Adjoints

Algebraic properties of the adjoint. Null space and range of the adjoint. The matrix of T* is the conjugate transpose of the matrix of T.

[KennyKim]

this is very informative. So it seems that the inner product is the main connector of the matrix A and its adjoint. Conceptually I'm starting to think of adjoint similarly to inverse, except that inverse is for matrix multiplication and adjoint is for inner product.

[HotPepperLala]

BUt why does one want to look at <Tv, w> in the first place?

[johnteran]

I believe this is that famous picture that gil strang put in his linear algebra book with 4 parallelograms.

[xanderlewis]

6:25 Why does this work? Because of uniqueness in the Riesz Representation Theorem? Fixing u in U, v |—> <v, (ST)*u> and v |—> <v, T*(S*u))> define identical linear functionals on V, so the vectors in the second slot are equal. (This reasoning is also used in an earlier proof.)

[sayanjitb]

Dear sir, what are the differences between Hermitian conjugate operator and adjoint operator? TIA

[kyang]

Is the complex inner product space inclusive of the real (just as real numbers are a subset of complex numbers) or are they distinct ones?

I‘m a bit confused because the proof of corollary 7.3 (T is adjoint if and only if <Tv, v> in R proved by deducting <Tv, v> from its conjugate and equating this with <(T-T*)v, v>) seems to make perfect sense in real inner product space too, implying if <Tv, v> in R (which is a given in real inner product spaces) then all T in real inner-product spaces are self-adjoint; relating it to proposition 7.4: if T is a self-adjoint operator on V such that <Tv, v> = 0 for all v in V then T = 0.

Then in the exception case where T(x, y) = (-y, x) (90 degrees rotation) <Tv, v> = 0 in R (self-adjoint) then infer T = 0 which is contradicting... but I just couldn't see why the proof of 7.3 cannot be applied to real inner product spaces as well...

[qbtc]

The adjoint is the inverse multiplied by a scalar - the determinant of the original matrix M.

[edwardhartz]

In your proof, how come you assumed immediately that null(T*)=(range(T))perp without proving that they are both contained in each other first?