Prove Eigenvalue λ=0 is Only Solution to Ax=0 for All x

[AsprngMathGuy]

Hey everyone,

I have a problem with over thinking things quite often, so I once again need help haha.
How would you go about proving this:

λ=0 is the only eigenvalue of A $\Rightarrow$ Ax=0 $\forall$x

Any help would be appreciated!
Thanks

 

[robertsj]

Are you sure it's true? The matrix
$$A = \left [\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right ]$$
has $\lambda = 0$ with multiplicity 3. Multiplying $A$ by the ones vector does not, however, yield the zero vector.

Perhaps you left out some constraints on $A$?

 

[chiro]

Hey everyone,

I have a problem with over thinking things quite often, so I once again need help haha.
How would you go about proving this:

λ=0 is the only eigenvalue of A $\Rightarrow$ Ax=0 $\forall$x

Any help would be appreciated!
Thanks

Hey AsprngMathGuy and welcome to the forums.

The result can be shown from the definition of the eigenvalue problem where you have:

$Ax = λx$ which implies $(Ax - Iλx) = 0$

If all your λ's are zero then the above reduces to $Ax - 0*Ix = 0$ which gives us $Ax=0$. This only uses the definition of the eigenvalue problem and you simply plug in the value for λ to get your equation.

 

[chiro]

Are you sure it's true? The matrix
$$A = \left [\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right ]$$
has $\lambda = 0$ with multiplicity 3. Multiplying $A$ by the ones vector does not, however, yield the zero vector.

Perhaps you left out some constraints on $A$?

That is not correct.

You have to change your matrix to this:

$$A = \left [\begin{array}{ccc} -λ & 1 & 1 \\ 0 & -λ & 1 \\ 0 & 0 & -λ \\ \end{array} \right ]$$ when you factor in the -λI term.

It still gives you zero (like you said above), but substituting in λ in your equation will give you Ax = 0.

You have to remember that you are solving an eigenvalue problem, and in doing this you want to usually assume your A is a non-singular matrix. In your example your A is a singular matrix and this will be useless.

The idea with an eigenvalue problem is that you want to find what x gets 'scaled' in a linearly dependent way from your matrix A, which will help you decompose it into the eigenvectors by analyzing where the scaling happens.

 

[HallsofIvy]

Perhaps Chiro is misunderstanding the question. A, given by robertsj, is, in fact, an example of matrix having only A as eigenvalue but such that Ax is not always 0.

IF there exist a basis for the vector space consisting of all eigenvectors (a "complete set of eigenvectors" so that A is diagonalizale) then Ax= 0 for all x. But in general, there may not exist such a set of eigenvectors. There will be some subspace such that Ax= 0 for all x in that subspace. In robersj's example, that would be the subspace of vectors <1, 0, 0>.

 

[chiro]

True HallsofIvy. Can't see the point for doing an eigendecomposition when you have no eigenvectors or can't find them.

 

[robertsj]

chiro: you're absolutely right if A were to be invertible, but that would be an unspecified constraint on A. As for assuming a nonsingular matrix, I work all the time with a method dealing with singular operators.

 

[chiro]

chiro: you're absolutely right if A were to be invertible, but that would be an unspecified constraint on A. As for assuming a nonsingular matrix, I work all the time with a method dealing with singular operators.

What kind of problems? Just curious.

 

[DonAntonio]

chiro: you're absolutely right if A were to be invertible, but that would be an unspecified constraint on A.


*** Not only "unspecified constraint" but also nonsensical: a (square, of course) matrix is singular iff zero is one of its eigenvalues, so

in your problem A must be singular and, in fact, nilpotent.

DonAntonio ***



As for assuming a nonsingular matrix, I work all the time with a method dealing with singular operators.

...

 

[AsprngMathGuy]

Thank you so much for your input. I actually misread the problem slightly. At the beginning it states to prove OR disprove the problem. So of course my mind goes right to attempting to prove it haha. Thanks!