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Question (10 pts) LetA= [: -] Ficd all the eigenvalues and corresponding eigenvectors;...

Question

Question (10 pts) LetA= [: -] Ficd all the eigenvalues and corresponding eigenvectors;

Question (10 pts) Let A= [: -] Ficd all the eigenvalues and corresponding eigenvectors;



Answers

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{ll}2 & 0 \\0 & 2\end{array}\right]$$.

This problem asked us to find Ivan values and Aiken vectors of given matrix. You can do this by funding the characteristic polynomial, which this inviting the determinant of a minus land of times. I That would be the determinant of 10 minus lambda zero negative. Eight. Native 12 to minus lambda 12 80 96 minus Landau. Okay, solving this out. We get 10 minus lander times two months. Slammed other times a night of six months. Lambda minus eight times, two months. Lander times negative eight. And then the other two for, um, the subtraction are zero spoke due to the zeros in the Matrix Southern. This out a bit more we can get. Um well, we find that we get negative 60 months for Lambda Plus Lambda Square plus 64 0 sorry. We need to include the two months slammed up minus it becomes a plus 64 to minus slammed up, which simplifies to because you can add the 64 in yet for minus four lambda plus lander squared times two minus lambda, which is equivalent to tu minus lander. Cute. Then with that, that means that our Eigen value is two with an algebraic multiplicity of, um three. So this means that the Matrix could have or this AG value could have three Eigen vectors associated to it, but a maximum of three. So solving for the Eigen vectors, we confined this by finding anyone slammed I and the final director X that gives the zero vector so plugging in to, yeah, 10 minutes to wanna leave blanks for zeros for save time. Negative. Eight. Um native 12 to minus two 12 and then eight. Negative. Six mines, too. Times the Eigen Vector X that is equivalent to eight negative eight and 12 0 12 Negative eight Over here. Eight. When we add the zeros to this room, then with gash in elimination, if you add the top row to the bottom row, that gives us eight negative 12 8 Then zeros on both these bottom row's times X, which is equivalent to if you divide by four. That gives us two negative 32 all zeros against here times the vector A B C shaped equal 000 and without weekend fronts literally Independent solutions, ABC to the system of Earth to the Single equation to AG minus three b plus two C is equal to zero and she saw this. You'll find that the two linearly independent ah solutions you'll find from this are going to be like 101 and three 20 meaning that this adding value has geometric multiplicity of two. So this is our final solution for Agon vectors and our Ivan value is what it to

This problem gives a matrix and asked us to find Eigen values and Eigen vectors. We do this by finding characteristic polynomial which is equal to the determinant of a minus. Lambda I and set it equal to zero. That will give us the matrix one minus lambda 00 00 minus land a negative 101 negative. Lambda, Take the chairman of that will give us, uh, lambda squared times one minus lambda plus one minus lambda, which simplifies to Lambda squared plus one and one minus lambda. And we set that equal to zero, which, give us gives us that our guys are equal to one and plus or minus I In order to sell for the Eigen vectors, we need to find vectors X such that a minus lamed I times that Vector X is equal to zero vector. So first, starting with land equal to one we plugged that into a minor, slammed I and it will give us the matrix. 000 zero Negative one negative. One 01 negative. One times the erector x one. We can do gash elimination or to simplify the calculation. So what I'm going to do is add the, um, second line to the third line that gives us we'll have all zeros on top. Still zero night of 11 and then we have zero and negative 20 times x one which dividing by two on the bottom row and dividing Yeah, divided by two On bottom row, we saw zero here zero night of 11 zero 10 times X one And then one more step. I'm going to add the, um, bottom road to the second row. Sweet balls, years on top. Then 001 and 010 times a one B one C one should equal 000 Since here we see that B one and C one have to equal zero. Then we can find that are Eigen. Vector X one is equal to 100 Next, we'll do land equal to five. But land equal toe I we get the matrix one minus I 00 zero Negative. I minus one and 01 modest. I times the Eid wrecker Vector x two. We can simplify this. I'm going Teoh, multiply the, uh, second row by I and add it to the third row. That gives us one minus I 000 minus one. And on the bottom row, we have all zeros. Times A to B to C two should equal 000 then solving this system. See that a two has two equal zero and that ah b two is going to have to equal Ah, odd or so items B two is gonna equal C two. So that will give us our Eigen vector x two to equal zero I want Lastly, we're gonna do land equal to negative. I playing that in to a Muslim, I we get one plus I 00 zero I negative 101 I times x three again performing gash In elimination We can multiply the second line by I and subtracted from the third row, and that gives us the same on top seeming second round. And then we have all zeros again on the third row times a three b three C three, which is equal to 000 and solving for this system, we see again that a three has to equal zero and then we can find that B three is equal to negative. I and C three is to one. And that will give us our third Eigen vector, which is equal to zero Negative. I won, and those are our final answers.

This problem gives us this matrix and acts us to solve first Eigen values and Eigen vectors. You do this first by finding its characteristic polynomial, which is found by taking a determinant off the Matrix minus lambda times, the identity matrix that will give us the determinant of the matrix five minus slammed, uh, along the diagonal and zeros everywhere else, which I'm just gonna leave blank those roll zeros that gives us the polynomial five minus lambda cute and setting. That equal to zero gives us that our Eigen values equal to five with an algebraic multiplicity of three due to the Cube. So then we'll use land equals five. So plug this into a mind slander times I to find the I director X That gives us zero factor. So playing in land equal to five, that will give us any tricks with zeros everywhere. So, as a results, we will have in order sulfur this system. Um, this wagon space will span three vectors, and those three vectors will be our Eigen vectors. They're 100 010 and 001 of which are literally independent from each other, giving us that since our final answer

This problem gives us a matrix and asks us to sell for Eigen values and Eigen vectors. We do this by funding characteristic polynomial, which is equal to the determinant of a minus. Lambda Times I This will give us the determinant of the matrix one minus lambda zero to 03 minus Lambda. Negative too. Zero to negative one minus lambda, which gives us one minus Lander times three minus lambda times Native one minus lambda plus four times one minus Linda This will simplify out to be lambda minus one cute meaning that are Eigen values going Thio going to equal one with our algebraic multiplicity of three. In order stole for the Eigen vector We need to find a vector x such that a minus lambda I times that Vector X is equal to this year a vector So taking land equal to one We plugged us into the matrix a minus land I That gives us one minus 102 03 Minus scorn Negative 202 native one minus one times the vector X simplifies out to be 002 zero to negative 20 to negative two times x Then we can perform Gash nomination to simplify this calculation. So what I'm going to do is add the second wrote to the third Rome. It gives us 000 say same secondary stays the same under the third road comes 200 times x Don't take this extra step if you want of Dividing wrote to and Row three by two First row stays the same. Second rope come 011 and there're O'Quinn's 100 times a B C just to exceed the components. Just going equals 000 Solving out this system, we can see that needs to equal zero, and B is going to have to equal negative C s. So that would be illegal one and seal equal negative one. And that gives us the collector 01 end of one. Since this is the only linearly independent solution there being no other Eigen vectors, which just means that land equal to one has geometric multiplicity of one. And that's our final answer


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