in-exercises-9-16-find-a-basis-for-the-eigenspace-corresponding-to-each-listed-eigenvalue-aleftbegin

Question

In Exercises $9-16,$ find a basis for the eigenspace corresponding to each listed eigenvalue.
$$
A=\left[\begin{array}{ll}{5} & {0} \ {2} & {1}\end{array}ight], \lambda=1,5
$$

Dharmendra Jain · Accepted Answer

allowing once. On this given question, the value off a metrics is 50 to 1. Where to find the value of agony. Space for the Lambda Value one and five. The first of all, for Lambda One, the question is a minor size. It was did you hear? I used Identity Matic identity metrics. So it becomes five minus 1201 minus one. It just request to 5. 34 0 to 0. Documenting metrics will be Jesus! Jesus! Right, So here we can see that. Excellent. It&#x27;s a quest to judo and x two is free. So the solution can be written in the form off X equals two x two Judo one. Here. This metrics. Jovan, is I gonna space from the definition? Okay. Yeah. Next we&#x27;ll solve for Lambda is it was 25 So the equation will become a minus five. I is a quest of judo. Hence five minus five, 201 minus five. Which is a cluster judo. Judo do minus four. Next step is to write documented argument and metrics. That is judo 20 minus four. Gino Gino Has this leads to the equation two X one minus four. Next Tuesday it was 20 or X 26. Questo works, too. Yes. The solution can be written in the farm off X equals two x two no one here. This metrics to one is dying in space full for Lambda value of five.

Amy Jiang · Answer

Okay, let&#x27;s start with Plan B is equal to one. We have a minute. I that&#x27;s just equal to 64 native, three and negative be writing are augmented matrix We gets 64 general, another three to connect to zero. That reduces to one to over 30000 So our solution, or actually, excellent. It is equal to over to over three x two, and we see that X two is free. So argon space corresponding to Lambda is equal to one is looking to over three and one. Okay, Now let&#x27;s look at Amanda&#x27;s equal to five. We get stay minus five. I is equal to 24 negative three. And the Group six. Okay. What is this? Releases? 212 zeroes up. So we can say that our idea of space for landed a good five would be negatives. Two and one

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?

Amy Jiang · Answer

Yeah. Okay. So let&#x27;s compute anyone. It&#x27;s for I. It&#x27;s 10 nine for another two minus for girl so forth. That gives me six negative nine for negative six. Okay, if you don&#x27;t make it. And just so he gets six negative nine mil for negative six. So Okay, so we have that X one tentacle to actually be over two x two. So you can say the X two is free. So general solution X one x two is equal to three over two x two and x two Pull out our ex too, to your next two times. Three over two and one. This is Ah, Are you in space?

Jim Van Der Valk Bouman · Answer

or it&#x27;s so in this question, were asked to find that the basis for Dragon space off a Metrix a course falling through the given I can value lender instead. Now the dragon space off, eh consists of those vectors v say bees x y such that eight times being it&#x27;s the same as value London country. So in our case, Lambda Legal Stand. So what this means is death for minus two minus tree nine times Liken vector Next way is the same as dragon value. Then times it&#x27;s better this way. So we have multiplied this matrix with the vector X y and what we will get is that&#x27;s, uh, four times X minus two times why instant approval and in the bottom row will give us minus three times X less nine times. Why? So this has to equal, uh, just then times after x y one of the road of a fetus, then X, you&#x27;re being rude. Then why? So these are the, um, patricians alive after to be in the wagon space so we can solve these equations and see what the requirements are next on. Why are so if you bring all the xs through the left side, another wise to the right side. What we get on the dobro is that minus six times X Master equal two times why? And on the bottom row we find that mine of three times X. That&#x27;s what equal just one thing. Why what we see is at the top row. It&#x27;s actually the same requirement, the same equation that&#x27;s the bottom row, just multiplied by two. So this in practice only gives one requirement on X and y Maybe that&#x27;s minus three times X. That&#x27;s what you call why. And if this holes for vector, X y and of this in our own space. Um so to find the basis for his ID in space, we have to find one example off such a vector that satisfies this question on then any multiple that back there were also lying died in space. So it&#x27;ll be a basis. Um, so there&#x27;s multiple correct answers, but one easy example would be after miles one, three. This this advice, our equation. And so it lice it. I guess base, um, it is in fact, the basis for our space. So this is the answer. Various minds 13

Rakvi . · Answer

So we want to find basis for Reagan spaces Forgiven values off lander. Right? So just street off the bed. Let&#x27;s scale clip. Let&#x27;s stay agree to be extra next to extreme And for any Lambda just lets us calculate what these what? This equality gives us the conditions for extra, extra extra. So our is or one on our last rule is is x one x two x three on this is Lambda explode Lived a extra No, no XLT. Okay. No, The first equation is just four times x one less x tree. It calls Landra him sex room. So this gives us that extra is a quartile Lambda minus four games X one. Okay, second equation is negative. Two times X one close extraordinaire equals under dames. Extreme. This gives us negative door dames. Exxon, is it quartile? Lambda minus one times extract. Oh, okay. And the third equation is negative. Two times x one. My 63 equals lander times extreme. This gives us negative two times. X one is a quarto lambda minus one time 63 So combining these two equations combining the 2nd and 3rd equation. You can see that. Okay, so let&#x27;s plug in specific values off land and see what we get. So when Lam Days equaled one first equation tends you that extreme sick weirdo Negative three off excellent and minus two. Excellent is zero. So that gives you X one is zero and tired. Equation also tells you that. Excellent. Zero. Okay, so if excellent zero the next 30 Okay, on extra can be anything. So you&#x27;re Agon. Victor. Look like zero extra deal. So one basis basis, which consists off. One element for this is, uh, Zito. Okay, so that&#x27;s your case for Lambda equals one. So the second cases when Lambda is equal, Toto. Okay, so when lambda is in Quito this go back to our equations. First equation tells you that X three is equal to negative toe affects. Want if you look closely at Equation three that tells you that negative to off X one is equal to x tree, which is basically our first equation. OK, so 1st and 3rd equation gives you the same thing. And then the second equation is that negative bull affects one has a little extra. So you&#x27;re Agon. Victor&#x27;s look like Well, you have extra and the next reason called Works too. And then Excellent. Is it, Gordo? Negative. Extra divided point. Okay, so one basis would be like basis. That consist off. One element can be negative. One over to one and one. So a third and final cases when Landry&#x27;s accord with three. So when land is equal to three, just look back at our equations. So the first equation is simply XT resicore do negative off x one. Uh, negative to off X one is a Kordell two off extra and third equation is negative. Two off X one is a core do two off extreme with the same as the first equation. And the second equation tells you that negative. Excellent. His acquittal fixed. Okay, so here is your excellent extra sick or the negative affects on an extra Z quarter. Negative. So basis is a single victor. And you can take this Toby one negative one. Okay,

Problem

In Exercises $9-16,$ find a basis for the eigensp…

Problem 9 Easy Difficulty

In Exercises $9-16,$ find a basis for the eigenspace corresponding to each listed eigenvalue.
$$
A=\left[\begin{array}{ll}{5} & {0} \\ {2} & {1}\end{array}\right], \lambda=1,5
$$

Answer

Eigen space for eigenvalue $\lambda=1$ is $\left[\begin{array}{l}{0} \\ {1}\end{array}\right]$
Eigen space for eigenvalue $\lambda=5$ is $\left[\begin{array}{l}{2} \\ {1}\end{array}\right]$

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

Linear Algebra and Its Applications

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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Eigen space for eigenvalue $\lambda=1$ is $\left[\begin{array}{l}{0} \\ {1}\end{array}\right]$Eigen space for eigenvalue $\lambda=5$ is $\left[\begin{array}{l}{2} \\ {1}\end{array}\right]$

Linear Algebra and Its Applications

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Eigen space for eigenvalue $\lambda=1$ is $\left[\begin{array}{l}{0} \\ {1}\end{array}\right]$
Eigen space for eigenvalue $\lambda=5$ is $\left[\begin{array}{l}{2} \\ {1}\end{array}\right]$

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

Linear Algebra and Its Applications