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Problem 1 : determine if the following transformationsare diagonalizable.1: T(a, b) = (a, 2a + b)2: T(a, b) = (a + 3b, 4a + 2b)3: T(a, b, c) = (3a + 2b − c, a...

Question

Problem 1 : determine if the following transformationsare diagonalizable.1: T(a, b) = (a, 2a + b)2: T(a, b) = (a + 3b, 4a + 2b)3: T(a, b, c) = (3a + 2b − c, a + 4b − c, a + 2b + c)

Problem 1 : determine if the following transformations are diagonalizable. 1: T(a, b) = (a, 2a + b) 2: T(a, b) = (a + 3b, 4a + 2b) 3: T(a, b, c) = (3a + 2b − c, a + 4b − c, a + 2b + c)



Answers

Let $$A=\left[\begin{array}{ll}a & b \\c & d\end{array}\right]$$ Show that (a) $A$ is diagonalizable if $(a-d)^{2}+4 b c>0$ (b) $A$ is not diagonalizable if $(a-d)^{2}+4 b c<0$ [Hint: See Exercise 29 of Section 5.1.]

Were given a matrix A and the basis be where B is the vectors b one B two and were given a linear transformation. T from our tools are too, which is defined by T F X equals X. In part, they were asked to verify that the Vector B one is an Eigen vector of a but that a is not diagonal Izabal. So this is pretty easy to verify. It's an idea. Nectar of a will simply calculate a Times B one. Now we're given that a is the Matrix column. Vector is one negative 113 and that be one is the vector 11 performing matrix multiplication we get one plus one is two and negative. One plus three is also to which is equal to two times 11 which is in turn equal to two times be one. So clearly the one is an Eigen vector of a with an Eigen value of two. Now to show that is not the like diagonal Izabal we need to find the Eigen values of a So we have that the characteristic polynomial of a this is going to be the determinant of a minus. I a minus, Lambda, I I mean, so this is going to be one minus Lambda times three minus lambda minus negative one or plus one. So we get Lambda squared minus four Lambda plus three plus one is plus four. This could be factored as Lambda minus two squared. So we see that a has the Eigen value. Lambda equals two with multiplicity, too. That's its only Eigen value. Now we have that The matrix A minus two I Well, this is the Matrix. Negative 11 negative one one. This implies that the Eigen space corresponding to do the Eigen value to is one dimensional. So it follows that hey is not diagonal Izabal in part B whereas to find the bee matrix for tea. So even though we see the A is not diagonal Izabal, it is still possible to find a bee matrix 40 on an example from the book we want to take the Matrix p to be the matrix whose column vectors RB one and B two. Then we'll have that the bee matrix for tea. This is going to be PM verse ap notice that t is not necessarily diagonal here. And to calculate p inverse First notice that P is the matrix 1154 So using the formula for in verses of the two by two matrix, we have one over the determinant of P. So this is four minus five is negative ones. We have negative one times and then we'll have DE which is four, followed by negative B, which is negative five, followed by negative C, which is negative one followed by a, which is one times the matrix a a is 11 negative 13 and P is the matrix. 1514 So first multiplying a and P together and multiplying through the negative I get negative 45 one negative one times and then I'll have one plus one is two five plus four is nine negative. One plus three is too and the negative five plus 12 is seven, then multiplying again. I get negative. Eight plus 10 is too negative. 36 plus 35 is negative. One tu minus two a zero and nine minus seven is too. So I end up with the matrix to negative 102

Okay, so for problem 22 were first given Matrix K, which is diagnosed Technol Izabal. So that means by definition we can have a have any murderball metrics say ass times as he first time says is De were where these are diving a matrix. Excuse me. Now also, by the definition off similarity then were given that be similar to a So we also have, um t inverse times. Hey, times t she's bi. Of course, T has to be a murderer matrix. Now we first soft a by multiplying I'm on applying t from the left hand side of right hand side so that mrs, uh, sorry. Second question. We must have a which is tee times be I'm stealing verse. Now we put the D. C. Based compression into our first question. No, we have asked members times t hey, t inverse times asked. Okay, Now here. Uh, yeah, I I missed the Dagenham matrix D on the right hand side. Now here, uh, let's take you to be a scene first times t so that, um our yeah, so our you immerse will be t t inverse times has. Okay, So since these two matrices are all in murderball, so you has to be in veritable so we can rewrite our expression here. D to be you times B times Ewing burst. So we've shown that be, by definition, bees.


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