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4. (10 points) Let A be an m X n matrix with rank(A) =r. If P and m X m and n X n matrices, respectively:are invertible(a) Prove that rank(PA) =T by proving that PA...

Question

4. (10 points) Let A be an m X n matrix with rank(A) =r. If P and m X m and n X n matrices, respectively:are invertible(a) Prove that rank(PA) =T by proving that PA and A have the same nullspaceProve that rank( AQ) by proving that AQ &nd A have the same nullspace. Hint: Recall that rankA = rankAT

4. (10 points) Let A be an m X n matrix with rank(A) =r. If P and m X m and n X n matrices, respectively: are invertible (a) Prove that rank(PA) =T by proving that PA and A have the same nullspace Prove that rank( AQ) by proving that AQ &nd A have the same nullspace. Hint: Recall that rankA = rankAT



Answers

Let $A$ be an $n \times n$ matrix with rowspace( $A$ ) $=$ nullspace( $A$ ). Prove that $A$ cannot be invertible.

Suppose we had on Matrix A. That's him by in So it has m rose and in column. So M in columns. Let me just emphasize this year has m number of rows in him Rose and in Cohn's in calling. And suppose this major is a was such that the column space of a was equal to the null space of a no what if the null space of a, well, the null space of a is all the set containing all the vectors x all the vectors x such that a times a vector acts is equal to the zero factor. So it's the set of all vectors x such that a Times X equals zero vector. So one vector that will always be in the know space of any matrix is that the zero vector the in by 10 factor because the zero vector times A is going to be equal to zero by how matrix mortification works. But another thing to keep in mind here. Another thing to keep in mind is how matrix made more application works. In this particular case, matrix multiplication is only defined if X is this has dimension in by one X must have to mention in by one. It must have the number of the same number of rows as a has columns, since A has in columns X must have in rows. Now, in this video in this video, with all of that in mind, we're gonna prove that if the confidence of a is equal to the normal space of a, then this implies this implies that em is equal to in. So this is what we're going to prove prove If Colin space of a sequel to the Northwest of a then that implies it in is equal to end. You might already already see why this is true. But let me let me let you know of a nice little trick that or maybe not not even a trick, but a nice, sometimes strategy or initial approach to prove problems like this. If you don't see an initial direct path to proving this, if you don't he up if it's not obvious to you and it's OK if I wasn't obvious to me as well. But if it's not obvious at first sight a direct proof of a direct, direct way of proving this statement then I encourage you to consider the proof by contradiction. So what I mean by that is consider what would happen if M is not equal to end. Suppose M is not equal to end. So in let's let's try to derive a chain of implications. Let's try to derive a chain of implications that leads us to a contradiction. And if we're led to a contradiction, If we were led to a contradiction, then that means that this supposition cannot possibly be true. Because if we were to suppose that M is not equal to end, then that would imply a contradictory statement. So am cannot possibly be equal to end. So therefore him must must equal in. So let's consider what happens when him is not equal to end. Let's first what what subspace does The column space of a lie in If a is a riel has really number entries. A has really number entries. What subspace. There's a column space of a lion. Well, the cost base of A is the span of the contractors, and since there are in rows, each column vector has em in trees. So the column space of a is a subspace off our in because each vector in the common space of a is the linear is a linear combination of the column vectors of the Matrix A and each of the column vectors in the Matrix A have em real number entries. So any linear combination of a bunch of vectors that have em real number injuries will also have a real number entry. So each factor in the con space of a we'll have em. Leo number entries are therefore each veteran. The confidence of A is in our M the set of all vectors having him Rio number injuries. Now what subspace doesn't know space live in the no space, the no space. Well, pause a video and try to think about what subspace been North based lives in. So I'm assuming you've had a go at it. The no space lives in our in. Why does the North base live in our in? Because the nose basis defined to be the set of all vectors acts such that a times acts is equal to the zero vector. So all the factors in the set X have in rial number in trees, all vectors in this set X have in Rio number interest because the matrix multiplication right here is on Lee defined if X has the same number off entries as there are columns in a so X must have in real number injuries. So check this out. What happens if the column space of a is a self face of RM? Well, this is always gonna be true in the no space of ese subspace are in. This is always gonna be true for an M by N Matrix. But what happens if we were to suppose that him is not equal to end? Well, we we were given the fact that the column space of a is equal to the no space of a. That means that every vector in the column space of A is also in the north face of a and every vector in the north face of A is also in the column space of a But check this out. This can't possibly be true. Our statement that we were given can't possibly be true if the column space of a lives in our em and the north face of eight lives and are in because the concepts of a lives in a completely separate vector space than the north face of a they live in. You can think of these two vector spaces is completely different universes when in is not equal to em. These are two different universes, so we can see that the column space if they can't share any vectors in common with the north face of a if M is not equal to end. But that implies that were contradicting roots groups. That implies that we're contradicting this statement right here. So in cannot possibly be not equal to end him must be equal to end, because otherwise we would arrive at this contradiction right here.

Hello there. Okay. So for this exercise we got a matrix define as follows is Q. A. Is a fine as a polynomial sort of Hussein of coefficients and the powers of the matrix A. To the end up to A 20 which is just identity matrix. So we had this matrix Q. A. And where A. Is uh N by N. Matrix. So now what we need to show is that if we find, well actually we have here a similar matrix B. Such that is equal to p members ap the similarity condition then the matrix Q B can be written as the yeah P inverse Q. A. P. Okay, so we need to show this relation here. Two little start. So first let's consider Q B. There's equals to A. N. Be in Plus eight N -1 B. and -1 plus the A. One B plus a zero. And identity matrix fuck. Then we're going to replace with the condition here. So we got that Q B will be close to a N P inverse A. P. And did you hear we're taking the end power? We know that be to the end is equal to we know that be to the end. It's going to be equal to P. To the the inverse A N. P. You're going to use these these property. So stay to the end will be just P. E. To the minus two PM birth to the to the N P Plus eight and -1. The members here is a And -1 p plus A one. The inverse A P Plus A zero on the Identity Matrix. And here we can take out all the P because P inverse is distributed on all the terms here. So we can take out as a common factor and even we can put it here on the identity because the identity matrix can be written as PM burst B. So we can take out that and we end with P m verse A N A n P plus a ny one A n minus one P. And so one A P plus zero P. And then we did the same with P. So we can take out the P. Because he is distributed on all the terms. So that means that we obtain here Pete inverse A N. A. To the N power plus A. And minus +18 The n minus one power plus a one A plus a zero data entity metrics P. P. And that's it. Here we got what we got in the middle is just Q. A. So the result is that this as opposed to the P. Embers Q of a matrix Q A B. Great. Now we need to show the following if a ZR diagonal Izabal matrix, that does that imply that QA the matrix is also diagonal, Izabal. And to prove this, we need to know that if A. Is the original Izabal, mhm Then K times A. Is also diagonal size. Because there's just this killer, it's just multiplying so it doesn't affect the if it is or not ever and if A. Is diagonal Izabal then a power of A is also signalized. And we have seen this property actually we use this fact to calculate the powers of metrics. So eight to the eight to some power is if is even realizable. They then eight to some power is also diagonal. Izabal and the same happen if we multiply some killer and Q. A. Is A. Some of this kind of matrices, hours of A. And multiplying by some scholars. So if a signal Izabal then this whole these Manami als we can take like kind of Manno meals, each term that is some on this matrix Q. A easier and relatable so to A is also devon allies.

Hello. Let's consider. We have two matrices A and B. That are similar matrices of order and buy in. Now if A and B are similar, then there exists some infertile matrix P. Such that is equal to p inverse times B. P. So if we multiply by P then we end up with P times A is equal to well P times P inverse times bp um which is just equal to well um the P A. Here we get this is equal to while p times p inverse is just the identities. Just just the identity times Bp which is just equal to well Bp Right, so um the rank of P A is equal to the rank of BP such that the rank of a P is equal again to the rank of Bp as P is convertible. Therefore the rank of A is equal to the rank of B. And according to the rink um nobody theorem we have that row of N plus N of M is equal to the dimension of em here, row of M represents the rank of matrix M and N of M represents the nullity of em. This implies that row um A plus the nullity of A is equal to the dimension of A. And we have that the at row of B plus the nullity of B is equal to the dimension of B. And as the rank of A is equal to the rank of B. We can substitute row of a equal to rho of B. Therefore we have that row of A plus melody of A is equal to mention B. And row of B plus melody of A to mention um dimension A. So what do we get then? Is that dimension um of a minus? The nullity of A. Is equal to the dimension of B minus the nullity of B. So we get that N minus the nullity of A is um equal to N minus the melody of B. And from there, um We get that nobody of A is equal to the melody be. So, therefore, um similar matrices have the same rank, um and nobody yes.


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