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

[vish_maths]

Let T be an operator on the vector space V and let λ₁, ... , λ_n 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, let's 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 .

Can you please advise why this paradox exists ? Am i making a mistake somewhere ?
Thanks

 

[AlephZero]

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

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

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).

 

[Fredrik]

I don't understand the bra-ket notation,

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

< 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.

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.

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) = 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 ? )

 

[Fredrik]

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

 

[vish_maths]

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).

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

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.

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

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 .

 

[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 .

 

[Fredrik]

<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 .

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.

 

[micromass]

<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 .

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

 

[AlephZero]

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

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!

 

[vish_maths]

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.

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

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!

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