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We are asked to find the polar decomposition of the operator $T \in \mathcal{L}(V)$ with matrix: $$\mathcal{M}(T)=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$

I know that the polar decomposition is of the form: $$T=S\sqrt{T^*T},$$ where $S$ is an isometry and $\sqrt{T^*T}$ is a positive operator.

I also know that: $$\mathcal{M}(T^*T)=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.$$

The eigenvalues of this operator are: $\lambda=\dfrac{3\pm\sqrt{5}}{2}.$

I am not sure how to calculate $\sqrt{T^*T}$ or $S$. Any help would be appreciated!

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Write $T^*T=PDP^t$, $P$ is given with the eigen vectors, D is diagonal with eigen values

then $\sqrt{TT^*}=P\sqrt{D}P^t$ where $\sqrt{D}$ is diagonal with square roots of the eigenvalues

$S=TP\sqrt(D)^{-1}P^t$

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  • $\begingroup$ This doesn't work. $\endgroup$ – Atsina Nov 30 '18 at 21:27

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