explain-why-a-2-times-2-matrix-can-have-at-most-two-distinct-eigenvalues-explain-why-an-n-times-n-ma

Question

Explain why a $2 	imes 2$ matrix can have at most two distinct eigenvalues. Explain why an $n 	imes n$ matrix can have at most $n$ distinct eigenvalues.

Rakvi . · Accepted Answer

on N by N Matrix. So you have a 11 and then you have a lemon. And similarly, you have B and limb and e and and everything that was in between. Okay. Uh OK, so these are not These are that&#x27;s just your key. And then you have lander times I and you&#x27;re subtracting this room, which is just lambda all the way up to London and times that them zero zero, everything else is you. Okay, so this matrix is just a 11 minus lambda, and then you have 812 even. And and then you have a one. A tow minus. Lambda it and I should have bigger bracket. But I think you get the idea and then eight and minus one. Okay. And you&#x27;re solving you take the determinant of this and million matrix, and then you&#x27;re solving for Lambda. Okay, so when you dated, return on what is the leading home? The leading dome or the term of the highest degree? Okay, so you would take product off, so you will take product off. All these storms died totems. They will come in your characteristic for a normal, and then you&#x27;re solving it trade. So term off. The highest degree will be minus one to the power and Langer. And okay, plus all the other times. So you&#x27;re characteristic polynomial is off Big end. Okay. And a degree of polynomial can have at most end this thing routes That&#x27;s for never Did he remove algebra? Okay, a degree. And paranormal cannot have more than investing groups. And these rules are basically your Eigen values. So when so when n million metrics cannot have more than end distant back and values, OK, And if you taken, do we do, then you to weigh two matrix can have at most two distinct taken village, OK?

Sam Sohn · Answer

Okay, so we are given. That&#x27;s the determinant. Ah, mhm. Minus lambda. I is equal to Lambda. One minus lambda landed too. My Miss Lambda And so forth until Lambda in minus Lambda. And we can see that if if Lambda equals zero than the determinant of a because the land of eyes just going to be the zero matrix is going to be equal to Lambda one minus zero times when the tu minus zero all the way until land a n minus zero. In other words, this is equal to land the one times land a to times all the way until land the n which happens to be the product of the Eigen values. Yeah, we&#x27;re done.

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.

Heather Zimmers · Answer

in this problem. We&#x27;re talking about how a row echelon form is not unique. And the easiest way to explain why is just to show it with an example. So let&#x27;s take a simple matrix that&#x27;s in row echelon form. Suppose it has elements. 123 012 So that&#x27;s in Rochel. Inform and suppose I decide to take it further into reduced row echelon form. So what I&#x27;m going to do is take twice row two and subtract it from row one. So I&#x27;m gonna write that as negative two times row two plus row one, some changing room one. So that&#x27;s going to give us 10 negative one for the first row and 012 for the second row. So these air, both in row echelon form one, happens to be in reduced row echelon form. But technically they&#x27;re both in row echelon form, so they look different, but their solutions are the same. And this is just one of many, many different examples that person could come up with to show multiple Rochel on form matrices have the same solutions

Sneha Ravi · Answer

So this question. We&#x27;re asked to explain why two by two matrix will not have an inverse if the roles for columns are all zero. So let&#x27;s take the case where one of the columns is zero. So if we have a column being zero, the determinant, so let&#x27;s call this matrix m one. So the determinant off metrics M one is zero right, which is a T minus B c. Same way if we have one of the rules being zero and we call this matrix them too. The determinant off M two, which is a D minus species, equals to zero. So we know that the inverse off the Matrix. So let&#x27;s call this matrix two by two matrix m. So the inverse of the matrix is given by the Formula One over determinant off the matrix T minus B minus e mhm. And we basically see that if one off our rows or columns, columns or rows zero, we see that the determinant off the matrix is zero. And so if the determinant is zero denominator, here is zero and one hour zero does not exist. And so that&#x27;s why The Matrix will not have an inverse So just to write this out, if column air rules offer two by two matrix and chains, well, zeros, then the determinant off the Matrix is zero. And if determinant is zero and the inverse of the Matrix does not exist and it is who? Singular matrix. So that&#x27;s why two by two matrix will not have an universe if the columns or rows contain zeros, because the determinant zero and when the determinant zero inverse does not exist. And that&#x27;s the answer to that question.

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

Problem

Construct an example of a $2 \times 2$ matrix wit…

Problem 23 Hard Difficulty

Explain why a $2 \times 2$ matrix can have at most two distinct eigenvalues. Explain why an $n \times n$ matrix can have at most $n$ distinct eigenvalues.

Answer

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

Linear Algebra and Its Applications

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Linear Algebra and Its Applications

Lily A.

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Vectors Intro

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

Lily A.

Anna Marie V.

Heather Z.

Caleb E.

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

Linear Algebra and Its Applications