II.3) (*) Determine if the following are true or false and...

II.3) (*) Determine if the following are true or false and think of a brief explanation of why that is the case. (i) Let $T: V \to V$ be a linear operator and assume that there is a basis $B_V$ such that $M(T, B_V)$ only has zeros on the diagonal. Then $T$ is not invertible. (ii) Let $V$ be a finite-dimensional inner product space and $T: V \to V$ an operator. Assume that there exists a basis $B_V$ such that $M(T, B_V) = M(T, B_V)^\dagger$, i.e. the matrix representing $T$ is equal to its conjugate transpose, then $T$ is self-adjoint. (iii) Let $V$ be a finite-dimensional inner product space and $T: V \to V$ an operator. As- sume that there exists an orthonormal basis $B_V$ such that $M(T, B_V) = M(T, B_V)^\dagger$, i.e. the matrix representing $T$ is equal to its conjugate transpose, then $T$ is self- adjoint. (iv) Let $T: V \to V$ and $S: V \to V$ be two operators, if $M(T, B_V) = M(S, B_V)$ for some basis $B_V$, then $T = S$. (v) There exists an invertible operator $T: V \to V$ on a finite-dimensional vector space such that there exists a basis $B_V$, such that $M(T, B_V)$ is not invertible.

Transcript

00:01 Okay, so what we want to show now is that given a linear operator on a finite dimensional in a product space, it's a joint always exists and is unique and is linear.

00:15 Okay, so what do i mean by that? so consider any vectors, you and v, inside the finite dimensional inner product space, big v, and given a linear operator t, well, the joint is defined in the following way.

00:32 So consider the inner product of tu and v, well, that is always going to be equal to the inner products of u and t star of v, where t star is the adjoin of t.

00:47 So this equality that i've written down is the definition of the adjoint.

00:52 Okay, so in order for this definition to be well defined and makes sense, we need to show that such a t star exists is unique.

01:02 And is linear.

01:04 Okay, so the first part that we will show is that it exists and is unique.

01:14 Okay, in the second part we will show it is linear.

01:18 So how do we go about showing that something like this is existing and that it is unique? well, the key observation that we will need is the facts that if we consider a mapping from the input vector u to the interprone, over here on the left.

01:37 Okay, i'm going to write this down here.

01:41 Well, this mapping that i've just written down is a linear functional.

01:46 Why is that? well, note that the input is a vector, and that the output inner products is just a scalar.

01:58 Right? so this is a functional.

02:00 Why is it linear? well, note that t is linear and that the inner product is linear, so the whole thing is linear.

02:07 So therefore, this mapping from you, to the inner product that is written down here is a linear functional.

02:15 By a classical theorem regarding linear functionals, we know that there exists a unique b star, okay, inside this inner product space, such that when we evaluate this linear functional at the input vector u, okay, and the evaluation is just equal to t of u inner products would be, when we evaluate this linear functional that is always going to be equal to the inner product of the input vector which is you and this unique vector v star okay so by the theorem we know that such a fee star must exist and that such a fee star is unique now the trick here is to just set okay they enjoin t star to be a mapping that maps any vector v to v star.

03:12 Okay? so now because such a v star by the theorem is existing and it's unique, therefore the t star, the adjoin, must also exist and must also be unique.

03:26 Okay.

03:26 And that is the existence and uniqueness result that we want.

03:31 Okay, so let me refresh the whiteboard here and let's talk about linearity of the adjoin operator t star.

03:44 So usually the way that we go about proving linearity of an operator is we will consider a linear combination as an input to that operator and we'll want to break down, okay, that linear combination.

03:58 And if we can do that, then we can show that the operator is linear.

04:04 So how do we go about showing something like that? and what do i mean in particular by breaking down the linear combination? well, consider the following.

04:13 So we have some t -star, and the input is going to be the following linear combination, where a and b are just arbitrary scalers, and v -1 and v -2 are just arbitrary vectors inside the inner product space.

04:29 Well, what we want, you know, like to show is that such a term over here is equal to a -t -star v -1 plus b -t -starr.

04:40 Star v2, right? so this is what i mean by breaking down the linear combination.

04:46 So you can pull out the scalar vectors and break down the addition.

04:49 If you can show that this right here, okay, is equal to this right here inside the inner product, then we're done.

04:58 Okay, so how do we go about showing that? well, we just have to do some symbolic manipulation.

05:03 So first, by the property of, by the definition, really, of their joint, we have the following.

05:10 We can just move everything.

05:14 We can just move the t to the first argument.

05:16 Or then by conjugate homogeneity and by linearity, we can rewrite this in the following way.

05:29 Right.

05:29 So why is this? well, there's a bar and b bar because of conjugate homogeneity.

05:35 And we can break down the addition because the inner product is linear.

05:41 Okay.

05:43 So now we can reapply the definition of the adjoint to put to put back the t into the second argument.

05:52 So we have something like this, right? so this is just applying the definition of the adjoin.

06:09 So now we can just combine all the inner products into one big inner product and put back in the scalers via homogeneity.

06:18 Then we get this one big inner product...

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