# Why this paradox in calculating eigen values for T*T ?

1. Apr 2, 2013

### vish_maths

Let T be an operator on the vector space V and let λ1, ..... , λn be it's eigen values including multiplicity .

Lets find the eigen values for the operator T*T then ( where T* refers to the adjoint operator . <u,v> denotes inner product of u and v )

< Tv , Tv > = < λv, λv >
= λλ° < v , v >
= |λ|2 < v , v>

( where λ° is the conjugate of λ . )

=> <T*T v, v> = < |λ|2 v, v >

=> T*T has an eigen value |λ|2 for the same eigen vector v which T possesses.

This can be inferred because :

Suppose f is a linear functional on V . Then there is a unique vector v in V such that f(u) =<u,v> for all u in V.

( A linear functional on V is a linear map from V to the scalars F , As sheldon axler pg 117 says it ) The above argument must imply then that : T*T v = |λ|2 v ( Doesn't this imply v is an eigen vector of T*T ? )

now, lets consider a matrix $$M(T) = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$$

[ 3 1 ]T is clearly an eigen vector with eigen value = 2 .

(T* T ) =\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix} multiplied by \begin{bmatrix}
1 & 3 \\
0 & 2
\end{bmatrix}$$= \begin{bmatrix} 1 & 3 \\ 3 & 13 \end{bmatrix}$$

M(T* T ) upon multiplication with [ 3 1 ]T should produce a vector equal to

4 [ 3 1 ]T

however , \begin{bmatrix}
1 & 3 \\
3 & 13
\end{bmatrix} multiplied by [ 3 1 ]T = [ 6 22 ]T .

Thanks

Last edited: Apr 2, 2013
2. Apr 2, 2013

### AlephZero

I don't understand the bra-ket notation, but I think

is wrong.

The eigenvectors of a matrix and its (conjugate) transpose are different in general. Your proof seems to assume they are the same. If $Tx = \lambda x$, then $x^*T^* = \lambda^* x^*$, but that doesn't make $x^*$ a (right) eigenvector of $T^*$. (It is a left eigenvector of $T^*$, of course).

3. Apr 2, 2013

### Fredrik

Staff Emeritus
There's no bra-ket notation there, just inner products, absolute values, and the definition of the adjoint operator.

Why would $\langle T^*Tv,v\rangle=\langle|\lambda|^2v,v\rangle$ imply that $T^*Tv=|\lambda|^2 v$?

I see that you tried to prove this, but the argument doesn't seem to make sense. There appears to be something missing from it as well. f(u)= what? Did you mean f(u)=v for all u? That actually implies that v=0.

4. Apr 2, 2013

### Fredrik

Staff Emeritus
Note that if we let w be any vector that's orthogonal to v, we have $\langle T^*Tv,v\rangle =\langle T^*Tv+w,v\rangle$.

5. Apr 2, 2013

### vish_maths

Thanks. <u,v> as fredrik pointed refers to the inner product of u and v . I get your argument regarding the adjoint operation.

Hi Fredrik,
Thanks for the answer. i have corrected it in the main thread , f(u) = <u,v> for all u in V.

I think the equation < Tx , y > = < z, y > => Tx = z if and only if < Tx , y > = < z, y > is valid for all y in V. ( w$\in$ V ) .

Lets consider the more general case of < w , y > = < z , y > , then w = z if and only if this relation is valid for all y $\in$ V.

Proof : Suppose there exists an another different quantity w≠z | < w , y > = < z , y >

Then : < w - z, y > = 0 for all y $\in$ V .
=> when y = w - z , then :

<w-z,w-z> = 0 => w = z .

6. Apr 2, 2013

### vish_maths

<T*T v, v> = < |λ|2v, v > -------- (1)

=> I guess , T*T has an eigen value |λ|2 for the same eigen vector v which T possesses , if and only if (1) is valid for ALL v $\in$ V .

7. Apr 2, 2013

### Fredrik

Staff Emeritus
If it holds for all v, I think the correct conclusion is that T*T is equal to $|\lambda|^2I$ where I is the identity operator. But more importantly, as you know, you have only proved that the equality holds for a specific v.

8. Apr 2, 2013

### micromass

Staff Emeritus
It is clearly not true for all $v\in V$ since you took a very special $v$, namely an eigenvector.

9. Apr 2, 2013

### AlephZero

Fair comment, but (as an ex-mathematician turned engineer) I never use the < > notation for inner products and have to think at least twice what "adjoint operator" means

But in any case the problem here is with the math, not with the notation!

10. Apr 2, 2013

### vish_maths

Got it. Didn't keep into consideration all v in V have to satisfy the condition . Thanks !

Last edited: Apr 2, 2013