use-a-property-of-determinants-to-show-that-a-and-at-have-the-same-characteristic-polynomial

Name: use a property of determinants to show that a and at have the same characteristic polynomial
Uploaded: 2019-08-07
Duration: 39 s
Channel: Numerade
Description: Use a property of determinants to show that $A$ and $A^{T}$ have the same characteristic polynomial.

Question

Use a property of determinants to show that $A$ and $A^{T}$ have the same characteristic polynomial.

Amy Jiang · Accepted Answer

Okay, so it&#x27;s that determine it&#x27;s a minus for them that I and take the transpose of inside do get terminates of a minus Lambda I transpose. What does that give me? That gives me determinants of a transpose minus lambda I currents pose, but I transposed in this eye so they get the 10 minutes of 80 is minus. Sam, don&#x27;t buy.

Robert Daugherty · Answer

Section 32 Problem number 53. So in this section they give me a being an inverted matrix and a B is you go to a C they want to use to terminates to show that b is equal to see. I think this question is a little bit, you know, formed, um, to show that b is equal to see if I&#x27;ve got a B is equal to a C. Um, I just multiply on the left side, a inverse since as an in vertical matrix. And so what this gives me is the identity matrix Times be is equal to the identity matrix. Dempsey therefore, B is equal to see. So that&#x27;s how I get the matrices convertible. If I used the term in its I can show that that a terminated B is equal to see. So with the determinants, I know that if this is my condition, I know the terminate of a B must equal the determine it of a C. And this is the determinant of a times. The determinant of B is equal to the determinant of a Thames. A determinative See, since a is in vertical. I know that this first determinant of a is not equal to zero. So that means that the determinant of B is equal to the determinant. Oh, see.

Lucas Finney · Answer

for this problem. We wants to prove to statements. When is that given a matrix? A. That eight times its transposed is a symmetric matrix. The second is that given Major sees a, B and C that the transpose of the product of those three matrices is equal to see transpose sense be transposed onto a transpose. So first looking, answering part A. We know that using the index definition of the Matrix product that a times a transposed the I Jeff element of that is going to be the some from K equals one upto end, assuming it&#x27;s a m by n matrix of elements ai que times the k j element of a transpose. But we know from definition that for a transposed, it&#x27;s i J element is equal to the J. I&#x27;ve element of a so setting that as an aside, that means that we can rewrite this sum up above some from K equals one up to end ai que as so we need to reverse the indices that we have here. So it&#x27;ll be a i k Times element a J K. Now, the next thing that we can dio, you can recognize that these two elements are just scale er&#x27;s unlike matrices scale er&#x27;s do commute so we can reverse their order there. Hagel&#x27;s one up to end of a J K A. I k. And this here is the same thing as of the J IFE element of a times a transpose that&#x27;s proving the first statement Now from be here we have a times, b, times C and we want to find the i J element of its transpose. So the idea of element of the transpose of ABC first of all, going to be same thing as the J IFE element of a times B times C the J I have elements of a times B times C so we&#x27;re going to have to do a do two consecutive sums here. So, first of all, we want to have some from K equals one up to end of a I K. Sorry. In this case, since we have the indices reversed, that should be a um yeah, sorry. That should be a J k times the element A B c element k i BC element K I we can write as second here that element there. It&#x27;s going to be the some from l equals one up to n both these air up to end Still of be elements K l times see Element Uh, l I now we can bring the sums to be next to each other equals one upto end times These some from l equals one up to end So you have a double summation there a j k b k l c l i. So now what? We can dio way we recognize him. I know I didn&#x27;t got a little bit lazy with using the square braces, but these air referring to the elements of the matrices we can recognize is that each of these elements they&#x27;re being multiplied together. Those are all scale ear&#x27;s so we can rearrange the order of that multiplication cake was one up to end times some from l equals one up to end of C L I B k l A J K. The next thing is that these elements here no element cli. Actually, I&#x27;ll write my summation Summations here with the limits of the summation unchanged CLI is the same thing has element I l of C transpose and B k l is the same thing as element. Okay. Oops. That should be for be transposed there. Moment l k times a transpose element k j which is equal to I will say, that was K. That was l some over k of C transpose I l Actually, I&#x27;ll use l as the outer summation here some over l of see transposed element I l times the some over k of be transposed element L k times a transpose element KJ, which is equal to the sum over l of C transpose element I l times second here be transposed a transpose element l j which is equal to element i j of c transposed be transposed a transpose, thus proving the second statement.

Robert Daugherty · Answer

so three to problem number 26 were given a three bathroom matrix, were asked to find its determinant and then use properties of determinants to find the determinant of the inverse and negative three times the original matrix A. So let&#x27;s just go ahead and do, ah, row reduction to make this work. So to clear this three, you could do negative three time Row one and add it to wrote to. So to find this determinate, you&#x27;re going to have one minus 12 and then negative three plus 30 negative three times negative won. His three plus one is four negative three times to his negative. Six plus four is negative two. Now I need to clear this particular element here. So in order to do that, I can multiply negative 1/4 times a second row So one minus one to 04 minus two zero. Multiply that four by negative. 1/4 you get negative one at it to the bottom row you get a zero multiply negative two times negative 1/4 um, and you get 1/2 1 half plus three is seven halves. So the determinant is going to be one times four times seven house, which is what, 28 over to which is 14. So the determinant of A is equal to 14. We know that the determinant of a inverse is one over the determinant of a So the determinant of a inverse it&#x27;s gonna be 1/14 and we know that, um, the determinant of minus three A. It&#x27;s gonna be minus three to the n. I think that a little bit clearer minus three to the end times the determinant of a were in is the dimension of this matrix. So in our case in is equal to three. So this determinate is going to be negative. Three cubed times 14 which is negative 27 times 14. Multiply that out and this is negative 300 and 78. So again find the determinant of the original matrix used to terminate properties. To find the determinant of these two variants

Nicholas Barvinok · Answer

we have the Matrix, a B zero seeing call it A with the property that is not see we want to show that day squared equals P D squared p inverse, where d is a diagonal matrix with the Eiken values that Anna&#x27;s Dagnall and P is a matrix with the eye in vectors of a condom in his columns. Well, we note that if a if a equals p d p inverse a squared, multiplying by the same thing again gets us exactly p d squared p inverse. So it suffices to show that a equals p d the inverse. In fact, it suffices to show that a equals PDP universe where D is just some diagonal matrix, and that&#x27;s it, because then automatically, it will d will have the Eiken values of a on It&#x27;s Diagonal and P will be made up of column vectors being the Eiken vectors of a. So To do this, all we need to show is that is diagonal izabal, or that which is to say that a has too, because the two by two matrix too linearly independent Eigen vectors, for which it suffices to show that a has two distinct wagon values. So let&#x27;s find the Eiken values of a by taking determinant of a minus lambda times. The identity this determinant is a minus. Lambda C minus lambda. The ion values were the roots of this equation, so we can see the roots are land equals A or C is not equal to see, so these are two distinct Aiken.

Nez Nikoo · Answer

Okay, Um, we have a three by three matrix matrix A I wrote here that matrix a right. Here&#x27;s a one a two, a three and transport of the matrix. We have to prove that the determinant of thes two are equal. So the way you do it is basically, you should know what the transport of the Matrix is, if not, go to the previous of videos. But basically, you get the row making my column, and vice versa on you could see this year. So I&#x27;m gonna find a determinant of this matrix. So I&#x27;m going to just repeat a one. The one see one here A to me too. See to. And if I go through finding the three by three determinant, I&#x27;m gonna basically right here is going to be a one be to C three plus a to B three C one plus a three b one see to minus a three. The two see one minus a one b three see to minus a to B one, C three. So that&#x27;s a determinant for for a for transports were going to do the same thing here. We&#x27;re going to write a one A to a three b one be to 33 and the determinant for this is going to be a one B two c three. You see already there&#x27;s a pattern, right? Plus the one see to a three plus See one A to maybe three minus See one. Me too. 83 minus a one C two 33 and minus B one a to C three. So in order for this to to be equal, all the elements, all that has to be equal. So as you see here, a to B to C three is the same as a to B to C three, the B one. No different color B one C two a three is the same as B one. See to a three with these two. Are you cool? And then see one a two b three c one a two B three. Right here a three. B two c one one goes here Well, changing color again. A one B three c two. These two are equal and B one C three a two piece to vehicle. So what that tells you is that a We prove that a determined over A is equal to determine it off a transpose when the Matrix in this case isn&#x27;t three by three. Okay, that&#x27;s it. A little messy, but actually very good problem. Thank you.

Problem

In Exercises 21 and $22, A$ and $B$ are $n \times…

Problem 20 Medium Difficulty

Use a property of determinants to show that $A$ and $A^{T}$ have the same characteristic polynomial.

Answer

see the proof

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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see the proof

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Heather Z.

Kristen K.

Joseph L.

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

Linear Algebra and Its Applications