Question

Q8. (a) Let y = [4, 8) and u = [3,1]. Write y as the sum of a vector in span{u} and a vector orthogonal to u. (b) Show that if U and V are n x n orthogonal matrices, then so is UV.

Answer 1

(a)
Observe that the underlying vector space of the discussion is $\mathbb{R}$ ².
Considering the vector u = [3,1], observe that the orthogonal complement of span{u} is
$span(u)^\perp$ = { v $\in$   $\mathbb{R}$² : v.u = 0 } = { [a,b] $\in$   $\mathbb{R}$² : 3a + b = 0 }.
In other words, the orthogonal complement of span{u} is the one-dimensional subspace spanned by the vector
v = [-1,3].

Since  $\mathbb{R}$² has dimension 2, and {u,v} is an orthogonal set, hence {u,v} is a basis of  $\mathbb{R}$². Consequently, y CAN be written as a linear combination of u and v. Observe that, y = [4,8] = 2.[3,1] + 2.[-1,3] = 2u + 2v

Since span{u} and span{v} are subspaces, hence 2u $\in$ span{u} and 2v $\in$ span{v}.
Thus, y = [6,2] + [-2,6] , where [6,2] $\in$ span{u} and [-2,6] is a vector orthogonal to u, is the required expression.



(b)
Suppose that U and V are n x n orthogonal matrices.
Then U.U^T = U^T.U = I = V^T.V = V.V^T where I is the n x n identity matrix.

Now, (UV)(UV)^T = UVV^TU^T = UIU^T = UU^T = I = V^TIV = V^TU^TUV = (UV)^T(UV).
This shows that UV is an n x n orthogonal matrix.

This proves what was asked for.

Infact note that, if a matrix U is orthogonal, then UU^T = I = U^TU implies that U^-1 = U^T, and therefore, U^-1 is orthogonal as well. This shows that the set of n x n orthogonal matrices is a subgroup of the General Linear Group GL_n of all n x n invertible matrices.