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Problem 5_ Let A be an m X n_ m < n. Show that A has Fank m if and only if there exists indices 1 < j1 < jz < jm < n such that the m X m matrix co...

Question

Problem 5_ Let A be an m X n_ m < n. Show that A has Fank m if and only if there exists indices 1 < j1 < jz < jm < n such that the m X m matrix consisting of the j1, j2: Jm columns of has non-zero determinant .

Problem 5_ Let A be an m X n_ m < n. Show that A has Fank m if and only if there exists indices 1 < j1 < jz < jm < n such that the m X m matrix consisting of the j1, j2: Jm columns of has non-zero determinant .



Answers

Prove that if $A$ is an $m \times n$ matrix and $D=$ $\operatorname{diag}\left(d_{1}, d_{2}, \ldots, d_{n}\right),$ then $A D$ is the matrix obtained by multiplying the $j$ -th column vector of $A$ by $d_{j}$ where $1 \leq j \leq n$

Hello there. Okay. So For this exercise we need to prove the theory. 8.5. Okay, for that. We're going to consider that a It's going to be an end square matrix. And we need to prove that the following statements are equivalent. So we have here these three statements and all of them are equivalent. Okay, so basically we have that if A is in veritable than equivalent to say that this homogeneous system has only zero solution are and that the determined of A is different from zero. Okay, so We're going to prove first that the first one is equivalent to the 2nd 1. Okay. That means that if A is convertible that is equivalent to say that these are my genius system has only the the resolution. Okay, so basically here we have that if a convertible we're going to show first this part this direction of this double implication. So we have that ive a is convertible, then there exists of course inverse matrix of A. Such that eight times a inverse as opposed to a inverse A equals to the identity matrix. Now, that means that we can multiply to the homogeneous system. We can multiply to the left. Bye. The members matrix. Right. But here we have a members and A. And we know that that's equal to the identity matrix. So we have here the identity matrix time X is equals two. The zero vector. Because any matrix multiply to the zero vector is going to be the zero vector. And here basically what's saying is that the solution is going to be the zero factor and that's the only solution. So with that we proved this direction Okay. Of the double implication. Okay, now we need to prove the other direction of the implications. And for that we're going to assume that this system has only there's a resolution. Okay. But it has only the zero solution. We know that in order to to solve the system, we need to consider these extended matrix. We're here. We put this this extended matrix. Then by real operations we arrived to the reduced asian form and that should be equal to the zero vector here. Right. And this extended matrix corresponds to the To say that the only solution is the zero vector. Okay, but here happens something here we're saying that A is row equivalent to the identity matrix. And we know that if a matrix A is roll, reduce it to the identity matrix, then a is convertible. Then there exists the members. Okay. Or in other words a days convertible. And with that with the proof and we have that the first Is equivalent to the 2nd 1. Now we need to to to prove that the first yes equivalent to the third one. Okay, so did the statement say that if a is convertible is equivalent to say that the determinant of a Is different from zero. Uh huh. Okay. So this is not that hard because basically if a is in veritable that implies that a is row equivalent to the identity matrix and we know that the determinant of the identity matrix is different from zero. And furthermore, we know that if the determinant of the matrix we have that a matrix B is. So this is going to put here parenthesis. If B is wrote equivalent, do some matrix A. Then we have the determinant of B is equals to zero if and only if the determinant of A is equal to zero. Okay, so if the determinant of the identity matrix is different from zero and we have that the identity matrix and a R row equivalent. That means that the determinant Of a is different from zero. These matrix we have determined equals to zero if and only if as these results say if the matrix to which this is royal equivalent is equal, determines equals to zero and this is not the case. Okay, so we have that the determinant is different from zero. So that showed this first direction of the equivalence. Now we need to show the other direction. Okay, let's say that if the determined of a is different from zero, then our matrix is inverted. Okay, So we have that this determinant of the matrix is different from zero and then we know that there exists an inverse matrix that is equal to the joint matrix of a divided by the determinant of a For a different from zero. Okay, so if the determinant is different from zero, then there exists this kind of this inverse matrix that satisfy the condition that eight times a inverse is supposed to a inverse A, which is opposed to the identity matrix. So therefore that means that A is convertible, and with that we in the proof that the first Is equivalent to the 3rd statement, and therefore the theorem, 8.5 has been proved.

Don't just the problem here we want to argue are a Is that many independent? So So there The corner of a linear independents is equivalent to say a s incredible. So it is actually a is inevitable if, in the only if agents pose a it is also you, Bertel. So they said speakers If we see this terrorist in so the so so here for this demolition If a the colon is they're Carlin's of a it's linear independence. Then we know a is you're irritable. So if a is the convertible and so we also know a transpose is inevitable So this is because the rink of a it's ricotta rink of transposed So yes, so they have the humors so safe. But he knows the U S of a transport be ancient spell beavers and the humorous of a will be a uber's. So we say a transpose Huber's and a universe we'll be the humors of a transpose eight. This is because it's their time. Together, it's echo to a uber's in a transpose universe, a transposed in a and become to a transpose i A. If you go to a inverse, a issue could you buy. So this means it is the inverse matrix of a chance was eight. So agents pose a is convertible, so for two this direction so we know a chance. Close A. It's the Invincible due to their rink knowledge t serum. We know that the rink of agents pose a trust that not a team of aid you This chance was a It's Nico to in the hair agents will agency in veritable So we know this rink. It's because you in so the knowledge he should be sure that you come to narrow So we know if agents pose a eggs. If you co 20 you have to have X You cut your hero, right? They were going to consider the now space of eighth. So for the now space of a so it looks like it's satisfies a X equal to zero. So we choose any experience now space okay thing than now of age. So this aches had a property. A ex coaches there. So also, if we transposed, there's we're gets shows X transferal agents bill should also be there. Oh, if we time that to them together we will get X transport a gentle a X if you cut through there. Mm, No, I don't want to do that into this is enough. So if X it's lifting this now space a X is equal to zero and we times a transpose to two this vector. So we know agents will A X is equal to zero since we already know if there now space for a transpose a is contents in the their oh, so we can derive that extra butte there. So if exit zero then we means this means this now space for any the X in the snow space We can deduce that X is zero. So this means there's now space only contend content zero So according to their well so well, I don't want to do that. So for any So we know there no space here, so only condemned zero. So the knowledge t for a it's their so called into the rank nal ity. Mr Room. We know the rank of a it's in. So here we know the the colon, the colon that's a his denia independence

So for this problem were given a Matrix A, which is an end by P Matrix matrix d just a diagonal matrix with the one up to D n along its diagonal. And we are asked to show that d a the product of the two it's found by multiplying the I throw vector of a by D I where one is less than or equal to I is less than or equal to end. I note here, that's for the diagonal matrix the i J. The element is just going to be d I times the chronicler Delta So that is going to the same thing as D. J here. Sorry, I should put that outside of the square braces. That's the same thing as D I J. So now we can look at this from the point of view of looking at the index form for the matrix product. So d a element i j is equal to the sum from K equals one up to n uh de element I times. Hey, Element K. J. We know what d element I K is going to be. That is going to be one second. Here was one of two end this D I k is going to be de i Delta I k times the k j eighth element of a But now, because we have this chronicler Delta here, this is going to make this some b are each term in the summer B zero whenever I isn't the same thing is K or K isn't the same things I so because that kills off all the terms, were just going to have the results. It's going Teoh d i times Delta II which is just going to you one times a element i j or that's going to be d I names element A Hi, Jay. So now if we consider what the, um I throw of this d a matrix is going to look like So we have d a is equal to, you know, have one dot, dot, dot, dot dot i dot, dot, dot going along s so we will have de i a. I won. Then we'll have d I a I to and so on up to d I a i p. Now, since we have these d eyes along each element of this road factor, this is the same thing as doing the row operation of just having d I times that row of ai one a i to up to a i p But this thing here, this row vector that we're considering that is the same thing as the I throw of our starting matrix. Hey! And so that proves the that proves the statement.


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