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Show that a linear operator $mathrm{A}$ on a finite-dimensional vector space $mathrm{X}$ is invertible if and only if it is one-to-one or onto....

Question

Show that a linear operator $mathrm{A}$ on a finite-dimensional vector space $mathrm{X}$ is invertible if and only if it is one-to-one or onto.

Show that a linear operator $mathrm{A}$ on a finite-dimensional vector space $mathrm{X}$ is invertible if and only if it is one-to-one or onto.



Answers

Use matrix algcbra to show that if $A$ is invertible and $D$ satisfies $A D=I,$ then $D=A^{-1} .$

Okay, So Okay, so in this program, if a caused to a block can be ah petitioning into a blocks of matrices B 00 on See now first direction If e and C our convertible that muse, beamers and the seamers exist, then we can directly construct yet the a inverse to be immersed seamers and 00 So if it before because dropped another men traces like this these metrics is thievery is okay because we can verify that eight times a humorous because toe humorous times a equals two and I grieved images. So this is one direction. So we label is by one for the other direction if a it's incredible. So that means a universe exist and the Immers has the same dimension is a them use. We can petition a humorous by X y z w in the X and the B has the same size and w and the sea have the same size. And since we petitioned both a numerous conformity so we can do the multiplication eight times a humorous equals to the 00 c times x y z tabouk that gives us p x b y Sesay, see Bablu that way. We already know that a inverse eight times a humorous equals two a night in the Matrix And, uh, we can petition abuse. I think it metrics by this. And by comparing both metric mentions this we know that be X equals dough I the CW because toe I So this is one direction and we can also we can do the same thing for a m Rose time say, sees a MERS 10 say's also we had any matrix It gives us x times p equals toe I and the tablet times see caused by. So our conclusion is B and the sea uh, also in vertebral because this is exactly the definition off. You're worth ability in the night proof this statement if they can be petitioned like this. So a is irritable if, and only if both being a c I multiple

Okay, So our question is, let s be a finance set. Okay? Finance it in a victory space victory space. We with the property that every X in V has a unique representation. Okay. Unique representation as a linear combination off elements off. Yes. So we have to show that, as is a basis off we Okay, so we know that we know that as is finite dimensional. Okay, is finite dimensional. So we can say that this is it. And dimensional ticket and dimensional. So on this elements are as I and for their values, less than it was toe end and greater than it was 21 Okay. And for that on bet Visa victory space. So this will be fine. So each each eggs and we has a unique representation. Okay? Unique representation as a linear combination off as a linear combination off combination off s. And as to s three. And so on s and Okay, so where definition as an as his spending set for V okay. And in particular, that is also means that zero has a unique representation. Okay, I tried to understand that Nick represent representation, okay? And currently we cancer. That this killer C one C two C three and so on. C and the victory questions. Okay. In the world of the factory question, it will be C one s one plus he two as two plus C three s three and so on. C n s m equals to zero. You know that from this. So this will have a unique solution. And the travel unique solution. And the solution is Jiro on zero. It means thistles, the travel solution. And we know travel solution is always the solution. Okay? And it is the only solution. So it means, by definition, it means as one as two s and so on s and is linearly dependent. Okay, because it has travel solution and only solution. So it is linearly independent. Okay, So any as Iselin independent spending set that does, by definition, definition, as is a basis off movie. Okay? And we can say that as it was to s equals two. We want we do. We three we and you can save as one as two s three s and is a spending set off. We okay? And these are linearly dependent. So the definition as he is a basis off we it is prude. Thank you.

We begin with the observation that the columns of a are literally independent, even a leave the rose off they transposed ordinary, independent. That's simply because the rows of a transport are the columns of a because that's how transports works. Now let's find an elementary metrics, eh? That reduces a transpose so hee, a transposed, is reduced by Rose. And we know that the elementary mattresses are in veritable. So let's keep that in mind. Okay, Now we show that the columns of a and therefore, well, the rose or transposed our little independent by contradictions. So we assume they're they're not. So if the roles of a transfers are not literally independent, there means that when we reduce a transport by Rose and we look at the very last row So let's say the end raw off e a transposed, we get a zero. So these are absurd hypotheses, and we show that if we assume that this is the case, then we have a contradiction and therefore the rolls off they transposed must be literally independent. Now with these absurd assumption, we look at the last So the end raw off the square metrics e a transpose A. But we know that the role of a product is the last throw off E a transpose That motive lies, the medics say, but now by Apophis is the last drop off E a transpose is just zero the multiplies they therefore and so we get zero. So the last throw off the m by n matrix e a transpose a zero. But that means that e a transpose a is not in vegetable because it has an entire row of zeros. But I remember that he was in elementary metrics. So Hee itself is in vertebral and that means that the problem lies in a transpose a so a transpose, a square matrix which is not in vertebral. But this is against our assumptions because we assumed to begin with that a transpose a is in vegetable. And so the only problem is are absurd. I bought it Is that the rose of a transpose are not linearly independent

In problem 19. We won't support that. The metrics a equals the multiplication off two mattresses Q and R were que is m multiplied by in metrics on the Are is in multiplied by in. We want to show that the columns off eight are linearly independent A s columns off nearly independent, then almost the inverted. It's go in the proof. If a has a linearly has linearly independent columns, which means if we multiply it by a victim X and we want to solve this equation for X because the metrics A has linearly independent columns, the solution would be on Lee Trivial Solution. The only Trevor solution. It has the victor off zeros. This is trivial as a trivial solution. The only solution is a trivial solution. Let's make a equals Q R. Then we'll get you are but deployed by X equals zero by applying the're, um seven for this equation, we get the North are X the norm of our X equals the normal of Q. The play by our Robert X equals normal zero, which is zero, which means we have or X equals zero and because X is has only a trivial solution which concluded from this step, this second step on this the third step. And as our X equals zero, we can conclude that X has only a trivial solution from the in veritable metrics serum. Because our is square metrics and X has only trivial solution. This means it's inverted or is invert. Then we start by X equals zero, which she gives which it gives this fact. X has only trivial solution. Then we make a equals Q. Or that's the third step is toe a ploy theorem seven to get our X equals zero. And we used the information from step one and or is a square metrics. And by end and from the in vertebral metric, the're, um we have our is involved.


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