Suppose that A is a real-value 2*2 matrix. For any two vector v,w Suppose that A

Question

Suppose that A is a real-value 2*2 matrix. For any two vector v,w

Suppose that A is a real - valued 2 times 2 matrix. For any two vectors v, w R2, define the "A - product" of v and w as follows: ( v, w)A = (A v) . w. On HW#3, you studied this object for a specific value of A. Show that for A = [2 1 1 2], this A - product is an inner product, as defined in class. In general, suppose A is a real - valued symmetric matrix. Is this always an inner product? If so, prove it; if not. create a counterexample, and come up with conditions on A that will insure that this A - product is an inner product.

Explanation / Answer

There is a theorem : "this A-product tis an inner product <=> A is definite positive", we will prove it in (b)

I'll write <x,y> instead of x.y for inner product, this is more readable here.

(a)

i) bilinearlity is evident, since inner product is bilinear

ii)

<w,v>_A=<Aw,v>

= (Aw)^Tv using definition of the standard inner product ( that is <x,y> = x^Ty )

= w^TA^Tv = w^TAv (A is symmetric)

=<w,Av> (definition)

=<Av,w>((Inner product is symmetric)

=<v,w>_A, so it's symmetric

iii)Let's write that using v = (x y)^T :

Av =

2x+y

x+2y

<Av,v> = (2x+y)x + (x+2y)y = 2x^2+2xy+2y^2 = 2 ( (x+1/2y)^2+3/4y^2 ) , so <v,v>_A >= 0

And if <v,v>_A =0

Then x+1/2y = 0 and y=0 => x=y=0

so v=0

Hence this is an inner product.

(b)

Take A =

1 0

0 0

Av =

x

0

Av.v = x^2

So <v,v>_A = 0 => v=(0,y), so we violate one property here.

The condition that is needed is that A is definite positive, when A is definite positive we have v^T A v > 0 and =0 only if v=0

And of course v^TAv = <v,Av> = <Av,v> = <v,v>_A

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.