Question

(1 point) What is the matrix P-(P) for the projection of a vector b є R3 onto the subspace spanned by the vector a- ? 5 9 Pl 3 1 2 P21 23 - P32 31 What is the projection p of the vector b0onto this subspace? 9 Pl Check your answer for p against the formula for p on page 208 in Strang.

Answer 1

Given, a =
559
and b = $\begin{bmatrix} -7\\ 0\\ -9 \end{bmatrix}$

Now, the required matrix is :

al a

Here,

-5 9

i.e.,

25 25 -45 aa'=125 25-45 -45-45 81

and,

-5 aTa=[-5-5 9] 9

i.e.,

a a 2525 +81

i.e.,

a' a 131

Then,

25/131 25/131-45/131 25/13125/131 Р-1 -45/131 -45/131 81/131 -45/131

Therefore,
25 P11 = 131
, $P_{12}=\frac{25}{131}$ ,
ー45 Pa-131
, $P_{21}=\frac{25}{131}$ , $P_{22}=\frac{25}{131}$ ,
ー45 P2s131
, $P_{31}=\frac{-45}{131}$ , $P_{32}=\frac{-45}{131}$ , $P_{33}=\frac{81}{131}$ .

Again, $p=\frac{b^Ta}{a^Ta}a$

Here,

-5 9

i.e., $b^Ta=35+0-81$

i.e., $b^Ta=-46$

Then, $p=\frac{-46}{131}\begin{bmatrix} -5\\ -5\\ 9 \end{bmatrix}$

i.e., $p=\begin{bmatrix} 5*46/131\\ 5*46/131\\ -9*46/131 \end{bmatrix}$

i.e., $p=\begin{bmatrix} 230/131\\ 230/131\\ -414/131 \end{bmatrix}$

Therefore, $p_{1}=\frac{230}{131}$ , $p_{2}=\frac{230}{131}$ , .

-414 73 131