EGR 6013 Homework Set 1 (12-18)

— David Wagner 2009/09/16 05:40

Problem 12.

Show ax+by+c=0 through (x1,y1) (x2,72) is equivalent to $delim{|}{ matrix{3}{3}{ x y 1 x_1 y_1 1 x_2 y_2 1 ~} }{~|}$ =0

= x y_1 +y x_2 +x_1 y_2 -x_2 y_1 - y x_1 -x y_2

= (y_1-y_2)x + (x_2-x_1)y + (x_1 y_2 - x_2 y_1) = 0.

Let .
ax+by+c=0

Problem 14.

If [A][B] = [B][A], prove [A]^T[B]^T = [B]^T[A]^T.

and .
Finally, one I saw right away!

How can I show this? The transpose of [A] interchanges the rows and columns, that is the element in row i, column j in [A] is the same as the element in row j, column i in [A]^T. Thus a_ij = a_ji for symmetric [A].

Let [A] = [a_ij], so [A]^T = [a_ji] by the definition of transpose.
If [A] = [A]^T, then [A] = [a_ij] = [A]^T = [a_ji].
If [a_ij] = [a_ji], then [A] = [a_ij] = [a_ji] = [A]^T.

Problem 15b.

Given symmetric matrices [A] and [B], prove [C] = [A][B] is symmetric iff [A][B] = [B][A].

Let [a_ij] = [A], [b_ij] = B, and [c_ij] = [C].

[a_ij] = [a_ji] and [b_ij] = [b_ji].

[c_ij] = [C] = [A][B] = $sum{r}{}{a_ir b_rj}$ , and [B][A] = $sum{r}{}{b_jr a_ri}$ .

If [C] is symmetric, then [c_ij] =[c_ji], and

[A][B] = [C] = [c_ij] = $sum{r}{}{a_ir b_rj}$ = [c_ji] = $sum{r}{}{a_ri b_jr}$ = $sum{r}{}{b_jr a_ri}$ = [B][A].

Problem 16.

Problem 17(a).

Let [A] and [B] represent diagonal matrices of order n.

Prove that [A][B] is also a diagonal matirx.

Let [A] = [a_ir] and [B]=[b_rs] .

By definition, for i ≠ r.
When i=r,

$[a_ir][b_rs]=sum{r}{}{ a_ir b_rs}$ .

The only terms of the sum that are nonzero occur when i=r and r=s.

$[a_ir][b_rs]=sum{r}{}{ a_rr b_rr}$ .

Thus, the only nonzero terms of the matrix product are those on the diagonal.

Problem 17(b).

Prove that [B][A] = [A][B].

$[a_ir][b_rs]=sum{r}{}{ a_ir b_rs}$ , $[b_sr][a_ri] = sum{r}{}{ b_sr a_ri }$ .

Diagonal matrices are symmetric: a_ir = a_ri and b_rs = b_sr . (In fact, all but the diagonal elements when i=r or s=r are zero, and the diagonals have equal row and column indices pointing to the same entry when reversed.)