chapter-7-will-focus-on-matrices-a-with-the-property-that-ata-exercises-23-and-24-show-that-every--2

Question

Chapter 7 will focus on matrices $A$ with the property that $A^{T}=A$ . Exercises 23 and 24 show that every eigenvalue of such a matrix is necessarily real.
Let $A$ be an $n 	imes n$ real matrix with the property that $A^{T}=A$ . Show that if $A \mathbf{x}=\lambda \mathbf{x}$ for some nonzero vector $\mathbf{x}$ in $\mathbb{C}^{n},$ then, in fact, $\lambda$ is real and the real part of $\mathbf{x}$ is an eigenvector of $A$ . [Hint: Compute $\overline{\mathbf{x}}^{T} A \mathbf{x},$ and use Exercise $23 .$ Also, examine the real and imaginary parts of $A \mathbf{x} . ]$

Michael Jacobsen · Accepted Answer

In this example, we have a matrix a that contains entries that are real numbers and this square of size and buy in. We&#x27;re going to assume that a transpose is equal to a and that a X equals Lambda X, where we have X is a non zero vector containing entries from the set of complex numbers. So with all those assumptions, what we&#x27;re going to do is the following show that Lambda is in our meaning that it&#x27;s a real number. And as per the hints provided, we already know that X transpose times a X is also a real number, no matter what the vector X happened to be, so it&#x27;s use the information to start out will compute X transpose ex congregate transpose times a X and by substitution. We know X equals Lambda X. So let&#x27;s make that substitution. Here we&#x27;ll have ex con trick it transpose times Lambda X, and because land is a scaler, we can let that commute so that we have altogether Lambda Times the congregate of X transpose times X Then we know already that the contra kit of X transpose times a X is in our and we claim that this is following X congregate transpose Times X is also in our despite the fact that X comes from the set of and two pools of complex numbers. Let&#x27;s give some justification to the last portion. Let&#x27;s say let X be equal to Z one through ZN where these are now complex numbers. Then the congregate of X transpose Times X would be equal to the contra gave Z one times z one plus all the way through the conjure git of zn times e n and Z I conjure git Times e is in our four all I will for any complex number we have that its contra git times itself is purely a real number. So we know two things Now we know that ex contra get transpose times a X isn&#x27;t is in our and we have no, by the last calculation, we see then that ex contra get transpose Times x is also in our So what does this mean? If we go back to this equation, this is purely a real number Now this is purely a real number And if Lambda was a complex number, then there&#x27;d be no way that this would be purely riel. So the conclusion that were forced to write is that this shows that Lambda is in our so all together. Knowing that a transpose equals a a X equals Lambda X is enough to show tell us that if X is not zero lambda must be a real number. We might wonder, where did this assumption come into play? Well, that was used to prove the hint here. Now the second thing we want to do here is to show that the real part of our Eigen Vector X is and I can vector of a to do this. Let&#x27;s start out by letting X be equal to at Vector Z plus w I where z and W are in or in. So if we take a vector X from CNN, we can always decompose it in this way where the two Vectors NW came from our end. So now the exes decomposed. Let&#x27;s write down something quick. So our equations were going to deal with are going to be eight times X which is now by substitution eight times z plus w i Well, this would be a Times Z, but recall that Z is also the real part of X, then we&#x27;ll add to that eight times. W i. But W is the imaginary part of X times. I Let&#x27;s look at this equation from a different point of view. We can also calculate a X equals lamb Dax provided by this equation here. Then, if we substitute at this stage, will have Lambda Times, z plus W Y in place of X. And now if we distribute, it follows that will have Lambda Times Z. But recalled Z is the real part of X so will express it in this way, plus Lambda Times W where w is the imaginary part of X times I now with the equation we&#x27;ve written, we have equality that goes here. And the only way that these two expressions can be equal is if they&#x27;re riel and imaginary parts are equal. And this forces this quantity to be equal to each other. And so we have now this conclusion this implies the A times The real part of X, which is a non zero vector, is equal to Lambda Times, the real part of X, and this shows that the real part of X is an Eigen vector of a as required

Michael Jacobsen · Answer

in this video. We&#x27;ve been provided with the Matrix A and we&#x27;re told it&#x27;s in our and by in which is just a way to say it&#x27;s a square matrix of size and by end, and all of its values are in the set of real numbers. Next, we&#x27;re also told that a transpose equals a. This is a way of saying that the Matrix is also symmetric, and finally, we&#x27;re going to pick a vector x from RN and define a new vector. Q. We see displayed below. This is a worry some way to define a vector because we have a vector times of matrix times a vector and that could well be undefined. So let&#x27;s do a quick dementia analysis to see what this quantity even is. First. If X came from our end, then X transpose, whether we congregate it or not, is going to be of size one by N, since it will have now one row within entries. Now the Matrix A, which comes next. We&#x27;re told already that it&#x27;s of size and buy in, and so far so good Notice that these inner dimensions match and so this portion is well defined Next we multiply by a vector X so more and more interesting things continue to happen. And since X comes from our end, it&#x27;s of size and buy one because it will now have en rose. But a single column for a column vector then noticed. This is still well defined because we next have in and one here for that matching dimensions. But one thing that&#x27;s very interesting is the one and one here. It tells us that the result of Q equals X bar transpose times a X is an element in our so it&#x27;s a real number or you could call it a scaler if you prefer. So let&#x27;s see what happens with this particular scaler. Well, we said it&#x27;s in our but we should be a little bit careful. Let&#x27;s say a little bit more generically. It&#x27;s a complex number in C. Then let&#x27;s see if it really is a real number, which seems plausible. Let&#x27;s calculate que transpose or excuse me que con? Trick it. By definition, we have to congregates the entire quantity here. So this is what it would look like First export transposed times a X, and were conjugating everything. And this expression Let&#x27;s break this down step by step, where we know the first thing that could happen is if we consider the product here by association, we can rewrite the congregation as first take X contra git conjugate, but that gives us just X backs. Then we take the Conjure git of eight Times X, which would look like this open. And don&#x27;t forget the transpose operator on X, so that&#x27;s the set up so far. Next, let&#x27;s break down the conjugal operator into what&#x27;s inside. The parentheses will become X transpose. Then we&#x27;ll have a congregate times ex con jacket. But a conjugal is a kind of a suspicious quantity because notice we said a came from our and by end, meaning all the numbers or all the entries in A are real numbers. And this means the congregate of a is just simply a so it&#x27;s right that in on our next step, we now have X transpose. We have just the matrix A in place of ex con a con jacket. The ex itself can be a complex number, so we have to leave the congregate above it. Now where to go from here? We&#x27;re trying to show that que con trick it is equal to queue so that it&#x27;s a real number but also noticed that we can do the following The entry here we have is in C for a particular skit scaler. And that means since we&#x27;re dealing with the scaler, we can transpose this entire thing and still get equality. Let&#x27;s write that and break drop down a line and right X transpose times eight times x bar. Then again, a transpose here works because this is a scaler in si or ah, one by one matrix. Next thing we could do is use the transpose operator. It&#x27;s going to transpose first This expert we see here and it becomes the first factor we take the transpose of a So there&#x27;s a transpose here. Lastly, we take the transpose of X transpose, but that just gives us X back open. Don&#x27;t forget the bar here that came from this first transposed. So this is very similar to when we took the conjure get of the conjure git on a first step the transpose of a transposed Just give this back X Now where to go from here? Notice. We&#x27;ve never used this assumption that a transpose is equal to a and that&#x27;s suddenly very relevant here. Let&#x27;s make that our next step. This is now equal to the congregate of X transpose than in place of a transpose. We can put a and then Times X, but notice. We&#x27;ve just matched this format here. So this is now equal to Q, which by our long chain of equality and by the transitive ity of Equality Q Bar or the Contra Give Q is equal to Q. And that implies that not only are we in C, but we can be more specific. Q is in our in other words, it&#x27;s simply a real number.

Runpeng Li · Answer

Okay, so for problem, Time to work, even of compacts matrix. And we&#x27;re given the argon victor acts and the wagon value Lunda. So we want to show that mute times eggs where Mew is a complex scaler is a mug in victor off. So first we know because this is Ah, this is actually sent again. Vector and lumber is having a very So we have eight times x is London Times X. Now we need to consider the vector u x so here. Now, we have eight times more bags since mu Excuse me since Marissa skater we can put me we brought away. So we have you times a X but we know a access equal to numb Dax. So we have from you times lambda. Thanks. And hiss. If we take let&#x27;s say Alfa to be be immune times, Lunda. And why to be the new vector off leaving time sex. Then we have a relationship like this eight times. Why? Because Alfa climbs Why? That means health is I can value uh, a with again victor. Why? So that means mu times acts. He said, not give Victor. Okay,

Chris Trentman · Answer

were introduced to the singular value decomposition method of matrix factory ization. So suppose that a is equal to u times d tend to be transposed where u and v r n by n matrices with the property that you transpose u equals I and that v transpose v equals I. And where d is a diagonal matrix with positive entries. Sigma one through sigma n on the diagonal were asked to show that the matrix is in vertebral and to find a formula for a inverse. Now we have that both u and V transpose are square. So the conditions on U and V imply that you and the transpose are both in vertical matrices by the in vertebral matrix. The&#x27;re, um so it follows that you inverse equals u transpose and we have that be transposed in verse is simply ve Now The diagonal entries of D are non zero. Since none of these entries or zero it follows that D is an in vertebral matrix, and we have that the inverse of D. It&#x27;s a diagnose metrics as well civically with entries Sigma one in verse through sigma and in verse on the diagonal. So it follows that a is a product of convertible matrices. So it follows that a itself is in vertebral. Not only is a convertible we have the a inverse is the inverse of you d be transposed and this is going to be be transposed in verse The inverse universe which is equal to the de in verse um, verse and these air Easy to our Sorry you transpose since you universe is u transpose and these air easy to compute

Ryan Williams · Answer

The first thing we&#x27;re gonna do here is use the fact that we know that in this equation we&#x27;re gonna have two parts to the general solution show. This will be at the limit. Misty approaches infinity of some first solution X one of team plus X two of tea. I mean, you can use the fact that the limit you can apply the limit to two things being added together to write this as the limit of the first term, plus the limit of the second term. And we can write out a general form for the two solutions that we know that will have, which will be the limit. Misty approaches infinity of some constant times E to the first Eigen value time some constant Eigen vector. And I&#x27;m gonna write his ex one with no indication of tea to make sure that its first I conductor plus the limit as to approach infinity of some second solution just some constant times. Another Agon value time. Some other Eigen vector and we could use the fact that the Eigen vector and the constant are constants and don&#x27;t depend on teas. We can bring them outside the limit. Isis of being those values times the limit. But being toe Linda one I&#x27;m see plus C two times x two times the limit as t Purchase infinity of each of the Lambda Tomb T and we can write at Write Lambda when Lambda two as some general number. Specifically some number A plus B times. I all times t so a complex number since we can have complex values. So we each of the 80 and you to the i times Bt plus c two x one times the limit As to your approaches Infinity I mean to the C team and needing to the I times DT where again C and D are just numbers and we can separate this limit since it&#x27;s a product of things involving t into the limit as t approaches infinity of the 1st 1 we just needed 80 times the limit. Best you purchase infinity of the 2nd 1 needs to the i times Bt plus C two time second wagon, my conductor times that when it has to purchase infinity of the first term need to the c t times of limit. As to your perch in community, the second term need to the I times DT and we can use and we want to know when this is equal to zero. Since that&#x27;s the expression we started with. We know that if you want to know that the limit as tea parties invent any of a solution equals zero, what are the restrictions on are constants A B C D. Well, we know that this is also torrey e to the i times. Some number will be the coastline of that number, plus the sign my time to sign of that number. And there&#x27;s no point at which this will be directly zero. This will just postulate so they can. They don&#x27;t really have to be any restrictions on B or D. These will just postulate. But if a is positive in the limit as t approaches infinity of usual positive overtimes t would blow up to infinity. It wouldn&#x27;t go to zero In the other case if e if over here see, is negative has he goes infinity? This will just go to zero. So if a or seeing are positive in our solution goes to infinity. But if a and C are negative and our solution goes to zero This is not what we want. So therefore A and C, you have to be less than zero in order for the limit of our solution to go to zero. So therefore, our Eigen values, which are a plus i b. They asked me Lesson zero and for I other Eigen values C plus times D. So you have to be less than zero. Or, in other words, Eigen values that they have a negative real part.

Problem

Let $A$ be a real $n \times n$ matrix, and let $x…

Problem 24 Hard Difficulty

Answer

$A x$ and $\lambda x$ are equal, their renl parts are also equal. $\rightarrow A u = \lambda u$ meaning that real part of $x$ is eigenvector of $A$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Heather Z.

Kayleah T.

Michael J.

Joseph L.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 5

Video Transcript

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

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$A x$ and $\lambda x$ are equal, their renl parts are also equal. $\rightarrow A u = \lambda u$ meaning that real part of $x$ is eigenvector of $A$

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Kayleah T.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications