1. ## 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)|| $\displaystyle \leq$||x|| for all x$\displaystyle \in$ 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 $\displaystyle \in$ V?

2. Suppose that T is a projection such that ||T(x)|| $\displaystyle \leq$||x|| for x $\displaystyle \in$ V. Prove that T is an orthogonal projection.

2. Originally Posted by studentmath92
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)|| $\displaystyle \leq$||x|| for all x$\displaystyle \in$ 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 $\displaystyle \in$ V?

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

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 $\displaystyle y\in (\ker(T))^\perp$ with $\displaystyle Ty\ne y$. Then $\displaystyle Ty = y + (Ty-y)$. But $\displaystyle Ty-y\in\ker(T)$ (because T is a projection). By Pythagoras' theorem, $\displaystyle \|Ty\|^2 = \|y\|^2 + \|Ty-y\|^2$, and so $\displaystyle \|Ty\|>\|y\|$.