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Let A and Bbe similar n X n matrices. Show that A and B have the same characteristic polynomial. (Hint: suppose that S is an n X n invertible matrix s0 that A = SBS...

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Let A and Bbe similar n X n matrices. Show that A and B have the same characteristic polynomial. (Hint: suppose that S is an n X n invertible matrix s0 that A = SBS-1 Note that A AIn SBS-I AIn = S (B AI,) $-1 Take the determinant; and use property of determinants (a theorem in Section 4.2)) What does the previous part imply about the eigenvalues of Aand B? What does the first part imply about the trace and determinant of A and B? 31 126 Use the previous part to show that that 10 is not similar t

Let A and Bbe similar n X n matrices. Show that A and B have the same characteristic polynomial. (Hint: suppose that S is an n X n invertible matrix s0 that A = SBS-1 Note that A AIn SBS-I AIn = S (B AI,) $-1 Take the determinant; and use property of determinants (a theorem in Section 4.2)) What does the previous part imply about the eigenvalues of Aand B? What does the first part imply about the trace and determinant of A and B? 31 126 Use the previous part to show that that 10 is not similar to 21 21 ~10



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Prove that if $A$ are $B$ are $n \times n$ matrices with the same Jordan canonical form, then $A$ is similar to $B$.

Hello there. Okay so for this exercise we need to show that the matrix A and a transpose have the same Eigen values. Now how to prove this. Well basically we need to show that the determinant off. We need to show that they have the same characteristic polynomial. Or in other words that they the determinant of land i minus A is equal to taking the determinant of lambda I minus eight transposed. Okay, now this is not as hard as it looks like. Okay, so let's see what's going to be this determinant. This determinant is going to be the finest follows minus lambda. Sorry is lambda 9811 812 until 81 and Right and here we are going tough. 8-1 λ -8-2. Did it 8 to 10 And so on until eight and 1. I am too. And the last term is lambda minus eight and A N. N. Now what happens if we transpose a? Basically the diagonal is going to remain in variant the terms on the diagonal doesn't change. So we're going to have the following determined lambda minus a 11 London minus 8 to 2 lambda minus A and M. So so far it looks good. Now what happened here is that now this role is going to corresponds to this column here. So now we have a 12 that that that a one N. Here 8-1 that 8 to 10 and so on. And here we have eight and 1. Now you observe something here of interest. And is that this made this determinant here is the matrix here is equal to the transpose of this matrix. Okay. And we know that the determinant of lambda I of A B is equal to the determinant of the transpose. And that's what we have in this case. Actually. We can prove this easily but I wanted to show you this because if you don't understand because well you have the transpose is equal to the same determinant. You can observe that if you pick this first road to calculate the determinant using the miners is equivalent to pick this first column in these matrix. To calculate the determinant be determinant using the miners is equivalent to to the to do these two operations. Or in the other way that you can observe this is do you suggest the properties of the transpose? Lambda I minus A and then you transpose these matrix. We know that we can distribute this. So we're going to have lambda Identity transpose -8 transposed. But the transpose of the identity is equivalent to the identity itself. So we have lambda I minus a transposed. So we have that this the transpose of this term here is equal to plumbed up I minus a transpose and that means that the determinant of lambda I minus A is equals to the determinant of lambda I minus A transposed. But we know that this is equal to the determinant of lambda I minus A transposed. These two matrices have the same characteristic equations. Therefore both have the same Eigen values. Now we need to show that that's a part B. So we need to show that A and a transpose neither not half the same Megan space. It's important because you may confuse they say, well these two matrices have the same meghan values, therefore they have the same again space and that's not true. And for that We're going to pick one. Counter example. Okay, 111 example of when where it gives is clear and this is a two way to case. So the A matrix is going to be a B, C and D. Right? And the matrix transposed will be the matrix A, C, b D. Right? No, let's assume that these two matrices have the two. I can follow slumber one, I'm number two. Or generically Lambda, I where I goes from 1-2. Now, we know that for this kind of matrices A These the Eigen values of A Are defined as -7. A minus lambda. I. Okay, but let's see what happened with a transpose. When we pick a transpose we need to take the wagon value. Love the eye. They've entity minus A transposed should satisfy the following. A homogeneous system. A transpose This in the matrix notation is equivalent to have a lambda minus than the I minus a minus c minus B and lambda, I minus D and extended matrix is defined as follows. Right? If we apply the following row operation the second row, we were going to sign B times the frizz, ro minus plus lambda I minus A, the second row. And by doing that we obtain the following matrix lambda I minus a minus c, zero on zero. Now from here, you can observe that the Eigen values for these matrix. So the Eigen vectors for a transpose are going to be are going to be defined as by minus C an a minus Bluhm that I which are totally different from the Eigen vectors of the matrix A. That are defined as -1 8- Lambda i. So they don't have they don't need to have the same angle space is not necessarily true.

Hello today have been approved that if A and B are and by our major season a convertible than the nullity of a B, people would only be, and that equals finality of the A. First, we're gonna prove that, uh, the north face of the is a subset than all space of a B. So no space of the subset of a B and we're gonna define X as the null space. Therefore, if we have a B Times X equals zero, then we can take this, multiply it by using red A n birth And because a and verbal, we know that a times a inverse equals the identity matrix. So you have the identity matrix times be thanks equals zero. And I think times the identity matrix is itself. So the times equals zero. And this tells us that B is indeed a subset of a B. Next, we want to do sort of the reverse and show that the null space of a B is contained within B. So let's do yeah, times are no space equals zero. So on the converse side, we can multiply both sides by a Let's use red for that. Go back a little bit. Must fly by a So this gives us baby X equals zero. So this tells us that the null space of a B is contained within B therefore therefore three and be have the same null space. Or, in other words, the milady of a B equals banality of B. So now let's go down and use our rank. Nobody here to try to investigate this a little deeper. So we have our rank melody here, um, using well. So if we do this for the Matrix A B, the rank of a B plus humility of a B equals the number of columns of a B and similarly, the rank let's be plus can only be equals the number of columns A B But we also know that the rank of a B going back over to the left equation the rank of a B is the same thing as writing the dimension of the column space. Oh baby. And that humility is the same as the dimension of the North space with a B. And that's the same shirt. The same is true for B, so we can therefore rewrite what we have in terms of the nullity of a B. So let's do that. So the dimension in green will be that agree. So the dimension of the null space of a B equals the number of columns of a B minus the dimension of the column space a baby. And similarly, the dimension of the column space sorry dimension of the North face would be equals. The number call would be minus the dimension of the column space of beef. Now, we proved above that no space, um, of a B and equals ethanol space of B or in other terms, that the ability of a B equals finality of beef. So this term equals you, Mr. So we can have these quick equations equal to one another such that the number of columns with a B minus the drink of a B equals the number of columns of the minus. The ranking be so we know that these are both and by n matrix meter seas. So the number of columns of a B is the same as the number of columns of B. Well, why is that? While a and by N e matrix time in and buy in the matrix again, a bee matrix is still an n by N matrix. Therefore, the number of columns a baby is equal to the number of columns of the. Therefore, the rank of A B is equal to the rank of B, so let's move a little lower and move on. So let's assume now, going back to read. Let's assume now that the rank be in the equal to the rank would be transposed Then, using been all space from before, this would tell us that a transposed be transposed times in old space. Pretty cool euro were. That's a different form of our A B basically zero. So a transfer to be transposed all space equals zero. Multiply that whole thing by the A inverse transpose fans we get be transposed of the null space equals zero. So this tells us similarly from before that be transposed shares the same doll face shares the same goal space as a transposed be transposed. They know pence be transposed and eight transposed be transposed, have the same rank for the same reason we having the proof above for why the rank of 80 equals the rank of beef. So now we have to move on. So the rank, uh they transposed be transposed could be written Such that is the rank of the A transposed which equals the rank of the A Therefore therefore, the rank of be a equals. The rank of the tea because we see here that be transposed is the same as a translator be transposed and transpose beast transpose is the same as the rank of B A for the rank of be a seems the rank of be transposed which is the same as the rank of B which then is also the same from our before Up here as the rank of a B have all the columns of the same and all the ranks are equivalent Then that means that our final solution is such that the Nell etc of be a equals finality as the which equals and nobody of a B proving our original statement now just seem out so that everything can be seen in full. That's it

Hello today have been approved that if A and B are and by our major season a convertible than the nullity of a B, people would only be, and that equals finality of the A. First, we're gonna prove that, uh, the north face of the is a subset than all space of a B. So no space of the subset of a B and we're gonna define X as the null space. Therefore, if we have a B Times X equals zero, then we can take this, multiply it by using red A n birth And because a and verbal, we know that a times a inverse equals the identity matrix. So you have the identity matrix times be thanks equals zero. And I think times the identity matrix is itself. So the times equals zero. And this tells us that B is indeed a subset of a B. Next, we want to do sort of the reverse and show that the null space of a B is contained within B. So let's do yeah, times are no space equals zero. So on the converse side, we can multiply both sides by a Let's use red for that. Go back a little bit. Must fly by a So this gives us baby X equals zero. So this tells us that the null space of a B is contained within B therefore therefore three and be have the same null space. Or, in other words, the milady of a B equals banality of B. So now let's go down and use our rank. Nobody here to try to investigate this a little deeper. So we have our rank melody here, um, using well. So if we do this for the Matrix A B, the rank of a B plus humility of a B equals the number of columns of a B and similarly, the rank let's be plus can only be equals the number of columns A B But we also know that the rank of a B going back over to the left equation the rank of a B is the same thing as writing the dimension of the column space. Oh baby. And that humility is the same as the dimension of the North space with a B. And that's the same shirt. The same is true for B, so we can therefore rewrite what we have in terms of the nullity of a B. So let's do that. So the dimension in green will be that agree. So the dimension of the null space of a B equals the number of columns of a B minus the dimension of the column space a baby. And similarly, the dimension of the column space sorry dimension of the North face would be equals. The number call would be minus the dimension of the column space of beef. Now, we proved above that no space, um, of a B and equals ethanol space of B or in other terms, that the ability of a B equals finality of beef. So this term equals you, Mr. So we can have these quick equations equal to one another such that the number of columns with a B minus the drink of a B equals the number of columns of the minus. The ranking be so we know that these are both and by n matrix meter seas. So the number of columns of a B is the same as the number of columns of B. Well, why is that? While a and by N e matrix time in and buy in the matrix again, a bee matrix is still an n by N matrix. Therefore, the number of columns a baby is equal to the number of columns of the. Therefore, the rank of A B is equal to the rank of B, so let's move a little lower and move on. So let's assume now, going back to read. Let's assume now that the rank be in the equal to the rank would be transposed Then, using been all space from before, this would tell us that a transposed be transposed times in old space. Pretty cool euro were. That's a different form of our A B basically zero. So a transfer to be transposed all space equals zero. Multiply that whole thing by the A inverse transpose fans we get be transposed of the null space equals zero. So this tells us similarly from before that be transposed shares the same doll face shares the same goal space as a transposed be transposed. They know pence be transposed and eight transposed be transposed, have the same rank for the same reason we having the proof above for why the rank of 80 equals the rank of beef. So now we have to move on. So the rank, uh they transposed be transposed could be written Such that is the rank of the A transposed which equals the rank of the A Therefore therefore, the rank of be a equals. The rank of the tea because we see here that be transposed is the same as a translator be transposed and transpose beast transpose is the same as the rank of B A for the rank of be a seems the rank of be transposed which is the same as the rank of B which then is also the same from our before Up here as the rank of a B have all the columns of the same and all the ranks are equivalent Then that means that our final solution is such that the Nell etc of be a equals finality as the which equals and nobody of a B proving our original statement now just seem out so that everything can be seen in full. That's it


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