Real inner product spaces and self adjoint linear transformations

[Kate2010]

Homework Statement

Let V be a real inner product space of dimension n and let Q be a linear
transformation from V to V .
Suppose that Q is non-singular and self-adjoint. Show that Q⁻¹ is self-adjoint.
Suppose, furthermore, that Q is positive-definite (that is, <Qv,v> > 0 for all non-zero v
in V ). Show that the eigenvalues of Q are positive. Deduce that there exists a positive
self-adjoint linear transformation S from V to V such that S² = Q.
Now let P be a self-adjoint linear transformation from V to V . Show that S⁻¹PS⁻¹ is self-adjoint. Deduce, or prove otherwise, that there exist scalars f1, . . . , fn and linearly independent vectors e1, . . . , en in V such that, for i, j = 1, 2, . . . , n:

*This is the bit I can't do*
(i) Pei = fiQei;
(ii) <Pei, ej> = dij fi;
(iii) <Qei, ej> = dij .

Homework Equations

The Attempt at a Solution

So I'm coming back to maths after a few weeks break for Christmas. I have managed most of this question, it is just the last section that I'm unsure how to attempt. I assume I use previous parts of the question, but I'd be really grateful if someone could give me a push in the right direction as to how.

 

[mighty2000]

Thank you for your post. I am happy to assist you with this problem. From the given information, we know that Q is a non-singular and self-adjoint linear transformation in a real inner product space V of dimension n. This means that Q has an inverse, which we will denote as Q^-1, and that Q is equal to its own adjoint, or Q*=Q.

To show that Q^-1 is also self-adjoint, we can use the definition of the adjoint operator. Let v and w be two vectors in V. Then, we have:

<v, Q^-1w> = <Qv, w> (since Q is self-adjoint)

= <v, Qw> (since Q is linear)

= <v, Q(Q^-1w)> (since Q is self-adjoint)

= <v, w> (since Q^-1 is the inverse of Q)

Therefore, we have shown that <v, Q^-1w> = <Q^-1v, w>, which means that Q^-1 is self-adjoint.

Next, we are given that Q is positive-definite, which means that the inner product of Qv with itself is always positive for any non-zero vector v in V. This can be written as:

<Qv, v> > 0 (for all non-zero v in V)

Since Q is self-adjoint, we can also write this as:

<Qv, Qv> > 0 (for all non-zero v in V)

Now, let λ be an eigenvalue of Q and let v be its corresponding eigenvector. Then, we have:

<Qv, Qv> = <λv, λv> = λ<Qv, v>

Since <Qv, Qv> > 0 and <Qv, v> > 0, we can conclude that λ must be positive. Therefore, all eigenvalues of Q are positive.

Using this information, we can now deduce the existence of a positive self-adjoint linear transformation S from V to V such that S^2 = Q. This is because we can define S as the square root of Q, which would satisfy both conditions of being positive and self-adjoint.

Moving on to the last part of the question, we are given a self-adjoint linear