in-exercises-21-and-22-matrices-are-n-times-n-and-vectors-are-in-mathbbrn-mark-each-statement-true-2

Question

In Exercises 21 and $22,$ matrices are $n 	imes n$ and vectors are in $\mathbb{R}^{n}$ . Mark each statement True or False. Justify each answer.
a. The expression $\|\mathbf{x}\|^{2}$ is not a quadratic form.
b. If $A$ is symmetric and $P$ is an orthogonal matrix, then the change of variable $\mathbf{x}=P \mathbf{y}$ transforms $\mathbf{x}^{T} A \mathbf{x}$ into a quadratic form with no cross-product term.
c. If $A$ is a $2 	imes 2$ symmetric matrix, then the set of $\mathbf{x}$ such that $\mathbf{x}^{T} A \mathbf{x}=c$ (for a constant $c )$ corresponds to either a circle, an ellipse, or a hyperbola.
d. An indefinite quadratic form is neither positive semidefinite nor negative semidefinite.
e. If $A$ is symmetric and the quadratic form $\mathbf{x}^{T} A \mathbf{x}$ has only negative values for $\mathbf{x} 
eq \mathbf{0},$ then the eigenvalues of $A$ are all positive.

Jack Chen · Accepted Answer

case in this program, we have five difference. They&#x27;ve been the world who figure out their true for us. So a the state Manet&#x27;s force because x the norm off X equals two excellent square plus X to a square, for example Proud upto excess square If it&#x27;s that in dimensional vector, this isn&#x27;t exactly pathetic, for we&#x27;ve listen with the magic soul with the metrics I here which is that it. So that means this unknown operator define it by a cathartic phone like this where I closed toe an identity matrix off this one on Bagnall and the other entrance Hours zero and the B is force. There&#x27;s a counter example so P equals too high. Ihsaa counter example. So this statement is true if and only if p RPT a p equals toe a diagonal matrix. Otherwise this is force are, for example, if we just simply let a close to 1 to 11 So this is a real symmetric matrix and the quadratic form x t x transpose times every time. Sex because two x one squared plus two x one x two plus two x two square This is the cross product time here But if we just anybody use, um, I as our orthogonal matrix, we have exchange X transpose times I Jen&#x27;s post tons a times I Pums eggs and he also is still you cost two x one squared plus two x one x two past two x two square Busy. It doesn&#x27;t change anything here. So this is a counter example. Our CSO The statement sees those Air Force and a confident account Eggs employers. If we find out pathetic a phone Q X you close to eggs, a one square process to a square. And if they chose CE cross the minus one. The graph is an empty said stemming is true because this is just the definition of indefinite quadratic form. Um, the statement last time in he is also force because if ex chose post time say Tom six is always known positive, that means or Eigen value off a ah whole negative

Jack Chen · Answer

Okay, so in this problem, we have six statement. Um, we won&#x27;t verify they are true or force. Okay, So hot A It is true because this is the definition off this demetrick for, um, the quadratic form. The metrics of the apologetic phone has to be a symmetric magics for party. They say so. So sure, Because if the metrics, it&#x27;s a diagonal magics, something like this we used to battle magics for his import. Save this symmetric matrix. This attack on all metrics of quadratic form. Excellent x two times a comes x one x two um, were not content. And the cross product because equals two d one x one square pastie two x two square. So there&#x27;s no cross product. ERM, like, excellent x two in this function. Ah, see is true. And this is the definition off the principal excess. Um, and the 40 He&#x27;s also true because this is definition of the positive definite quadratic form and the U. S also true. So if for any symmetric matrix eight, we has has some idea values, for example, under 12 lambda and and they&#x27;ve off them a positive them is there exist of metrics p such that, um, peaches posed a p e closed toe a diagonal magics. I&#x27;m the one London and off the mat zero since off This are Londoners are positive them is for a knee X in our in x X Um, we defied Why to be X equals Dupee times. Why so x transpose a X equals Do why transpose peaches post a p times y and the equals two lambda one by one square purse lamb that to try to square up to London and Dwyane Square. So we already know that off this lambda eyes I&#x27;m positive. Mm Use X transpose t a x transpose X eyes, greater rico&#x27;s zero and the equality hose if and only if XY causes, you know, which satisfy the definition of the positive definite and that if it&#x27;s also true So we don&#x27;t that, uh, Chesky off factory ization of a symmetric metric. They close to an upper triangle of metrics. R t aren&#x27;t you supposed tops our So we&#x27;re eyes that upper triangular matrix with a positive diagonal entrance. So basically this is equivalent. This is a government statement. A Jew a is positive, definite, which is in textbook

Amy Jiang · Answer

Okay, so a is false because this is just the absolute value to determine it. He is also false cause determinants of a is equal to the determinant of 80. We notice from my term. Uh, but see, it&#x27;s joke. Murthy is false.

Sam Sohn · Answer

alright. So Part A is claiming that the determinant is the product of the diagnosed entries in the Matrix. So as a counter example, I am going to give you this matrix to one six three. Now the product of the diagonal entries is going to be two times three, which is equal to six, however, but however, the determinant is actually going to be a D minus B c, which in this case is going to be equal to two times three minus one time six, which is equal to zero. And since sex is not equal to zero, that gives part A is false because one counter example is enough to prove that a statement is false. Part B says that an elementary operation on a does not change the determinant. All right, so let&#x27;s look at, for example, the Matrix. Uh, let&#x27;s say 2110 and I&#x27;m going to say 2120 All we&#x27;re doing is doubling the second row right? The determinant for the original matrix is zero minus one, which is equal to negative one, and the determinant for the second matrix is zero minus two, which is equal to negative two and we can clearly see that the determinant did change. So this is also false. Now, Part C. This one is true and I&#x27;m going to link the proof for why that is true in the answer. But a simple way, Thio. A quick summary is we have the determinant of a times. The determinant B is equal to the determinant of a B. And that&#x27;s what we&#x27;re trying to prove. We&#x27;re going to split this into two cases. Case Case one is the determinant of a is equal to zero. This can also be either the deterrent B is equal to zero. Basically, we&#x27;re looking at the case where one of the matrices is not in vertebral one or more, I should say In that case, the product of a B is also going to be non in vertebral. So since a is non in vertebral, that means a B is none in vertebral, which means the determinant of a B is equal to zero. So that&#x27;s the easy part of the proof. And then we look at the quote unquote harder parts where the determinants up a and the determinant of B are non zero, as in We&#x27;re looking at the case where A and B are both in vertebral, in which case a can be expressed as a product of multiple linear operations on matrices. So let&#x27;s call them E one e to all the way up to E n. And these e to the I. Major cities are just linear operations, for example, multiple multiplying the one of the rose or swapping to other rose or adding a multiple of a row to another row. And we&#x27;re gonna look at all of those. So since we&#x27;re looking at a B, that&#x27;s gonna be either one, either the to all the way up Thio Eat the end Be and keep in mind the sense matrices are associative. We can write this as either the one times the product determined by eat it, too, times the product determined by either three times the product determined by and so on until e to the end. Times be time and let&#x27;s close up all the pro disease. Now, let&#x27;s assume either the end or rather either the I is a rose swab operation. Then the determinant of either I is going to be negative one, and when we swap the rows of the Matrix. The determinant is going to change the sign and that&#x27;s gonna be the same thing that happens with the E f and B if if the e ev en is a roast flop operation. So, in other words, the e of i B is gonna be negative of Yep, which is basically the same as the determinant negative to determine. Sorry. The determinants of either the I since that determined it is negative one times the determinant of beat. So we know it works there. Nice case to be is going to be if the either the I is a road product. So basically, that&#x27;s like saying Let&#x27;s say the forth road gets multiplied by five. And on Lee, that road changes. In that case, the heated I I just passed my real product by ah mm Let&#x27;s say then the determinant of either the I is going to be m. And if you multiply the role of a deterrent over matrix by m, the determinant of that matrix gets multiplied by m. So either I be ends up being am times the determinants of B which is equivalent thio the determinate up. Either the I times the determinant a b so nice it works there as well Part to see ISS. Either I is a linear operation of one row being added to another row. So we&#x27;re going to write that as are a to our A plus RB. In this case, the determinant doesn&#x27;t change it all. So the determinants of e f I is going to be equal toe one. And you also notes that if we&#x27;re just adding a multiple of a row into a different row, then the determinant stays the same. So R e f I B is just going to be the determinant of be itself which is the same as that the determinants of either the i times the determinant a b so nice, no matter what type of role operation we&#x27;re doing for either the I that it does not change the determinant in any unexpected way so we can see that the determinant of a B, which is determined by the all the determinant operations done by e one e one e two all the way to e n, which is the determinant of a times that determine its A B. So that does end up being the determinant a times, the determinant a b again. I recognize that this was not necessarily the best explanation. So I&#x27;m going to link the sites that help me with this proof now for party, it says, If landed plus five is a factor of the characteristic polynomial of a than five is an Eigen value of A and that is false. Because if land a plus five time something times, something is equal to zero and that means land of plus five. Being equal to zero is one of the solutions. And as you probably know, that means Lambda has to equal negative five not positive five. Therefore, party is false. Yeah, we&#x27;re done.

Bobby Barnes · Answer

right. So they want us to determine if these statements are going to be true or falls. Um, so this first one, it says essentially, that we have the zero vector and H. So this is a subspace of our in and this is going to be false because it doesn&#x27;t say anything about the other two properties that we need to have, which is it Closed under sums and closed under scaler multiplication. So false. Because we do not No. If actually, let me beat the Patriots. Bigger. Do not know if closed under some and scaler multiplication. Okay, so we have a there now for B. They say we have this set of factors, and the set of all linear combinations is a subspace. And so this one is going to be true, because if we just think about, um, what this is so it would be something like see one V one plus c two, b, two plus dot, dot dot all the way up to C p v p. So we can get the, um zero vector by saying all see, I equal to zero for every eye. So that&#x27;s fine. Um, we can get any scaler out of this by saying, see, hi equals zero except for some value. Um, but I is equal to and where it is, like the vector that we want. So that gives us the scale of property and three, some wolf. I mean, some just kind of comes from the definition. So some is, um, true by definition of linear combination linear combo. So that&#x27;s how we end up getting true for B. See, I believe this one is also true, but let&#x27;s go ahead and kind of draw it out. So if we have over here some Mbai in Matrix, remember what the knoll space really is? It&#x27;s saying, Well, what collections of vectors here output the zero vector. And, um, in order for us to multiply these we needed in by one matrix, which means we would have in entries here and then this is an element of our in. So then the null space has to be a subspace of our because it will have the zero vector on. We&#x27;ll also have where if we add two things by the linearity of matrix, that should be fine. And then if we just multiply this, that should also be fine. Um, well, actually, let&#x27;s go ahead and write all those out. So eso let&#x27;s say just so we can kind of be a little bit mawr exact right? So Ah, we have some matrix A with the vectors. Let&#x27;s just call it X Y elements of the knoll of a All right. So if we want to do the some property So let&#x27;s look at a Times X plus y Well, then this is going to be a X plus a y and by definition of both of these being in the null space, this is the zero vector plus zero vector which is zero vector. And so then that implies X Plus y is also an element of nor a Okay, so we have that one and then first scale er&#x27;s. So this is some. So that&#x27;s good. I guess we should have also did the zero vector. But that one&#x27;s kind of true just by definition. So this is zero element of Norway. Alright, so that&#x27;s good so far and then lastly we need these scaler. So we have a times c of X, remember axes so in the north space. But we can go ahead. And do you see? Actually me, Right This over here. So this is going to be C a X like that so we can apply community since he is just some constant. And then we&#x27;ll that c times zero vector, which is equal to zero vector. So then that implies C X is also in the knoll. So then the constant multiple or the scaler also holds. So, um, I don&#x27;t know if you really need to be that exact, but, I mean, it doesn&#x27;t really hurt now for D the column space of Matrix A is the set of solutions. A X is equal to be all right. And so this is actually going to be, uh, false. And the reason why this is false is because it&#x27;s not necessarily just for the solution of one. It&#x27;s supposed to be the span of all of the columns I&#x27;m sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that&#x27;s why this one isn&#x27;t necessarily true. Because it&#x27;s not talking about a solution set and then for E. So this is saying that B is the echelon form of a matrix A in the pivot columns of B form a basis for a And so this one, um, is also false Because it&#x27;s not the pivots of B, but the pivots of call a Arab the pivots of a instead. So this should actually say a here on beacon, actually kind of show that off on the side really fast. Eso Let&#x27;s come down here and do that, right? So let&#x27;s say we have some matrix here. And so this is kind of like you can use, like, a counter example as to why this wouldn&#x27;t be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don&#x27;t know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn&#x27;t just kind of tweak this so it does. But I believe this will give us that. Okay, well, if we look at it, this is a and then this is the echelon form. Be over here. Well, if we were to look at this, though, over here, this is saying if we take each of these are fourth position of our vector is always going to be zero. But over here, notice There are numbers here. So we wouldn&#x27;t even be able to get out the, like, original columns here. So because of that, um, we would need to use the columns for a as opposed to the columns of being So Yeah. So that&#x27;s kind of like the rationale behind. Why that? Why that ISS? Yeah. So s it looks like the first statements. False second statement. True, They&#x27;re true. And then the last two are going to be false.

Chris Trentman · Answer

were given statements and were asked the label them Sure false and to explain in part, a statement. Is that an n by N matrix that is orthogonal e Dagnall Izabal must be symmetric. This statement is true. This follows from here Um too from this chapter which says that and then by n Matrix A is or Thorgan aly diagonal Izabal if and only if a is a symmetric matrix part B The statement is if a transpose equals a in defectors U and V satisfy a U equals three you and a V equals four d then you dotted with V equals zero. This statement is true noticed that since a transpose equals a we have that is symmetric therefore by theorem one from this chapter it follows that Eigen vectors from distinct wagon spaces are orthogonal to each other. And we have that since a U is equal to three. You and eighties equal for V. We have that you and the are Eigen vectors of a associated with the Eigen vectors for Eigen values three and four, respectively. These are distinct. I&#x27;d in values so by fear and one it follows that you one is orthogonal to or use or Foggin all to be. And so it follows that the doctor product of U and V is going to be zero in part. C Statement is an n by n. Symmetric matrix has end distinct really Eigen values. This is false Indeed. An example in this section included a symmetric matrix a which did not have and distinct rial agon values. This was example three. From this chapter, a true statement would be that the end by in symmetric matrix has and realizing values accounting for Multiplicity. Statement. D says that for a non zero vector V in RN the matrix be transposed. It&#x27;s called a projection matrix. Need to make sure this is so. There&#x27;s actually attend. Um, this should be matrix vv transfers Kidding. And this is false. See why I recall the definition of a projection matrix we have. The projection matrix is a vector. Well, the matrix such that the projection matrix times a vector and are in, is the orthogonal projection of that victor onto the subspace spanned by you. And the reason this isn&#x27;t true is because we need you. Could be Ortho normal Eigen vector. Simply not the that This would be true if the Waas and a member, uh, a set of or so normal I in vectors of the Matrix A Not just any matrix, but a symmetric matrix, but this is not true.

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Exercises 23 and 24 show how to classify a quadra…

Problem 22 Hard Difficulty

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A. False
B. true
C. False
D. true
E. False

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

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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A. FalseB. trueC. FalseD. trueE. False

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Alayna H.

Caleb E.

Kristen K.

Michael J.

A. False
B. true
C. False
D. true
E. False

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

Linear Algebra and Its Applications