Let T be an operator on the vector space V and let λ(adsbygoogle = window.adsbygoogle || []).push({}); _{1}, ..... , 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, 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}.

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

Thanks

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

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