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We will now slow that M,(R) is an inner product space over R with a inner product known as the Frobenius iner product . (a) (2 points) Show that M,(C) is vector spa...

Question

We will now slow that M,(R) is an inner product space over R with a inner product known as the Frobenius iner product . (a) (2 points) Show that M,(C) is vector space.

We will now slow that M,(R) is an inner product space over R with a inner product known as the Frobenius iner product . (a) (2 points) Show that M,(C) is vector space.



Answers

Let $V$ be the vector space of $m \times n$ matrices over $\mathbf{R}$. Show that $\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right)$ defines an inner product in $V$.

So the salt this problem first, listen up the properties of our enterprise they were going to be using in our solution. So the first thing we need to show is one t of a us to be is equal to t of a plus B. So what this means by the definition, should offer a linear transformation is do you have a plus to be is equal to okay in approx u one comma a comma The enterprise A Are you one plus the inner product. You want to be come of inner product? Oh, be common you want. So by prop three after three we know that we can add the second components together and we can get this result you know, probably you one common a plus they're back. You want Congress be colonel their product a plus B comma you want but by Pete property too. Well, we know that this is actually equal to a call you one plus e commerce you want in a product of a plus b color you want it's probably too says if you have any robotic UV than in approximately equal to be you. And so now we comply by a P three. The It turns out we can add a and B and we get a plus B comma you one common in a product a plus B. Connor you want. And now we comply p two again To get the inner product. You will call it a plus B. Come on, it must be you're one which is, of course, an equal to the transformation of a plus B, which is what we want to show. So the second part we want show is that C times t of A is equal to TFC times A where a is a factor and see it's a concert. So by the definition, are transformation. C times t of a is equal to see time's in a product you want comma a drama. The Interfax, a college you want, which, by the dissolution we get See Time's you one comma A Let's see times Hey color you want by prompt one. This is actually equal to see Time's You want common? A coma? See, Time's a Colin you wanna but well by properly to weaken Flip, you will come in a Teoh having equal to their product of a you want. So this is actually see Time's a U one. Time out. See, Time's a Come on, you one. So then, by property three to me about how everyone we could distribute this to get And, of course, see times a condom, you one and then ah, by property if you to, you can flip it to get you won Thomas C. Times a comma. See, Time's a common one, which is, of course, equal to T C U times I.

It was supposed to show that this statement is true. This that you don't you is equal to the magnitude of you squared used three doing selector So we can say U is equal to some vector with, um if it's three dimensional vector that has a B and X y and Z components which I'm just gonna call a B and C there a b c, your elements numbers then you dot with you is equal to a B C with a B C, which is equal to a squared plus B squared C squared. We just multiply each component with perspective on self on the other vector. So at the day with the A to B to B to C with C which is equal to this where route of a squared plus B squared C squared all squared, which is just equal to the magnitude of you squared and we've proven the identity to be true

So we have these linear transformation tea from a vector space v T o r n, and we have, ah interpret well where the finding. I mean a product on V given by the brother of you against V is by definition to you dot tv. We remember that the dot there is in a product that standard in the product inside our end because the U. N TV are factors in our end and so we just need to check that he's in a product against your V defined in the vector space. Big B is actually the product, so we need to check the four actions. So what are we, the 1st 1? Which is the symmetry? So we want to check that V against you is by definition tv dot to you and these by the symmetry off the dot product in rn. This is TV at you. Sorry, not TV and all these by definition you against me. So dysfunction on V is symmetric they only to try that it's leaner. So x plus y against a vector V We want to check the disease X against being plus wagons being so we use the definition of the inner product. Well, these in a product t x plus y dot d y. And now again by linearity off T and by the charity off the dot product. In our end, this becomes tx dot tv bluff the y dot TV And now again, by definition, did his ex against v blast. Why against me Now for the third property we need to check it is or more genius So CX against V. It is, by definition, tee off, see axe dot tv. But now both t and dot product are linear. Well, that I'm a genius. So this becomes C t X dot tv and again, these by definition is c times X against me. And finally, we need to check that the dot product, while the inner product is positive definite off. We need to change X against ex. He's got their equal to zero so exciting. Sex is by definition, tx dot tx. And now because dot is another product on our end, we know that this is definitely great unequalled in zero. And also this can only be zero if t x zero because the season in the product So did your brother have something against itself can be zero only director itself. Zero. But now remember that tea is want one. So the only victory gets within the vector in V that gets maps to the zero vector. Our end zero is zero factor in we and that means that X must be zero. And therefore we have checked for actions showing that isn't a problem is indeed

So to solve this problem, I implicit out the properties of the enterprise that we're going to be referencing in our solution. So the first thing you do is one you need to show. T of a post TV is equal to t of a plus B where A and B are factors. So by definition of our transformation t of a us to you being is equal to the inner product you want common Be okay, comma the enterprise you two comma a plus the inner product you want comma, be comma the inner product. You too. Com A B So the thing to know is ah in the first component of each term. If we look at their inner product, first term of their inner product is both you want, which means my prop free. We can add these two terms in the second and in their second component while fixing them in the first component. And then it's also true for the second component of each of these terms. So what this means is that keep property three. We get, uh, enterprise. You want comma A plus B palla in approx you to come on a posting which is of course, equal to tee off a plus B. And so the second thing we need to show is that C times t of A is equal to t of C times A where a is a vector and C is a constant. So what this means, by definition, definition of transformation is that c times t of a is equal to see Time's the in a direct Monica you want common a comma, the inner product of you to come at now by P two proper to We know that, for instance, you want common A is equal to the enterprise A comma you want and the same is true for you to come in a. So this is actually equal to see times the inner product Yeah, a common you one comma see times, you know, product a Colin. You, too. And then by P to we could distribute the p one we could distribute. And so we end up with see, Time's a common you one and comma the proxy times a college you, too. And then by P to we could flip these in a product that 19 equality and we end up with you in a product you want Colisee times a come on in a product, you too gonna see times A which is, of course, equal to tee off, See times at.


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