In Exercise 29(c) of Section 6.5, we are to use the 3 x 3 matrix whose columns are the vectors w_1, w_2, w_3, respectively, in Example 4 of Section 6.2. This example is located in page 345, and the vectors are
The 3x3 matrix is A = ( w_1 w_2 w_3 ) where w_i are columns of A. So, based on what you know from part b) construct Q and R using the information about A.
Bernard A. Doria wrote: > In Exercise 29(c) of Section 6.5, we are to use the 3 x 3 matrix whose > columns are the vectors w_1, w_2, w_3, respectively, in Example 4 of > Section 6.2. This example is located in page 345, and the vectors are
Christopher Reeder wrote: > I know we talked about this in OH today. But, why is it again that > the adjoint of the inverse equals the inverse of the adjoint?
> thanks, Chris
Let V be a finite-dimensional inner product space, and let T be a linear operator on V. Suppose T is invertible. Then we have (T)(T^-1) = (T^-1)(T) = I. Taking the adjoint of each expression in this equation, we apply Theorem 6.11 (p. 360) to obtain [(T^-1)^*][T^*] = [T^*][(T^-1)^*] = I. But this means that there exists a function that is the inverse of T^*, namely, (T^*)^-1 = (T^-1)^*.
> Christopher Reeder wrote:
>> I know we talked about this in OH today. But, why is it again that
>> the adjoint of the inverse equals the inverse of the adjoint?
>> thanks, Chris
> Let V be a finite-dimensional inner product space, and let T be a
> linear operator on V. Suppose T is invertible. Then we have (T)(T^-1) =
> (T^-1)(T) = I. Taking the adjoint of each expression in this equation,
> we apply Theorem 6.11 (p. 360) to obtain [(T^-1)^*][T^*] =
> [T^*][(T^-1)^*] = I. But this means that there exists a function that
> is the inverse of T^*, namely, (T^*)^-1 = (T^-1)^*.
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