Thread: orthogonal projection

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)|| $\leq$||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.

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)|| $\leq$||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 $\|Ty\|>\|y\|$.