Just stuck on the last part of a problem, and the solution gives that:
If $[C(X)]^2$ = $X(X^T)-I$, X a 3x1 column vector, unit length, then I know that $XX^T$ is the orthogonal projection of vectors onto X.
But, why is $C(X)^4$ = $I-XX^T$? The solution doesn't explain, so I tried to work it out by squaring $[C(X)]^2$, but I'm not getting that answer.
Also, the solution interprets $I-XX^T$ as the orthogonal projection of vectors onto $X^\perp$. Is this sort of like the range of $XX^T$ contains all the vectors projected onto X, and then 1 - this range space is "everything else", namely the vectors in $X^\perp$? If my intuition is correct here, how do I make it precise? Or, perhaps it is just a definition and doesn't need any further clarification.
Thanks,
Here's one way to look at things: let $U$ be a matrix with columns $u_1 = X,u_2,u_3$ so that the columns of $U$ form an orthonormal basis of $\Bbb R^3$. Note that $UU^T = I$. We can write $$ X^TX= U\pmatrix{1&0&0\\0&0&0\\0&0&0}U^T $$ To show this, try the block matrix-multiplication $$ \pmatrix{u_1&u_2&u_3} \pmatrix{1&0&0\\0&0&0\\0&0&0}\pmatrix{u_1^T\\u_2^T\\u_3^T} $$
furthermore, $$ I = UIU^T $$ it follows that $$ I - X^TX = U\pmatrix{1&0&0\\0&1&0\\0&0&1}U^T - U\pmatrix{1&0&0\\0&0&0\\0&0&0}U^T\\ = U\left[\pmatrix{1&0&0\\0&1&0\\0&0&1} - \pmatrix{1&0&0\\0&0&0\\0&0&0}\right]U^T\\ = U\pmatrix{0&0&0\\0&1&0\\0&0&1}U^T \\ = u_2u_2^T + u_3u_3^T $$