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Using the result of Problem 9 , prove that the only matrices in $F_{n}$ which commute with all matrices in $F_{n}$ are the scalar matrices....

Question

Using the result of Problem 9 , prove that the only matrices in $F_{n}$ which commute with all matrices in $F_{n}$ are the scalar matrices.

Using the result of Problem 9 , prove that the only matrices in $F_{n}$ which commute with all matrices in $F_{n}$ are the scalar matrices.



Answers

$\begin{array}{l}{\text { If } A \text { is nonsingular, then }\left(A^{-1}\right)^{-1}=A} \\ {\text { (a) Verify this theorem for } 2 \times 2 \text { matrices. }} \\ {\text { (b) Prove the theorem for any } n \times n \text { matrix. }}\end{array}$

That's the show that this is some n by n matrix to square matrix. And it says that prove that if every element of a row or column matrix is small, the probe of the real receiving The determinant of the Matrix is multiplied by Tse. So if you think about it, what we're going to do is kind of imagine that some element of theirs let's say there's three is a three by three. We don't know what the elements are, but whatever, um, the determinant is based on going along one of these rose. So it doesn't matter how big the matrix is. It's always based on one of these rows, and then everything is is put down from there. So if you multiply each of these by sea that essentially what you're going to get is see, multiplied by the determinant of a because you're multiplying, each there rose by sea. Okay, so that's that's the basic premise of the proof. If you kind of think of it like this, see, times ay, ay. Jay is equal to C A I J. And then imagining that you're doing the determinant. Using this row, you're gonna get a number of different gonna get See, I Let's say we do it when I is one for a row So we do see eye one times J whichever j value it is, But they're gonna be a one a one, a one Everybody you have there oughta be multiplied by C so you can factor that sea out of your determinant. So that's the way that we can see that it's just gonna be see, Time's a determinative, eh? All right. And then it says, Imagine that you have in Matrix where all of so you've got your diagonal like this and all of the elements either above or below are zero. What happens with that? Is that all? If that happens that all of the values that get multiplied in our determinant become zero as well. So in other words you're gonna get let's say that the main dying relatives air A B C just for our purposes, it's gonna be a and then times whatever the determinant of of the stuff is that not including a but they're all gonna have zeros in them. So you're essentially going to get something with like 00 and then a number here. Let's say so. Let's say that, See? Okay, so you're gonna end up with just a product of the means you go because all of the other parts of this are gonna be zero, so I can probably go into more of a sort of a no more professional proof, but it's just a matter of understanding that as these Air zero and as you're going along, the determinant is a product of along this row taking this road out and having the rest of them included. And so it ends up being so that the the diagonal elements here and here, the diagonal elements they're ending, gonna end up being the ones along the middle, and the other ones are gonna be zero. Let me just show you an example, because I know I'm talking and you're probably thinking that guy's just blabbering about something, and I know what he's talking about. So let's let's say I have this ABC and then this is 000 and zero. Okay, so just as an example, this is a three by three, but the determine is gonna be a and then you're gonna get rid of this row in this column so you needn't be 00 c and then plus, and then everything else is gonna be zero, because the elements along this row are zero and zero. So the only thing that's gonna end up being is just any times the department of this and the determined of that is B times C minus zero, time zero. So we end up with just a Times B bidet seats. You could see it clearly true there. And that's the process of going. It doesn't matter what we ADM or elements or not, because, as we do, well, eventually I determine it. That's just the main diagonal, and everything else is gonna get multiplied by zero. So hopefully that makes sense. So those your two proofs

In this example, we have a three by three matrix a provided, and our goal here is to determine if it has an inverse for this particular case. I'm going to take the approach of role operations until it's reduced into echelon form. So to that end, let me make row to our first row, which is 102 row three. Let's make that the second row negative four negative 97 and the old first row could be our new third row 03 and negative five. Now, since we're only going for echelon form from this pivot here will eliminate this entry below, so this will be raw equivalent to first copy 102 then multiply row one by four. At the result to Row two will obtain zero negative nine and let's see eight plus seven, which is a 15 then the next row is 03 and negative five. Notice, however, that the last two rows air now suspicious. If we take negative three times the third row, then we would get to negative nine and positive 15. This means these rows are linearly dependent, and so if we make a new row operation it would be 102 copy zero negative, 9 15 and say Divide that row, Bite negative three or excuse me, divided by three and added to Row three that produce Onley zeros. So that tells us we have a pivot here and here. But the third column is not a pivot column due to this zero. So this tells us the A is not. Let's start with a verb. A is not row equivalent to the identity matrix I. And this implies that a inverse does not exist. If we did get a pivot here, then we would have concluded the opposite. That a inverse does exist because then it would be row equivalent to the identity matrix.


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