Proving (B+C)^2=B^2+2BC+C^2 with Matrix Properties

[EV33]

Homework Statement

If B=C^(-1)

Is (B+C)^2=B^2+2BC+C^2

Homework Equations

If A and B are (mXn) matrices and C is an (nXp) matrix, then (A+B)C=AC+BC

If A is an mXn matrix and B and c are nXp matrices,then A(B+C)=AB+AC

The Attempt at a Solution

(B+C)(B+C)=B^2+2BC+C^2

Then I decided to substitute B+C for D, but only one of them.

D(B+C)=B^2+2BC+C^2

DB+DC=B^2+2BC+C^2

Then I substituted the b+c back in for D

(B+C)B+(B+C)C=B^2+2BC+C^2

Then I get

BB+CB+BC+CC=B^2+2BC+C^2

Then from here I plugged in the C^-1 in for the B's in the two middle terms, which gives c time c inverse, plust c inverse times C, and they both are equal to 1, and 1 plus 1 is equal to two so I get

B^2+2+C^2=B^2+2BC+C^2

And then on the right hand side you can plug the c inverse in for b and you get 2 times c inverse times C which is just equal to 2.


B^2+2+C^2=B^2+2+C^2

which finall gets me back to

(B+C)^2=B^2+2BC+C^2

B^2+2+C^2=B^2+2+C^2

Did I screw up anywhere? The thing I am mainly not sure of is if I can make the substution like that.

 

[HallsofIvy]

Homework Statement

If B=C^(-1)

Is (B+C)^2=B^2+2BC+C^2

Homework Equations

If A and B are (mXn) matrices and C is an (nXp) matrix, then (A+B)C=AC+BC

If A is an mXn matrix and B and c are nXp matrices,then A(B+C)=AB+AC

The Attempt at a Solution

(B+C)(B+C)=B^2+2BC+C^2

Then I decided to substitute B+C for D, but only one of them.

D(B+C)=B^2+2BC+C^2

DB+DC=B^2+2BC+C^2

Then I substituted the b+c back in for D

(B+C)B+(B+C)C=B^2+2BC+C^2

Then I get

BB+CB+BC+CC=B^2+2BC+C^2

Then from here I plugged in the C^-1 in for the B's in the two middle terms, which gives c time c inverse, plust c inverse times C, and they both are equal to 1, and 1 plus 1 is equal to two so I get

B^2+2+C^2=B^2+2BC+C^2

And then on the right hand side you can plug the c inverse in for b and you get 2 times c inverse times C which is just equal to 2.


B^2+2+C^2=B^2+2+C^2

which finall gets me back to

(B+C)^2=B^2+2BC+C^2

B^2+2+C^2=B^2+2+C^2

Did I screw up anywhere? The thing I am mainly not sure of is if I can make the substution like that.

You can make any substitution you like! And your proof is correct though more complicated than needed.

For any B and C, (B+ C)^2= B^2+ BC+ CB+ C^2. Since we are given that B= C^{-1}, both BC and CB are equal to I. That is (B+ C)^2= B^2+ 2BC+ C^2 because both are equal to B^2+ 2I+ C^2.