a-is-a-7-times-7-matrix-with-three-eigenvalues-one-eigenspace-is-two-dimensional-and-one-of-the-othe

Question

$A$ is a $7 	imes 7$ matrix with three eigenvalues. One eigenspace is two-dimensional, and one of the other eigenspaces is three-dimensional. Is it possible that $A$ is not diagonalizable? Justify your answer.

Samuel Goyette · Accepted Answer

Okay, so we have a matrix A which has three again values all in the one and two and three and the first Eggen Spaces of Dimension to and the second Eggen spaces of that mentioned three. Or this sir, we only know that there&#x27;s one for sure with dimension to one for sure Would I mentioned three? I took the liberty of labeling them first at second. It&#x27;s not really important. And we wonder, is it possible that this matrix is not dying on advisable? And the answer is yes. It is possible. It is possible that a isn&#x27;t dia gone uh, realizable. So why is why is it But why is it possible that it isn&#x27;t diagnosable? Because if the let&#x27;s say that the third again space is of dimension to then we would have that dimension of the first in space close the dimension of the second Dragon space was that I mentioned of the third Dragon space. Okay, is two plus three plus two, which is seven a is a seven by seven matrix and we have seven literally independent again vectors. Therefore it is diagonal Izabal. So if we have this, then a is Ah diagonal sizable. But let&#x27;s say we have the something different. Let&#x27;s say the dimension of the third hag in space is only one. So for the third hag in value, we only get it. I mentioned one for the high in space. So in that case is the dimension of the first haggans spay. There&#x27;s the dimension of the second hag in space. Trusted, I mentioned of the third haggans space is two plus three plus one. Then we get six For the total I mentioned. That means we only have that we only have Ah, six. Then we&#x27;re the independent hag in Victor&#x27;s and a is seven by seven. Therefore, because six six is not equals two Selves six isn&#x27;t equal to seven. Pretty obvious. A isn&#x27;t Diagon Izabal. So, yes, it is possible that this maitresse is not diagonal Izabal. If the dimension of the remaining haggans spaces only one then our matrix isn&#x27;t I noted Sizable

Samuel Goyette · Answer

Okay, so we have Ah, a more theoretical question. So is a four by four matrix with three in value. The first Hagen dead Agon space associated with Linda one is a dimension once Orlando one has a multiplicity of one. The second Hagen value has ah, multiplicity off to And third value is isn&#x27;t set. So we don&#x27;t We have no information given. And the question is, is it possible that is diagonal sizeable? And the answer&#x27;s yes. Uh, it is possible that a is they&#x27;re gonna izabal that Ah a is diagonal Izabal because ah, one a day&#x27;s third hag in space Uh, I in space must have dimension. What must have dimension once is the other hygiene space have dimension one and to and we know that one and that I mentioned of each wagon Space must be at least one. But since a plus two equals three and four, the that means an a is fall by four that means three So this is the dimension of the first against based I mentioned of second hike in space. The dimensions of the third hike in space must be one since it This whole thing cannot be greater than four. And the dimension cannot be lesser than wanted, at least one and everything. And this equation here must not be greater than four. So this is the dimension of so I in space because it cannot be greater than one or cannot be lesser than one. So it has to be one instance the dimension of the 1st and 2nd and 3rd haggans space equal four, which is that dimension off matrix dimension off A. That means that a has for generally independent hanging victors, therefore okay, is good.

Amy Jiang · Answer

Okay, so we have that. There are 21 dimensional Agon spaces, so some of their dimensions is too. But our magics a is to be my dream. And based on this is that is not the egg. No likable.

Samuel Goyette · Answer

Okay, so we have a matrix A with two again values the one age in space of Dimension three and one again. Space of the I mentioned to what we want to know is a is a diagonal Izabal is a dyke on eyes. Ah, dancers that Yes, and we&#x27;re gonna prove it right now. So a is diagonal sizable. I&#x27;m just going to write this, but it&#x27;s not for diagnosed for diagonal. Sizable If we can write a in the form of p times d times P minus one with p, of course being an in vertebral matrix. Andy being a diagonal matrix of non zero entries. Let then no one and I&#x27;m going to be the A and values of a and let V one to V five be the egg and vectors of Hey, so how do we know that there&#x27;s five angle again? Vectors was pretty simple because we know that the first a green space is of dimension. And three So Dimension three, which means that it contains let&#x27;s save Iwan retool And the three now it could It is our dressed name. So you could write V three before be five. You want G 35 between before five. It doesn&#x27;t matter. They&#x27;re just name and the second Eggen Space contains V four and the five. Because it&#x27;s a dimension to so we know that there&#x27;s four. There&#x27;s five again values. There&#x27;s 45 again devalues five again vectors so we could right a like this. Ah, v one. So the first Colin is the one. The second Colin is V two, and the fifth column is V five. So that would be our matrix p. And how do we know that P is in vertebral? Well, it&#x27;s because ah vectors V one and V one v two V three v for five are the literally independent because their basis for each of their against face. So we it when you have a set of five men really linearly independent vectors and you put them in a matrix of Will you put them in the matrix five by five, as columns matrix with and then we&#x27;re in mineral e independent columns is convertible. Therefore, we know that P is in vertebral. And what should we put and d? Well, let&#x27;s say that we just have to say that Lambda wants fits with the first Hagen space. And then the two is the the pagan value for the second Hagen space. So we will just have a diagram matrix with five entries around. The war will appear three times, and we have them to which will appear to time. That matrix was if it&#x27;s small and we will have a bunch of zero uh, outside off, dead tired I know, uh, exhibit agree. But you should never right, Timmy tricks like this you you should just write the DIA guards and three. So Harry added, you have We know that a is diagnosable because it has five distinct Eggen vectors which are literal mineral e independent and you can write it in this four, so ap times D times P minus one

Rakvi . · Answer

So we just multiply the big the vector with the corresponding matrix and see if this is equal to a multiple of this vector or not. I love matrixes this and our victor is. And if you multiply, we get minus three times one, which is minus three plus four and second entries minus three Danish one plus 32 it turns for and this is a quarter to one. And the second injury is 29 which is not equal toe multiple off the vector one for so 14 is not an Eigen Victor for this metrics.

Amy Jiang · Answer

Okay, so let&#x27;s multiply. 367337565 With our given value one negative to one. You can see this is equal to *** too. For negative too. That&#x27;s just a multiple of our first, like, two times, 1 to 1 So we can see that this isn&#x27;t I connected, and we haven&#x27;t even value of

Problem

Show that if $A$ is both diagonalizable and inver…

Problem 26 Easy Difficulty

$A$ is a $7 \times 7$ matrix with three eigenvalues. One eigenspace is two-dimensional, and one of the other eigenspaces is three-dimensional. Is it possible that $A$ is not diagonalizable? Justify your answer.

Answer

cannot be diagonalizable.

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

Linear Algebra and Its Applications

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cannot be diagonalizable.

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications