Let $f_1,f_2,...,f_k$ be pairwise orthogonal nonzero vectors of $R^n$

Question

Let $f_1,f_2,...,f_k$ be pairwise orthogonal nonzero vectors of $R^n$. Show that $ P = \frac{f_1f_1^T}{f_1^Tf_1} + \frac{f_2f_2^T}{f_2^Tf_2} + ... + \frac{f_kf_k^T}{f_k^Tf_k}$ is a projection matrix.

So, my textbook shows me that the definition of a projection matrix is $ P = PP$ but how do I use that to complete this proof?

Answer 1

Just to simplify notation a bit, let $g_i = f_i / \lVert f_i \rVert$, so that the expression becomes

$$P = \sum_i g_ig_i^T.$$

Now, choose an orthonormal basis for $\mathbb{R^n}$ of the form $(g_1 \dots, g_k, e_{k+1}, \dots, e_n)$ for some suitable $e_{k+1}, \dots, e_n$. We are done, if we can show that $Pv = PPv$ for each of these basis vectors. If $v = e_j$ for some $j$, then

$$Pv = Pe_j = \sum_i g_ig_i^Te_i = \sum_i g_i 0 = 0,$$

and therefore also $PPv = P0 = 0$. If on the other hand $v = g_j$ for some $j$, then

$$Pv = Pg_j = \sum_i g_ig_i^Tg_j = g_jg_j^Tg_j = g_j,$$

and $PPg_j = Pg_j$.

Answer 2

We have $$ P^2 = \left(\sum_{i=1}^k \frac{f_if_i^T}{f_i^Tf_i} \right)\left(\sum_{j=1}^k \frac{f_jf_j^T}{f_j^Tf_j} \right) = \sum_{i=1}^k \sum_{j=1}^k \frac{1}{(f_i^Tf_i)(f_j^Tf_j)} f_i(f_i^Tf_j)f_j^T $$ Now, note that $f_i^Tf_j = 0$ whenever $i \neq j$, which means we can rewrite the above as $$ P^2 = \sum_{i=1}^k \frac{1}{(f_i^Tf_i)^2} f_i(f_i^Tf_i)f_j^T $$ now, show that this is the same as $P$.

Answer 3

You have the matrix

$$P:=\sum_{m=1}^k \frac1{\left\|f_m\right\|^2}f_mf_m^t\implies PP=\sum_{m,n}^k\frac1{\left\|f_m\right\|^2\left\|f_n\right\|^2}f_mf_m^tf_nf_n^t$$

But $\;f_m^tf_n=0\;$ for all $\;m\neq n\;$ , so you're left in the above doulble sum only with the summands where $\;m=n\;$ :

$$\sum_{n=1}^k\frac1{\left\|f_n\right\|^2\left\|f_n\right\|^2}f_nf_n^tf_nf_n^t=\require{cancel}\sum_{n=1}^k\frac1{\left\|f_n\right\|^2\cancel{\left\|f_n\right\|^2}}\cancel{\left\|f_n\right\|^2}f_nf_n^t=P$$