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Math Help - orthogonal projection

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    orthogonal projection

    Let T be a linear operator on a finite-dimensional inner product space V.
    1. If T is an orthogonal projection, prove that ||T(x)|| \leq||x|| for all x \in V. Give an example of a projection for which this inequality does not hold. What can be concluded about a projection for which the inequality is actually an equality for all x \in V?

    2. Suppose that T is a projection such that ||T(x)|| \leq ||x|| for x \in V. Prove that T is an orthogonal projection.
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    Quote Originally Posted by studentmath92 View Post
    Let T be a linear operator on a finite-dimensional inner product space V.
    1. If T is an orthogonal projection, prove that ||T(x)|| \leq||x|| for all x \in V. Give an example of a projection for which this inequality does not hold. What can be concluded about a projection for which the inequality is actually an equality for all x \in V?

    2. Suppose that T is a projection such that ||T(x)|| \leq ||x|| for x \in V. Prove that T is an orthogonal projection.
    For a (non-orthogonal) projection T and a vector x with \|Tx\| >\|x\|, you could take T = \begin{bmatrix}1&1\\0&0\end{bmatrix} and x = \begin{bmatrix}1\\1\end{bmatrix} in the space V=\mathbb{R}^2.

    For problem 2, if T is not orthogonal then the kernel of T is not orthogonal to the image of T. So there is a vector y\in (\ker(T))^\perp with Ty\ne y. Then Ty = y + (Ty-y). But Ty-y\in\ker(T) (because T is a projection). By Pythagoras' theorem, \|Ty\|^2 = \|y\|^2 + \|Ty-y\|^2, and so \|Ty\|>\|y\|.
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