in-exercises-13-16-define-t-mathbbr2-rightarrow-mathbbr2-by-tmathbfxa-mathbfx-find-a-basis-mathcalb-

Question

In Exercises $13-16,$ define $T : \mathbb{R}^{2} ightarrow \mathbb{R}^{2}$ by $T(\mathbf{x})=A \mathbf{x}$ . Find a basis $\mathcal{B}$ for $\mathbb{R}^{2}$ with the property that $[T]_{\mathcal{B}}$ is diagonal.
$$
A=\left[\begin{array}{rr}{0} & {1} \ {-3} & {4}\end{array}ight]
$$

Chris Trentman · Accepted Answer

So we&#x27;re given a transformation t in terms of a matrix A and were asked to find a basis be for our two with the property at the matrix of tea with respect to the basis be is diagonal and we&#x27;re given that the Matrix A in this case is the matrix with column vectors zero negative 314 This means if this is a in the characteristic polynomial of a is going to be the determinant of a minus slammed I So this is going to be negative Lambda Times for minus lambda and then minus negative three or plus three. This is the same as Lambda Squared minus four Lambda plus three which could be factored as lambda minus one times lambda minus three. So we see that since this is the characteristic polynomial, the Eigen values of a are lambda one equals one and Lambda two equals three. This is when the characteristic polynomial zero. So if lambda is equal to one well, we calculate that the matrix a minus slander I which is simply a minus. I This is going to be the matrix. Negative 11 negative three three. And so the equation a minus I times Vector X equals zero. This is equivalent to the system with the single equation Negative x one plus X two equals zero. So we have that X one equals x two, where X two is a degree of freedom. So a basis specter for this Eigen space, for example, we could take to be V one equals if we take X to be one 11 Now consider the egg in space or lambda is equal to three. In this case, we have re matrix a minus three I is given by negative 31 negative three one And we see that the mate the equation a minus three i times X equals zero is equivalent to solving the single equation. Negative three x one plus X two equals zero Or, in other words, we have that X one is equal to one third x two where X to of course is free. We have one degree of freedom. So, for example, if we take X two to be three, then we have that The vector V two, which is one third of three which is 13 is a basis Fekter For this Eigen space, it&#x27;s nice because it has thes whole numbers which are easy to work with. Now, from these two Eigen vectors, we can construct the Matrix P, which has V one V two as column vectors. And so this is the Matrix with column vectors 11 and 13 And we know that this diagonal eyes is a so by a previous the&#x27;re, um we have that the basis v one be to has the property the bee matrix of the transformation from X two x is a diagonal matrix.

Runpeng Li · Answer

Okay, so problem 16 again, It&#x27;s Ah, two by two matrix. And to find out of the Matrix, we just need to find out the, uh the young man is so that given Matrix is too negative. Six active. Wanted three. Now consider. Determine. End a minus. Lund, I He was zero. Now we have two minus lambda three minus lambda Nick of 6 to 1. No. Find a determine. End to be zero. So is to minus lambda. I was three minus lambda, minus six. So this is six minus two. Lunda minus three. Linda thus come to squared minus six, which is equivalent to Lambda Squared. Gives me minus five. Lunda, you go. Zero take out Linda. Then we have loved them. Minus five. So our first lycan battery is sterile and over. Sorry. I know I&#x27;m No. One. His cereal and our second wagon value is five. So our matrix D, which is also our bi matrix 0500

Runpeng Li · Answer

Okay, So for these problems, working work, even the Matrix A, which is five negative. Three, 27 And what? So you wanted to find the bee Matrix Matrix is just the deep dive on the Matrix, Which is he? Worse times? Eight times. So it&#x27;s just a thea. So if we re re re writers, it will be p times E times p in first. Okay, so So it&#x27;s just the wagon decomposition off matrix A. So Okay, so the first thing is to Okay, so for this problem, working work, even the matrix A, which is five negative three No, find the Eiken values. Okay, So for this problem for king work, even the Matrix A, which is five negative three elective seven and wants of once we have already. Once we have found the dragon value, then the family will be under death. We&#x27;re about to be on the diagonal off this matrix D, and this is sure to be a diagonal matrix. Then we know this d will be our bi matrix. Okay, so yeah. Okay, So for this problem, working work, even the matrix A, which is five negative. Three negative, seven and one. So you wanted to find a bi matrix matrix is just the deep dive on the Vectrix. Which is he? Worse times eight times people. So it&#x27;s just a Thea. So if we re re re write this, he will be p times digging value a minus lambda for this problem for King work. Even the Matrix A, which is five negative three elective, seven and one. So you wanted to find a bi matrix matrix is just the deep dive in the Matrix, which is he wears times eight times people. So it&#x27;s just a Thea. So if we we rewrite this, he will be p times d times pt in first, eh? To be zero now. So he&#x27;s five miners Lunda one minus lambda. It has seven and one. So you wanted to find a bi matrix matrix is just the deep dive in the Matrix, which is He wears times eight times people. So it&#x27;s just a thea. So if we re re re writers, he will be p times d times pt in first. So it&#x27;s just the wagon decomposition off matrix A. So Okay, so the first thing is to after three look. So you wanted to find the bee matrix. The matrix is just the deep dive into matrix. Which is he? Worse times? Eight times people. So it&#x27;s just a Thea. So if we re re re writers, it will be p times d times p inverse. So it&#x27;s just the wagon decomposition off matrix A. So, Okay, so the first thing is to find the icon value is subjective seven, which is serial no expended. We have my minister mda times one time said a number minus 21. So this will be lemma squared e times p in first. So it&#x27;s just the wagon decomposition off matrix A. So Okay, so the first thing is to find the Eiken values. So once we have already, once we have found the dragon value, then the family will be under death. We&#x27;re about to be on the diagonal off this matrix D, and this is sure to be a diagonal matrix. Then we know this d will be our bi matrix. Okay, so pretty I get, huh? So it&#x27;s just the wagon decomposition off matrix A. So Okay, so the first thing is to find the Eiken values. So once we have already once we have found the dragon value. Then the family will be under that. We&#x27;re about to be on the diagonal off this matrix D, and this is sure to be a diagonal matrix. Then we know this d will be our bi matrix. Okay, so pretty. I can value a minus. Some guy, um, minus six. Lunda has five minus 21 and find the Eiken values. So once we have already, once we have found the dragon value, then the album Valley will be under that. We&#x27;re about to be on the diagonal off this matrix D, and this is sure to be a diagonal matrix. Then we know this d will be our bi matrix. Okay, so pretty. I can value a minus some guy to be zero now. So he&#x27;s five miners Lunda one minus lambda, which is equivalent to once we have already. Once we have found the dragon value, then the family will be under that. We&#x27;re about to be on the diagonal off this matrix D, and this is sure to be a diagonal matrix. Then we know this d will be our bi matrix. Okay, so pretty. I can value a minus some guy to be zero now. So he&#x27;s five miners Lunda one minus lambda. Active three. I&#x27;m a squared minus. Excuse me. Uh, six Lambda minus 16 which is zero now. This is secret into Lambda, minus eight times Lunda plus two value a minus. Some guy to be zero now. So he&#x27;s five miners Lunda one minus lambda. Active three negative seven. Which is serial. No expended. We have my money. Standa Times one time said a number minus 21. So this will be lemma squared. So our Aiken values to be zero now. So he&#x27;s five miners Lunda one minus lambda Active three negative seven. Which is serial. No expended. We have my minister mda times one time said a number minus 21. So this will be lemma squared, huh? Of the one they said number two is connected to. So this gives our matrix D, which is the bi matrix 23 negative seven. Which is serial No expended. We have my man ist Amanda Times one time said a number minus 21. So this will be lemma squared. Um, minus six. Lunda has five minus 21 on DDE. That is negative. Seven, which is serial no expended. We have my minister mda times one time said that I love them minus 21. So this will be lemma squared. Um minus six. Lunda has five minus 21 and which is equivalent to aid connected to zero and zero.

Doruk Isik · Answer

Yeah, in this trouble were given a set up. So we&#x27;ll start solution by writing the organ directors for each polynomial. So for people in the quarter, Inspector Ho. But like this 101 said, the 1st 1 is the question. But these year old 2nd 1 is to one of you two. So using this Pete, too apartment that too perfect. You could be rich. Zero. What being? And for P three, that could be, um, that would be one one negative three. So then the augmented matrix phoned my bees. Corn factors would be 101 01 negative three and 11 Negative three. Now let&#x27;s write this over mental midgets in traditional echo form. And that is he could do warn. 00010001 is you can see. We have three columns and each column has a few bits elements. So three columns, three events elements at each column. So it means that these vectors Spector&#x27;s spend our fee. It means that the given Colonials no meals spin he two. Now in part, we were given 1/4 collector for a loan Jew, and that is next to 112 And whereas to find this bologna to using this So we know that cube of native one times people plus one p two plus thio three So Dunn will be negative one times ones You&#x27;re one plus one times zero wounding three waas Two times One wonder three And we find this to be one the next 10. And using this, uh, using the information that these numbers represent corruption tease. We can ride the roller milk you is one plus B t minus 10 t spring.

Amy Jiang · Answer

Okay, so we&#x27;re given land as you cook, too. Negative, too. So we have a plus two I That&#x27;s equal to 10 Negative. 11 negative. 30 for *** 13 1 plus choose. +00020002 What does that give me That gives me Freeze the whole look of 11 neighbor, Evelyn Ezell. And for negative 33. Now, let&#x27;s reduce this. Mister, this is 7 to 1. So they would have won over three one marjorie 01 They would have won over three and 000 So we have that X one physical to 1/3 x three and x 2 to 1 of three X trait that we have a doctor. It&#x27;s free. So our general solution. Okay, plus two. I got two x three times. One over +31 over three and one so we can see Aryan space for a good two. Is this over here?

Runpeng Li · Answer

okay for problem Problem 10. We have the leaner, faster mission maps from P three to Arthur are poor. Such that t of pulling nominal is equal to by taking different input to the Polo Meo. Just pee of connective three first p A connected one and p of one and p up three. Hey, So the first thing we need to show that is that, um to show this transformation is linear. Okay, so do that. My transformation. We can first check tee up. People ask you where panic you are all clean on yourself. Order to be so then, by our assumption, we can have a p rescue while we take the include to be elected three. And he ask you with the input elective born people ask you with include one and people ask you with your food three. Now, because we are adding these two point nominees so we can we can calculate the number Sweet that with this input on dad them together just just by by the rules, our tradition. So that&#x27;s he, uh, 93. I trust you collected three and PR. Negative one. Uh, you, uh, negative one and p of one less queue of one. And here, three askew on three. So by now we can separate this, uh, this vector with by a vector off Minami O P and the vector polynomial Q. So it turns out to be thio he that&#x27;s t l Q. So that&#x27;s the first part here. So the second part is considered skater Linda times people. So again, by our assumption, we still have Lunda cons. P off negative three Lambda SPF connective one, huh? Times p of one and Linda Times p of three. Excuse me. So we can take out the skater from our rector&#x27;s. So we have loved a in front of our specter and P uh, negative three. He, uh, connected one he won and ps three. So that turns out to be Lunda time. So? So we&#x27;re done for the first part because I our this relation and second relation. Then we can conclude that he&#x27;s a renewed transformation. Now, Part B. We need to find the matrix for tea relative to the basis won t t square t t cubed for p three and standard basis are for Excuse me. So now let&#x27;s, um defined business P start this is B to be given basis in our assumption, which is one two he squared and t Q. So we first calculate, uh, actually the tea with the input up won t t square and teeth cute separately. So first we have to be one which gives our vector all once and then we&#x27;re plugging t which gives Dr Inactive three negative one, one and three. Then we&#x27;re plugging our t squared. This gives Excuse me. Um 911 night, and at last we plugging our thank you. So we had a vector collective 27 negative. 11 and 27. So our metrics will be this band of these four vectors. So this will be he of this is B should be one collective three nine Next 27 one defective. 111 1111 and 139 27. And that&#x27;s it.

Problem

In Exercises $13-16,$ define $T : \mathbb{R}^{2} …

Problem 13 Easy Difficulty

In Exercises $13-16,$ define $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{x})=A \mathbf{x}$ . Find a basis $\mathcal{B}$ for $\mathbb{R}^{2}$ with the property that $[T]_{\mathcal{B}}$ is diagonal.
$$
A=\left[\begin{array}{rr}{0} & {1} \\ {-3} & {4}\end{array}\right]
$$

Answer

$b_{1}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right], b_{2}=\left[\begin{array}{l}{1} \\ {3}\end{array}\right]$

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$b_{1}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right], b_{2}=\left[\begin{array}{l}{1} \\ {3}\end{array}\right]$

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Kayleah T.

Caleb E.

Michael J.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications