Linear Algebra, Inner Product of Matrices

[rbpl]

Let M_2x2 denote the space of 2x2 matrices with real coeffcients. Show that

(a1 b1) . (a2 b2)
(c1 d1) (c2 d2)

= a1a2 + 2b1b2 + c1c2 + 2d1d2

defines an inner product on M_2x2. Find an orthogonal basis of the subspace

S = (a b) such that a + 3b - c = 0
(c d)

of M_2x2 defined by with respect to this inner product.


I know how to find the orthogonal basis, so I don't think I need any help with that, I'm only having trouble with the first part--showing that the two matrices define inner product. I have no idea where the 2's in 2b1b2 and 2d1d2 are coming from. I thought that it should be just a1a2 + b1b2 + c1c2 + d1d2, without the 2's.

 

[Dick]

There is more than one inner product you can define on a space. They just put the '2's there and want you to show this defines another inner product. Just look up the properties that an inner product has to have. The only one that requires some thought is to prove M.M>=0.

 

[rbpl]

let u=(a1 b1), v=(a2 b2), w=(a3 b3)
(c1 d1) (c2 d2) (c3 d3)

(a1 b1) . (a2 b2) = a1a2 + 2b1b2 + c1c2 + 2d1d2
(c1 d1) (c2 d2)

There are four properties that:

1. <u,v>=<v,u>

(a1 b1) . (a2 b2) = a1a2 + 2b1b2 + c1c2 + 2d1d2
(c1 d1) (c2 d2)

(a2 b2) . (a1 b1) = a2a1 + 2b2b1 + c2c1 + 2d2d1
(c2 d2) (c1 d1)

2. <u+v,w>=<u,w>+<v,w>

<u+v,w>=
(a1+a2 b1+b2) . (a3 b3)
(c1+c2 d1+d2) (c3 d3)
= (a1+a2)(a3) + 2(b1+b2)(b3) + (c1+c2)(c3) + 2(d1+d2)(d3)=a1a3+a2a3 + 2b1b3+2b2b3 +
c1c3 + c2c3 + 2d1d3+2d2d3

<u,w>+<v,w>
((a2 b2) . (a3 b3)) + ((a2 b2) . (a3 b3))
((c2 d2) (c3 d3)) ((c2 d2) (c3 d3))
=(a1a3 + 2b1b3 + c1c3 + 2d1d3)+(a1a3) + (a2a3 + 2b2b3 + c2c3 + 2d2d3)=a1a3 + a2a3 + 2b1b3+2b2b3 + c1c3 + c2c3 + 2d1d3+2d2d3

3. <ru,v>=r<u,v>

<ru,v>=
(ra1 rb1) . (a2 b2) = ra1a2 + r2b1b2 + rc1c2 + r2d1d2
(rc1 rd1) (c2 d2)

r<u,v>=
r (a1 rb1) . (a2 b2) = r(a1a2 + 2b1b2 + c1c2 + 2d1d2)=ra1a2 + r2b1b2 + rc1c2 + r2d1d2
(c1 rd1) (c2 d2)

4. <u,u> >= 0 and <u,u>=0 iff u=0

a) <u,u>=
(a1 b1) . (a1 b1) = a1^2 + 4b1^2 + c1^2 + 4d1^2
(c1 d1) (c1 d1)

now, whether a1, b1, c1, d1 are negative or positive does not matter since squaring makes them possitive, thus <u,u> >= 0

b) <u,u>=0 then u must be 0, since the only way <u,u>=0 when we have
(0 0) zero matrix and a1=b1=c1=d1=0
(0 0)

c) u=0 implies <u,u>=0
since (0 0) . (0 0) = (0 0)
(0 0) (0 0) (0 0)

For the terms that I highlated in red: 2as I right to put a four there? Or since we are just squaring a1, b1, c1, d1 there is no 4 or 2 in front of b1^2 d1^2.

 

[rbpl]

Since S=
(a b) such that a + 3b - c = 0
(c d)

We have that S=
(-3b-c b)
(c d)

Thus the basis for the Gram-Schmidt Orthogonalization is:
(-3 1),(-1 0),(0 0)
(0 0) (1 0) (1 0)

This gives us:

w1=
(-3 1)
(0 0)

w2=
(-1/10 -3/10)
(1 0)

w3=
(1/2 0)
(1/2 0)

I am not sure if it did w2 right, in particular, I am not sure if there are two terms that I have to multiply by 2 because of the earlier shown inner product. Here is how I got the w2 above:

(1 0) -
(-1 0)

(-3 1)(-1 0)
(0 0)(1 0) . (-3 1) =
------------- (0 0)
(-3 1)(-3 1)
(0 0)(0 0)

(-1 0) - (3/10) . (-3 1) =
(1 0) (0 0)

(-1/10 -3/10)
(1 0)

 

[Dick]

Well, no. For part 4a) you should have <u,u>=a1^2+2*b1^2+2*c1^2+d1^2. That's the definition of the inner product isn't it? And, for example, if you take u=[[-1,3],[0,0]], then <u,u>=(-1)^2+2*3^2=19, not 10. You have to keep the two's in there when you are computing this inner product.

 

[rbpl]

Yes, I understand it makes sense now.

So, I guess that in my w2 in the post where I used Gram-Schmidt, I schould have done this:

(1 0) -
(-1 0)

(-3 1)(-1 0)
(0 0) (1 0) . (-3 1) =
------------- (0 0)
(-3 1)(-3 1)
(0 0) (0 0)

(-1 0) - (3/11) . (-3 1) =
(1 0) (0 0)

(-2/11 -3/11)
(1 0)

rather than having (-3/10) in the place of red text.

 

[Dick]

Yes, I understand it makes sense now.

So, I guess that in my w2 in the post where I used Gram-Schmidt, I schould have done this:

(1 0) -
(-1 0)

(-3 1)(-1 0)
(0 0) (1 0) . (-3 1) =
------------- (0 0)
(-3 1)(-3 1)
(0 0) (0 0)

(-1 0) - (3/11) . (-3 1) =
(1 0) (0 0)

(-2/11 -3/11)
(1 0)

rather than having (-3/10) in the place of red text.

Right.

 

[rbpl]

Thank you for your help.