Solutions to Linear Algebra Problems

Let's now go to the solutions for the linear algebra problems in the earlier blog on basis, orthonormal basis. The problems and their solutions are as follows:

1) Analogous to Problem (1) in the examples, calculate the matrix if φ = π/6

We have:

F(E1) = cos(φ) (E1) + sin(φ) E2

F(E2) = -sin(φ)(E1) + cos(φ) E2


If φ = π/6:

F(E1) = cos(π/6) (E1) + sin(π/6) E2

F(E2) = -sin(π/6)(E1) + cos(π/6) E2

And we know: cos(π/6) = [3]^½/ 2 and sin(π/6) = ½,

therefore:

F(E1) = ([3]^½/ 2) E1 + ( ½)E2

F(E2) = (- ½)E1 + ([3]^½/ 2) E2

and the matric is:

([3]^½/ 2) ...... ½)

(- ½........([3]^½/ 2)


(2) Find an orthornormal basis for the subspace of R^4 generated by the following vectors:

A = (1, 1, 0, 0)

B = (1, -1, 1, 1)

C = (-1, 0, 2, 1)


For A we have:

A/ ‖A‖ = (1, 1, 0, 0)/ [1^2 + 1^2 ]^½ = (1, 1, 0, 0)/ [2 ]^½



For B we have:

B/ ‖B‖ = (1, -1, 1, 1)/ [1^2 + (-1)^2 +1^2 + 1^2]^½

= (1, -1, 1, 1)/ [4 ]^½ = (1, -1, 1, 1)/ 2


For C we have:

C/ ‖C‖ = (-1, 0, 2, 1)/ [(-1)^2 + (2)^2 +1^2 + 1^2]^½

so: C/ ‖C‖ = (-1, 0, 2, 1)/ [6 ]^½


(3) Find an orthornormal basis for the subspace of C^3 generated by:

A = (1, -1, -i) and B = (i, 1, 2)


For A we have:

A/ ‖A‖ = (1, -1, -i)/ [1^2 + (-1)^2 + (-i)]^½

But i is a complex number, i = [-1]^½ and therefore (-i)^2 = -[-1]^2 = 1

So:

A/ ‖A‖ = (1, -1, i)/ [1^2 + (-1)^2 + 1]^½ = (1, -1, i)/ [3]^½

For B we have:

B/ ‖B‖ = (i, 1, 2)/ [(i)^2 + 1^2 + 2^2]^½

= (i, 1, 2)/ [-1 + 1 + 4]^½ = (i, 1, 2)/ [4]^½ = (i, 1, 2)/ 2



4) Let V be a finite dimensional vector space over R, with positive definite product. Prove for any elements v, w in V:

‖u + v‖^2 + ‖u - v‖^2 = 2(‖u‖^2 + ‖v‖^2)

Hint: the solution uses bra-ket notation, but since this notation confuses the interface editor, we use [ ] brackets instead. (But readers should note to substitute the 'ket' > for ] and the 'bra' for the [ !

Then:

‖u + v‖^2 = [u + v, v + u] = [u,u] + [u,v] + [v,u] + [v,v]

But: [u,v] + [v,u] = [v,u]* + [v,u] less than or = to 2 ‖[ v,u]‖


‖u + v‖^2 = [u - v, v - u] = [u,u] - [u, v] - [v,u] + [v,v]

Whence:

‖u + v‖^2 + ‖u - v‖^2 = 2 [u,u] + 2[v,v] + 2[v,u] - 2[v,u]

= 2 {[u,u] + [v,v]}

but:

‖u‖^2 = [u,u] and ‖v‖^2 = [v,v]

hence:

‖u + v‖^2 + ‖u - v‖^2 = 2(‖u‖^2 + ‖v‖^2)