changed the title of this post to $L^2 = 1$ .

The title was very long before the change. I changed it from $L^2 = 1$ to $L^2 = I$ .