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[20 points] Prove the following statements for a 2 x 2 matrix A. (a) If A is symmetric, then its eigenvalues are purely real....

Question

[20 points] Prove the following statements for a 2 x 2 matrix A. (a) If A is symmetric, then its eigenvalues are purely real.

[20 points] Prove the following statements for a 2 x 2 matrix A. (a) If A is symmetric, then its eigenvalues are purely real.



Answers

(a) Prove that if $A$ is a square matrix, then $A$ and $A^{T}$ have the same eigenvalues. [Hint: Look at the characteristic equation $\operatorname{det}(\lambda I-A)=0 .]$ (b) Show that $A$ and $A^{T}$ need not have the same eigenspaces. [Hint: Use the result in Exercise 30 to find a $2 \times 2$ matrix for which $\left.A \text { and } A^{T} \text { have different eigenspaces. }\right]$

For this problem. We have been given a matrix. A. We've been told that a is both convertible and skew symmetric. Now, before we get any further, let's review what skew symmetric means. Ah, Matrix is skew symmetric if the transpose of the matrix equals the opposite of that matrix. Okay, so I just wanna get that definition in. So because we're gonna be referencing that a lot in this problem. Okay, so we have this matrix A It is in vertebral and skew symmetric, and we want to prove a few things. First, we want to prove that the inverse of a is also skew symmetric. So is this true? Well, if that's true, what we're really trying to show is that the transpose of the inverse of a equals the opposite of the inverse of a This is what we need to prove. So let's begin with the left hand side. Okay, so the transpose of the inverse Well, if you go back to the're, um 1.4 point nine, we can rewrite this. The transpose of the inverse is the same as the inverse of the transpose. So theorem 1.4 point nine allows us to rewrite this Now, from my definition, I know a skew symmetric. So the transpose of a is the opposite of a and I'm taking the inverse of that. That is exactly what I was trying to proof. So the inverse of a is indeed skew symmetric. Okay, Yeah. Now let's go to a second set of problems here. We're gonna have to matrices a and B that are both skew symmetric. So given that there's four pieces that we want to prove first we want to prove that the transpose of a is skew symmetric. So what we want to show is that the transpose is equal to the opposite. That's what we want to show. So like we did before, we're gonna do this pretty much for all these problems. We're going to start with the left hand side. Okay, Now, the transpose, um of a transposed transposed just gives me what? What? I start what I started with. So the transpose of the transpose of a is just a well, a is skew symmetric. So what I started with was also skew symmetric. Yes, that is true. Next, let's look at a plus B. Is this skew symmetric Well, if it is than what we want to show is that it's transpose is equal to its opposite. We want to show this. So let's start with the left hand side. Let's take the transpose of a plus B. Well, I can break that up into the transpose of a plus. The transpose of B A and B are both skew symmetric. So shut up. The transpose of a is the negative A The transpose of B is negative b and I can rewrite that like that. That is indeed what I needed to show. So the sum of a plus b is indeed skew symmetric. Okay, what about the difference? Is the difference of to skew symmetric matrices skew symmetric? Well, if it is, their transpose will be equal to will be equal to the opposite. That's what I need to show. So let's start with the left hand side and again I could break that apart. The transpose of a difference is the difference of the transposes. Yep. And that's gonna be a B. Apologies for that. Let's get that. Okay, now a is skew symmetric, So that equals negative. A B is also skew symmetric so that equals the opposite of B. And if I factor out in negative, that is indeed what I needed to show. So the difference is also skew symmetric and the last piece. What if I multiply K um Matrix A by a scaler, some some scaler Multiplication K Is this also skew symmetric? Well, I need to show that it's transposed equals. It's opposite. So start with the left hand side. I could break that apart and say, That's k times The transpose of a and the transpose of a is the opposite of a And that's exactly what I needed to show so that scaler multiplication also results in a skew symmetric matrix.

Were given a waitress. A you're s You're saying that sneakers speaker say is the general two by two real symmetric matrix A B B C where maybe in Sierra Leone's of real numbers theorem is that a hasn't been valued Most policy to defend our Maria is scaling matrix so group this year infections pose a as I down you lander one of multiplicity too. The characters of falling on the holiday is the determining a minus land that I, which is equal to Time's up matrix a modest lambda Be see my eyes lined up is equal to a minus Landed time See my assigned at minus piece Where and we write this out for we Lambda Squared waas minus a plus C Landau plus a C minus b squared. And because we know that has I know you went one policy too, Randall one. This is a bunch of nature's. The war is the only Ivan value. And so the characteristic calling on your land also has a form de some real number. Be times landed one minus lander. We received a complaint and land the one to this character's upon you. We get that home zero So it is nine and the online does are lander one Listening to we expand this polynomial. This is de times Landis wear minus two d Learned one plus you remember once where and since this is the same polynomial as war, you have that polymer meals are equal Prohibitions of the same terms are equal. This implies that he is the one negative to do. Round one is equal to Nadiya. They see finally you and the one squared is equal to a C minus the square terms of the scene scene issue. So using our first equation equals morning right Theater to vision says negative too. What I want equals negative it was and our third queen is now I'm going swear is equal to a C Nice piece where So we get the lander. One is equal to a plus c over to here for we have that a plus c What were to swear is equal to a C minus the square most part in both size that for and well, you swear out you he's where close to a C. We'll see square is equal to four C minus 40 square. This is equal to is where, minus two A. C plus. He's where you could make it for the square. So a minus C swear waas or be square quarters here. Since A B and C are all real numbers, this is a some, uh, squares of no members. So it's strictly great and equal zero, and this sums only 40 We have a pet times in my C equals zero and zero you supplies equals zero days. It will see eight. He some remember our then matrix A becomes our hero zero or or are times by two. By definition, is it scary? Our natures prove the other half of this statement Suppose that uses steam of your tricks Then there is some are in the real numbers such a is equal to are to Then we have that the characteristic polynomial a is the term a minus lambda I to is equal to substituting for a determinant are too bias like I to factoring out I to this is the same as ar minus learned up that to and we have that the determinants of a constant times that end matrix. This is going to be constant or minus Amanda Times the number of rows in the Matrix I to just two or to power the number one. It's like times that it turns make exciting. You know, the determinant of the matrix instantly. Once this ar minus Landis, where this is our characters department. And moreover, you know that the roots of this polynomial are the I in values meters A so that this equals zero. We get that our news Oh, a on Landau warm equals positive are with a multiplicity up to So therefore a has an idea and this includes the group.

The first thing we're gonna do here is use the fact that we know that in this equation we'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'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'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'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'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'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's no point at which this will be directly zero. This will just postulate so they can. They don'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'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.


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