Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:
[tex]Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y [/tex]
B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that [tex]||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2} [/tex]. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then [tex]||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||[/tex].
C) Consider [tex]S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2[/tex]. This set is orthogonal because [tex](0,2)\cdot(2,0)=0(2)+2(0)=0[/tex], but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in [tex]\mathbb{R}^n[/tex]. Then the columns of A form an orthonormal set. We have that [tex]A^{-1}=A^t[/tex]. To see this, note than the component [tex]b_{ij}[/tex] of the product [tex]A^t A[/tex] is the dot product of the i-th row of [tex]A^t[/tex] and the jth row of [tex]A[/tex]. But the i-th row of [tex]A^t[/tex] is equal to the i-th column of [tex]A[/tex]. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then [tex]A^t A=I[/tex]
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set [tex]\{u_1,u_2\cdots u_p\} [/tex] and suppose that there are coefficients a_i such that [tex]a_1u_1+a_2u_2\cdots a_nu_n=0[/tex]. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then [tex]a_i||u_i||=0[/tex] then [tex]a_i=0[/tex].
