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TerGalenaatBUZLEGAU branAlgeb--Ovevil"Stuius undtollon UnPAlcnamLet A be a n1 X n1 symmetric matrix Thenwhich of the following not true?Ais diagonalizable: b) ...

Question

TerGalenaatBUZLEGAU branAlgeb--Ovevil"Stuius undtollon UnPAlcnamLet A be a n1 X n1 symmetric matrix Thenwhich of the following not true?Ais diagonalizable: b) If 1 is an eigenvalue of A with multiplicity k, then the eigenspace of A has dimension k Some eigenvalues of A can be complex. All eigenvalues of aFC real:

Ter Galenaat BUZLE GAU bran Algeb-- Ovevil" Stuius undtollon Un PAlcnam Let A be a n1 X n1 symmetric matrix Thenwhich of the following not true? Ais diagonalizable: b) If 1 is an eigenvalue of A with multiplicity k, then the eigenspace of A has dimension k Some eigenvalues of A can be complex. All eigenvalues of aFC real:



Answers

Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ be a real matrix. Find necessary and sufficicnt conditions on $a, b, c, d$ so that $A$ is diagonalizable - that is, so that $A$ has two (real) linearly independent eigenvectors.

Okay, so for problem, Time to work, even of compacts matrix. And we're given the argon victor acts and the wagon value Lunda. So we want to show that mute times eggs where Mew is a complex scaler is a mug in victor off. So first we know because this is Ah, this is actually sent again. Vector and lumber is having a very So we have eight times x is London Times X. Now we need to consider the vector u x so here. Now, we have eight times more bags since mu Excuse me since Marissa skater we can put me we brought away. So we have you times a X but we know a access equal to numb Dax. So we have from you times lambda. Thanks. And hiss. If we take let's say Alfa to be be immune times, Lunda. And why to be the new vector off leaving time sex. Then we have a relationship like this eight times. Why? Because Alfa climbs Why? That means health is I can value uh, a with again victor. Why? So that means mu times acts. He said, not give Victor. Okay,

Okay, so we are given. That's the determinant. Ah, mhm. Minus lambda. I is equal to Lambda. One minus lambda landed too. My Miss Lambda And so forth until Lambda in minus Lambda. And we can see that if if Lambda equals zero than the determinant of a because the land of eyes just going to be the zero matrix is going to be equal to Lambda one minus zero times when the tu minus zero all the way until land a n minus zero. In other words, this is equal to land the one times land a to times all the way until land the n which happens to be the product of the Eigen values. Yeah, we're done.

Were given a matrix A A has elements to 213 Yeah. Well no that's in part they were asked to find all the item values and correspondent Eigen vectors of this matrix. Well, first we'll find the characteristic polynomial of A. So delta T. This is because this is a two by two matrix T squared minus the trace today. Oh okay. Times T plus the determinant today which is T squared minus five T plus four Which you can factor as T -1 times T -4. Now the roots lambda equals one and landed equals four of delta T. These are in fact the Eigen values of our matrix egg. Now let's find the correspondent Eigen vectors. So first we'll find the ones corresponding to lambda equals one. We'll subtract one down the diagonal of A. So we get the matrix M which is a minus, I beat a black man with it. And the corresponding homogeneous system m times X equals zero. Will yield the Eigen vectors belonging to land equals one. We have the matrix M is 1, 2, 1, 2. And to the corresponding homogeneous system is X plus two, Y equals zero, X plus two, Y equals zero. Which of course is just the same as X plus two, Y equals zero. So we have one degree of freedom and one of the non zero solutions is if we take X to be too and Y to be negative one. So for example, V one, the matrix two negative one is a non zero solution and therefore it's also an Eigen vector belonging to Eigen value lambda equals one. Now consider the Eigen value land equals four or subtract this down the diagonal of a. To get the matrix M. This is M is a minus four. I which gives us the matrix negative 2 to 1, negative one. This corresponds to the homogeneous system negative two, X plus two, Y equals zero and x minus Y equals zero, which corresponds to the system X minus Y equals zero, which has only one independent solution. So for example, you can take X and Y to both be one and therefore the two, which is the vector 11 is an Eigen vector. And then they crossed that belonging to the iron value. Lambda equals four. All right, then, in part B, we were asked to find a non singular matrix P such that the matrix D, which is P inverse times ap is diagonal, and also find the matrix p inverse, which we know exists because P will be non singular. Well, but PB the matrix whose columns are are Eigen vectors V one and V two. Then P is the matrix two negative 111 and the matrix D, which is P inverse. Ap Well, this is simply the matrix whose diagonal entries will be the Eigen values corresponding Megan values. One and four. Now mm key inverse. We can find simply using our formula for the inverse of a two by two matrix. So the determinant of P is two minus negative one, which is three. So we have one third times one -1, 1, 2, which is one third negative one third, one third two thirds. I love that. The answer. Just the right thing. Karma police in part C. Whereas to find A to the sixth power and F of a where F is given by the certain polynomial out of F. F T equals T to the 4th minus 32 huge minus 60 squared plus 70 plus three. That's nice. That's the type of rodeo. We'll use a diagonal factory Ization for a so we have that 8-6. Well, rearranging This is the same as uh p. d. p inverse to the 6th. Which we know is the same as P times d. to the 6th times p inverse. And plugging in. This is uh to one negative 11 times 100 So one of the fourth is 100 And then uh sorry, four to the sixth is 4096th. Real rodeo. Queer. Yeah, um mm Times the matrix. Uh One third negative. One third. One third, two thirds. And if you carry out these matrix multiplication, you'll get 1366 2230 1365 And 2731. Right? So this is the value of the age of the cysts. Now look at our function F. We see that f of one is equal to one minus three minus six plus seven plus three, which is positive two and F of four on the other hand, well this is negative one and therefore we have that. They once again, this is the same as F of P times D times p inverse, which is the same as P times F of D comes p inverse which is equal to P, which is the matrix to one negative 11 times F of D. Which is well, it's a diagonal matrix. We simply evaluate the diagonals that. So F of one which is 200 and F of four which is negative one times P inverse, which we found to be one third negative one third, one third and two thirds. And this simplifies after matrix multiplication, 2, 1, 2 negative, 1, 0. Shit. I've seen it happen literally three or 4 times that's F. Of a. Finally in part D were has to find a real cube root quote unquote of B. Which is well defined to be a matrix B. Such that be cubicles A. And B has real item values list about a half a dozen. Mhm. Put it out for I want them so much. He keeps saying birthday boy. Yeah. Happy birthday. Well we see that the Matrix 100 Hubert or four. This is a by our definition real Q. Root of our matrix D. Since clearly the diagonal matrices is just the matrix whose diagonal entries or products. And it has really good values. And therefore by rearranging a real key brute of A. E. Is equal to our matrix P times uh Eight times p inverse that were taking the key brute of it. So this is key times again distributing the power, the cube root. Sorry? So initially D so cube root of D tends to the universe. And so this is equal to matrix P, which is 21 negative 11 times 100 cubed root of four times one third negative one third one third two thirds. And if you evaluate this matrix multiplication helps if you pull out a third, if you get one third times two plus the cube root of four negative two plus two times the cube root of four negative one plus the cube root of four and one plus two times the cube root of four came up with. And and this is the real cube root of a or a real keeper today.

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.


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