Express in terms of $E$ a self-adjoint operator $T$ such that $T^2 = I+E$

Question

I was trying the following problem:

Let $V$ be a finite dimensional inner product space. Let $E: V \to V$ be an orthogonal projection onto some subspace of $V$. Express in terms of $E$ a self-adjoint operator $T$ such that $T^2 = I+E$

I could not make any attempt. Thanks in advance for help.

Answer 1

Suppose that $E$ is the orthogonal projection onto $X$, and let $Y$ be its kernel. For every $x\in X, (I+E)(x)=x+x=2x$. for every $y\in Y, (I+E)(y)=y$. Define $T(x)=\sqrt 2 x$ for every $x\in X$ and $T(y)=y$ for every $y\in Y$, $T^2=I+E$. Verify that $T$ is self adjoint.

Answer 2

Notice that an orthogonal projection $E$ satisfies $E^2=E$, and $I$ is the identity matrix. Hence, let $T:=\alpha E + I$, where $\alpha\in \mathbb{R}$. You have that: $$ T^2= \alpha^2 E^2 + 2 \alpha E + I= (\alpha^2 + 2\alpha) E + I.$$ Now you can compute $\alpha$. Lastly you can check that $T$ is self-adjoint since $E$ is selfadjoint and for $v,w\in V$ you have that $$\langle (\alpha E +I) v, w \rangle =\alpha\langle Ev, w \rangle + \langle v, w \rangle,$$ $$\langle v, (\alpha E +I) w \rangle =\alpha\langle v, Ew \rangle + \langle v, w \rangle.$$