All eigenvalues zero => zero map

[noospace]

I want to prove that if all the eigenvalues of a linear transformation T : V --> V are zero, then T = 0. I think this is obvious but I'm having difficulty putting it into words.

If all the eigenvalues of T are zero, then there exists a basis B for V in which $[T]_B$ is the zero matrix. Thus $[T(v)]_B=[T]^B_B[v]_B = 0\; \forall v \in V$.

I think I should show that the matrix representation of T in an arbitrary basis D is zero. We have from second year linear algebra $[id]^B_D [T]^B_B [id]^D_B = [T]^D_D$.
Thus

$[T(v)]_D = [T]^D_D [v]_D = [id]^B_D [T]^B_B [id]^D_B [v]_B = 0[v]_D = 0$.

But $v \in V$ is abritrary so $[T]^D_D [v]_D = 0 \; \forall [v]_D \in \mathbb{F}^n$ where $\mathbb{F}$ is the base field. Hence T = 0.

Is this correct?

 

[Hurkyl]

I want to prove that if all the eigenvalues of a linear transformation T : V --> V are zero, then T = 0.

This is not a theorem! There are at least two ways this can fail:

This matrix over the reals has only one eigenvalue, 0. (It has 3 over the complexes, of course)
$$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{array} \right)$$

This matrix over any field has only one eigenvalue, 0, of (algebraic) multiplicity 2.
$$\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$$

 

[Hurkyl]

By the way, you don't have to deal with that "arbitrary basis" stuff.

The zero linear transformation from V to W is the linear transformation such that, for every v in V,

Tv = 0.​

If you can find a basis so that the coordinate representation of T is the zero matrix, then it's easy to show that T is the zero transformation.

(Recall that two functions f and g are equal if and only if f(x) = g(x) for every input x)

 

[Hurkyl]

$[T]^D_D [v]_D = [id]^B_D [T]^B_B [id]^D_B [v]_B$

That doesn't look right: $[v]_D$ and $[v]_B$ are usually different.

Oh, but I think it's just a typo, because you switched back to $[v]_D$ in the next expression.


You could have done the basis calculation in a shorter way:
$$[T(v)]_D = [id]_D^B [T(v)]_B = [id]_D^B 0 = 0.$$
Other than your typo, I think you did the basis manipulations correctly.

 

[noospace]

Hi Hurkyl,

You're right, my theorem is a little bit too general.

Let T be an Hermitian linear operator.

Now I think it's okay.

And yes, that's a typo.

Btw, I don't see any other "easy" way to the proof based on what you said about equal functions.

 

[Hurkyl]

(caveat: this is only for finite-dimensional linear algebra. I'm not sure what generalizes properly to infinite dimensions)


That sounds fine; Hermitian matrices are diagonalizable over the reals, and I believe that is sufficient to prove your theorem. (BTW, you can do the proof by working directly with the diagonalized form!)

 

[Hurkyl]

Btw, I don't see any other "easy" way to the proof based on what you said about equal functions.

Starting from

For every v, $[T(v)]_B = 0 = [0]_B$​

it follows that

For every v, T(v) = 0​

and thus

T = 0.​

 

[noospace]

Yeah, but to get from assumption to therefore you need what I said right? Like l said it's obvious.

 

[Hurkyl]

Yeah, but to get from assumption to therefore you need what I said right? Like l said it's obvious.

It follows simply because the coordinate representation is faithful:
$$v = w \text{\ if and only if\ } [v]_B = [w]_B.$$
Of course, your way is correct too.

 

[Hurkyl]

(BTW, you can do the proof by working directly with the diagonalized form!)

*bonks self* I just realized this is exactly what you are doing when you said you could find a basis relative to which T's coordiante representation is the zero matrix!

 

[noospace]

Right, like saying $[\cdot]_B: V \to \mathbb{F}^n$ is injective. Thanks.