# Prove that ##\lambda## or ##-\lambda## is an eigenvalue for ##T##.

• Hall
In summary, the statement "If ##T: V \to V## has the property that ##T^2## has a non-negative eigenvalue ##\lambda^2##", means that there exists an ##x## in ##V## such that ## T^2 (x) = \lambda^2 x##. To prove that at least one of ##\lambda## or ##-\lambda## is an eigenvalue for ##T,## we can use the hint ##(T^2 - \lambda^2I)(x)=(T+\lambda )(T-\lambda )(x)= (T-\lambda )(T+\lambda )(x)=0## and consider the case ##T(x)\neq \pm \lambda x##
Hall
Homework Statement
If ##T: V \to V## has the property that ##T^2## has a non-negative eigenvalue ##\lambda^2## , prove that at least one
of ##\lambda## or ##-\lambda## is an eigenvalue for T. [Hint: ##T^2 - \lambda^2 = (T + \lambda I)(T - \lambda I)## .]
Relevant Equations
##T(x)= \lambda x##
The statement " If ##T: V \to V## has the property that ##T^2## has a non-negative eigenvalue ##\lambda^2##", means that there exists an ##x## in ##V## such that ## T^2 (x) = \lambda^2 x##.

If ##T(x) = \mu x##, we've have
$$T [T(x)]= T ( \mu x)$$
$$T^2 (x) = \mu^2 x$$
$$\lambda ^2 = \mu ^2 \implies \mu = \lambda ~or~ -\lambda$$.

Is my solution correct? The only thing which is quite not very well is that I assumed that there exists an eigenvalue ##\mu## and an eigenvector ##x## in ##T##.

What the hint in the question wants me to do?

PeroK
Hall said:
Homework Statement:: If ##T: V \to V## has the property that ##T^2## has a non-negative eigenvalue ##\lambda^2## , prove that at least one
of ##\lambda## or ##-\lambda## is an eigenvalue for T. [Hint: ##T^2 - \lambda^2 = (T + \lambda I)(T - \lambda I)## .]
Relevant Equations:: ##T(x)= \lambda x##

The statement " If ##T: V \to V## has the property that ##T^2## has a non-negative eigenvalue ##\lambda^2##", means that there exists an ##x## in ##V## such that ## T^2 (x) = \lambda^2 x##.

If ##T(x) = \mu x##, we've have
$$T [T(x)]= T ( \mu x)$$
$$T^2 (x) = \mu^2 x$$
$$\lambda ^2 = \mu ^2 \implies \mu = \lambda ~or~ -\lambda$$.

Is my solution correct?
No.
Hall said:
The only thing which is quite not very well is that I assumed that there exists an eigenvalue ##\mu## and an eigenvector ##x## in ##T##.
You assumed that the eigenvector ##x## to the eigenvalue ##\mu## is the same vector as the eigenvector to the eigenvalue ##\lambda^2## of ##T^2.## I do not see that this is necessarily true. And what if ##T## hasn't any real eigenvalues at all?

Hall said:
What the hint in the question wants me to do?
You started correctly: There is a vector ##x\neq 0## such that ##T^2(x)=\lambda^2 x.##

Now use the hint: ##(T^2-\lambda^2I)(x)=(T+\lambda )(T-\lambda )(x)= (T-\lambda )(T+\lambda )(x)=0##
and consider the case ##T(x)\neq \pm \lambda x## since otherwise we are done.

Can you find an eigenvector to ##\lambda ## or ##-\lambda ## anyway?

A good habit to avoid mistakes like this is by the notation: Always note dependences by indices or similar.

Example 1 (epsilontic):

For all ##\varepsilon >0## exists an ##N## such that ##|a_n-L|<\varepsilon ## for all ##n>N##.

should actually be

For all ##\varepsilon >0## exists an ##N(\varepsilon )## such that ##|a_n-L|<\varepsilon ## for all ##n>N(\varepsilon )##.

Example 2 (eigenvectors):

Assume ##x## is an eigenvector of ##T^2## to the eigenvalue ##\lambda^2,## i.e. ##T^2(x)=\lambda^2x.##

should actually be

Assume ##x_{\nu}## is an eigenvector of ##T^2## to the eigenvalue ##\nu:=\lambda^2,## i.e. ##T^2(x_{\nu})=\nu x_{\nu}=\lambda^2x_{\nu}.##

Delta2
I think a nice example (not quite right since the question is assuming real numbers) here is a rotation matrix in 2 dimension ##T(x,y)=(-y,x)##. ##T^2## is negative the identity so every vector is an eigenvector. But the eigenvector of ##T## is not obvious. It turns out ##-i## is an eigenvalue in ##\mathbb{C}^2##

Hmmm, I can't solve this unless it is given that ##ker(T)=\{0_V\}##.

Hall
fresh_42 said:
No.

You assumed that the eigenvector ##x## to the eigenvalue ##\mu## is the same vector as the eigenvector to the eigenvalue ##\lambda^2## of ##T^2.## I do not see that this is necessarily true. And what if ##T## hasn't any real eigenvalues at all?You started correctly: There is a vector ##x\neq 0## such that ##T^2(x)=\lambda^2 x.##

Now use the hint: ##(T^2-\lambda^2I)(x)=(T+\lambda )(T-\lambda )(x)= (T-\lambda )(T+\lambda )(x)=0##
and consider the case ##T(x)\neq \pm \lambda x## since otherwise we are done.

Can you find an eigenvector to ##\lambda ## or ##-\lambda ## anyway?
##
[T^2 - (\lambda I)^2] (x) = (T+ \lambda I) (T- \lambda I) (x) ##
##
T^2 (x) - \lambda^2 I(x) = (T + \lambda I) [T(x) - \lambda I(x)] ##
##
\lambda^2 x - \lambda ^2 x = (T +\lambda I) (T(x) - \lambda x)##
##
(T +\lambda I) (T(x) - \lambda x) = 0##
But without being given that ##(T + \lambda I)## is one-to-one, we cannot equate thus
##
T(x) - \lambda x = 0##
## T(x) = \lambda x##

(Changing the order in the very first equation to ##(T- \lambda I) (T+ \lambda I)## would give us ##T(x) = -\lambda x ##).

Delta2
Hall said:
##
[T^2 - (\lambda I)^2] (x) = (T+ \lambda I) (T- \lambda I) (x) ##
##
T^2 (x) - \lambda^2 I(x) = (T + \lambda I) [T(x) - \lambda I(x)] ##
##
\lambda^2 x - \lambda ^2 x = (T +\lambda I) (T(x) - \lambda x)##
##
(T +\lambda I) (T(x) - \lambda x) = 0##
But without being given that ##(T + \lambda I)## is one-to-one, we cannot equate thus
##
T(x) - \lambda x = 0##
## T(x) = \lambda x##

(Changing the order in the very first equation to ##(T- \lambda I) (T+ \lambda I)## would give us ##T(x) = -\lambda x ##).
Step by step. You are right, you cannot equate this.

We have ##(T-\lambda I)(T+\lambda I)(x)=0## and ##T(x)\neq -\lambda x.## (The other case is according.)
We need a vector ##v## such that ##Tv=\lambda v## and ##v\neq 0.##

Can you see what ##v## is? It is already written there!

Hall and Delta2
I think me and @Hall did same mistake, we expected the ##v## of post #7 to be equal to the eigenvector ##x## of ##T^2##, but ##v## can be different (can be the same though, it depends if ##T## is 1-1 it is the same ).

Hall and fresh_42
fresh_42 said:
Step by step. You are right, you cannot equate this.

We have ##(T-\lambda I)(T+\lambda I)(x)=0## and ##T(x)\neq -\lambda x.## (The other case is according.)
We need a vector ##v## such that ##Tv=\lambda v## and ##v\neq 0.##

Can you see what ##v## is? It is already written there!
##(T-\lambda I)(T+\lambda I)(x)=0##
$$T [ (T+ \lambda I) (x) ] - \lambda I [(T+ \lambda I)(x) ]= 0$$
##T [ (T+ \lambda I) (x) ] - \lambda [(T+ \lambda I)(x) ]= 0##
$$T [ (T+ \lambda I) (x) ] = \lambda [(T+ \lambda I)(x) ]$$

Eigenvector = ##(T+ \lambda I) (x)##, Eigenvalue= ##\lambda##.
(We must state that ##T(x) \neq -\lambda x##, else the eigenvector will be zero, which we don't want).
The second case:
$$T^2 - (\lambda I)^2 = (T + \lambda I)(T- \lambda I)$$
##
[T^2 - (\lambda I)^2](x) =[ (T +\lambda I)(T- \lambda I)] (x)##
##
0 = T [ (T- \lambda I) (x)] + \lambda I [(T- \lambda I) (x)]##
##T [ (T- \lambda I) (x)] = - \lambda [(T- \lambda I) (x)]##

Eigenvector = ## (T- \lambda I)(x)##, Eigenvalue = ##-\lambda##.
(with the condition that ## T(x) \neq \lambda x##)

fresh_42 and Delta2
Hall said:
##(T-\lambda I)(T+\lambda I)(x)=0##
$$T [ (T+ \lambda I) (x) ] - \lambda I [(T+ \lambda I)(x) ]= 0$$
##T [ (T+ \lambda I) (x) ] - \lambda [(T+ \lambda I)(x) ]= 0##
$$T [ (T+ \lambda I) (x) ] = \lambda [(T+ \lambda I)(x) ]$$

Eigenvector = ##(T+ \lambda I) (x)##, Eigenvalue= ##\lambda##.
(We must state that ##T(x) \neq -\lambda x##, else the eigenvector will be zero, which we don't want).
The second case:
$$T^2 - (\lambda I)^2 = (T + \lambda I)(T- \lambda I)$$
##
[T^2 - (\lambda I)^2](x) =[ (T +\lambda I)(T- \lambda I)] (x)##
##
0 = T [ (T- \lambda I) (x)] + \lambda I [(T- \lambda I) (x)]##
##T [ (T- \lambda I) (x)] = - \lambda [(T- \lambda I) (x)]##

Eigenvector = ## (T- \lambda I)(x)##, Eigenvalue = ##-\lambda##.
(with the condition that ## T(x) \neq \lambda x##)
Looks good. Now you only have to gather all possible cases: ##Tx=\lambda x\, , \,Tx=-\lambda x\, , \,Tx\neq \lambda x\, , \,Tx\neq -\lambda x## or reason with symmetry.

fresh_42 said:
Looks good. Now you only have to gather all possible cases: ##Tx=\lambda x\, , \,Tx=-\lambda x\, , \,Tx\neq \lambda x\, , \,Tx\neq -\lambda x## or reason with symmetry.
Okay, we can do that. But why do we need to look at only those cases? If we're not done in Post #9, then we can consider many cases for ##T(x)##.

Hall said:
Okay, we can do that. But why do we need to look at only those cases? If we're not done in Post #9, then we can consider many cases for ##T(x)##.
These are all possible cases. If ##Tx=\pm\lambda x## then we are done. And ##Tx\neq\pm\lambda x## is what you did in post #9. You mentioned yourself that we need the vector to be different from zero. So the case if it is zero is left. It is only an especially easy case.

Delta2 and Hall

## 1. What is an eigenvalue?

An eigenvalue is a value that, when multiplied by a corresponding eigenvector, results in the same vector with a scalar multiple. In other words, the eigenvector is only scaled, not changed in direction, when multiplied by its corresponding eigenvalue.

## 2. How do you prove that ##\lambda## or ##-\lambda## is an eigenvalue for ##T##?

In order to prove that ##\lambda## or ##-\lambda## is an eigenvalue for ##T##, you must first show that there exists a non-zero vector ##v## such that ##Tv = \lambda v## or ##Tv = -\lambda v##. This can be done by solving the characteristic equation for ##T##, which is ##det(T - \lambda I) = 0##. If ##\lambda## or ##-\lambda## is a solution to this equation, then it is an eigenvalue for ##T##.

## 3. What is the characteristic equation?

The characteristic equation for a linear transformation ##T## is given by ##det(T - \lambda I) = 0##, where ##I## is the identity matrix. This equation is used to find the eigenvalues of ##T##.

## 4. Can a linear transformation have more than one eigenvalue?

Yes, a linear transformation can have multiple eigenvalues. In fact, most linear transformations have multiple eigenvalues and corresponding eigenvectors.

## 5. How are eigenvalues and eigenvectors used in linear algebra?

Eigenvalues and eigenvectors are used to simplify and solve systems of linear equations and to understand the behavior of linear transformations. They are also used in many applications, such as in physics and engineering, to model and analyze systems.

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