let-each-matrix-in-exercises-1-6-act-on-mathbbc2-find-the-eigenvalues-and-a-basis-for-each-eigensp-6

Question

Let each matrix in Exercises $1-6$ act on $\mathbb{C}^{2} .$ Find the eigenvalues and a basis for each eigenspace in $\mathbb{C}^{2} .$
$$
\left[\begin{array}{rr}{4} & {3} \ {-3} & {4}\end{array}ight]
$$

Runpeng Li · Accepted Answer

Okay, so for problem six give the matrix is 43 negative. Three and four. So consider determined off a mine. Islam die. This will be for minus I three. Next week, three, four minutes. I he&#x27;s here now. By solving the you question, you find out London should be for plus or minus three. I okay, And to find out basis we first take London to before Waas three. I I&#x27;m sorry and this difference would be negative. Three i and three and neck of three and thank you. Three I now by I really applying the culture and damnation get the reduce station form. So the result would be I active one 00 So if we flooding our unknown vector excuse me are unknown vectored to be ex lax too. Then by the first roll we have on times X one minus x two will be zero. Now that is I time sex one would be x two. So our first base is will be one Excuse me, one and I And our second business will be one and negative. Hi. So

Michael Jacobsen · Answer

in this example, we have a two by two matrix say that&#x27;s provided medical. Here&#x27;s to find its Aiken values and corresponding. I get vectors to start off for the first part. To find Eigen values, we take the determinant of a minus lamb die or will be solving for the very bold Lambda as a determinant will have negative Lambda Negative eight in the first column than one end for my is Lambda in the second column, Then, to take this to determine it will first have the main diagonal multiplied, which is negative. Lambda Times four Mice Lambda and subtract the product of the off diagonal, which is negative. Eight. So we&#x27;ll have a positive eight so far. Then, if we simplify this, we obtain Lambda Squared minus four. Lambda Plus eight equals zero, and this is our characteristic equation. The solutions are Lambda One equals two to minus two. Lambda or excuse me to I and Lambda Two is that&#x27;s constant to with a plus two I. So we confined that using the quadratic formula, and the next thing we&#x27;re going to do is determine the corresponding Eigen vectors for each Eigen value. So let&#x27;s focus on Lambda one To start out, we can solve the system a minus lambda one, which is two minus two I times X equals zero vector. So let&#x27;s augment and say Our matrix is going to be distributing the negative sign in. In the first calm will have negative two plus two I and the negative eight below. In the second column, we have one. Then take four and at together the result of taking negative two plus two. I now we&#x27;ll have altogether two plus two. I they were augmenting with a zero vector, which I&#x27;ll place here. Now it&#x27;s due to operations as we ro reduce. I want to go to the second row, divide it by negative eight, then make it become the first row will have one. Then negative 0.25 plus 0.25 I and a zero. Then the second row is going to be the old first row, which is negative. Two plus two I one and a zero. Now we have a pivot that&#x27;s here and uncle is to eliminate this entry. So let&#x27;s start by copying the first row. It is a one negative points to five plus 50.25 I and a zero. And what we&#x27;ll be doing is multiplying Row one by two minus two i, and adding the results to the second row will obtain zero. And if we take negative points to five plus 50.25 I and multiply it by Tu minus two, I we obtain exactly negative one, and that combines with this entry producing a zero here. Then we have a zero for the last column where we augmented. So in the row one of our Matrix, which is in row reduced echelon form, we find that if we saw for the very bill, X one will be equal to the opposite of 0.25 plus 0.25 i times x two and X two is a free variable Salt. Just write X two equals X two Oh, it will have a negative here that was neglected to be copied from there. So now we need to describe what our first vector, Aiken vector V one, is that corresponds to Lambda One and we have some choices. If we decide to not deal with these decimal values we see here, we can allow X two to be any real number except zero since zero is never Nike in Vector. So is pick X two to be, say, a four. If we do that, then all together will get a one minus I. That&#x27;s a result of placing for here distributing as well as distributing the negative sign. Now we have to re piece repeat this row reduction to find the second I get vector, but there is a wonderful shortcut. That shortcut says that V two is the congregate of the one and the one conjugated would be one plus I and the contra gave a real number is itself. So what we&#x27;ve done is we found dykan values V one, V two and the corresponding Eigen vectors V one and V two. So these V one V two also form a basis for their corresponding Eigen spaces.

Michael Jacobsen · Answer

in this example, we have a two by two matrix, and we&#x27;re going to start out by finding what it&#x27;s Eigen values might be. First, In order to find Eigen values, we take the determinant of a my slammed I a My slammed. I is going to be the determinant of one minus lambda, then one below. And in column two, we have negative to within three minus lambda. So the determines would then become one minus Lambda times three mice Lambda plus two. And that&#x27;s our characteristic polynomial. This reduces to the following polynomial Lambda squared minus four Lambda plus five. And we said it equal to zero so that this is our characteristic equation. So far. Now, the solutions to this characteristic equation by use of the quadratic formula our Lambda one equals two plus I and Lambda two is always going to be a con. Trick it. So this will be to minus I. Now we have both Eigen values, and from here we&#x27;re going to look for the Eigen vectors. Let&#x27;s start with, say, Lambda One, and find its associate ID Aiken Vector. We&#x27;ll take a minus lambda one times identity matrix times a vector X set equal to the zero vector in our to So this is the equation that we&#x27;re going to solve. And the augmented matrix for this equation is going to be first go to this entry. It&#x27;s attract too. And I we&#x27;ll have altogether negative one minus I, then copy Negative to for the next entry for the next row will have one. Then, on this entry here, we do the same subtraction. Take three minus two minus I, and we&#x27;ll get altogether one minus I. And then we&#x27;re augmenting with the zero vector. So placed that here. Let&#x27;s begin row reduction. Well, for the row reduction, if we&#x27;d like, we can interchange rose one and two. So that row one reads one one minus I and zero. And then the next row will be negative. One, my Asai. Negative two and zero. Now we&#x27;re ready to begin a row. Operations. Well, we have a pivot here and now to eliminate the entry directly below we multiply row one by one, plus I at the results of row two. So let me start by copying row one. It&#x27;s one when my sigh and zero, Then we&#x27;ll have zero. And next we&#x27;re multiplying one my Asai with one Plus I well, one my side times one plus I. If you foil us out, you&#x27;ll see we have one plus one, resulting in a two which are adding to this entry. So we get a zero, then a zero below. And we knew we should get a zero because we&#x27;re on the right track if and only if, Since we have a Nike in value, this equation has nontrivial solutions for X. So we&#x27;re ready to start describing our solutions. So we can say from row one that X one is going to be equal to negative one minus I times x two. Then from row two, we see that X two is a free variable. So we can say that X two equals x two, then V one which corresponds to Lamda won the that as Eigen value is Eigen Vector where we can pick anything we like for ex too. If we pick it to be negative one, it will wipe out the negative Rex one making it somewhat simpler style picking negative one here. Then we have one minus I for the first entry. Now we can repeat this entire process with row reduction to get the second Aiken vector. Let&#x27;s use an important fact. Dykan values and the Eigen vectors come as congregate pairs. So V two has first entry, which is a congregate of one, My Asai, for one plus I and the contra get of the real number is itself. So this is the Eigen values here and here. And these are the corresponding Eigen vectors for this particular matrix.

Michael Jacobsen · Answer

in this example, we have a two by two matrix say that&#x27;s provided medical. Here&#x27;s to find its Aiken values and corresponding. I get vectors to start off for the first part. To find Eigen values, we take the determinant of a minus lamb die or will be solving for the very bold Lambda as a determinant will have negative Lambda Negative eight in the first column than one end for my is Lambda in the second column, Then, to take this to determine it will first have the main diagonal multiplied, which is negative. Lambda Times four Mice Lambda and subtract the product of the off diagonal, which is negative. Eight. So we&#x27;ll have a positive eight so far. Then, if we simplify this, we obtain Lambda Squared minus four. Lambda Plus eight equals zero, and this is our characteristic equation. The solutions are Lambda One equals two to minus two. Lambda or excuse me to I and Lambda Two is that&#x27;s constant to with a plus two I. So we confined that using the quadratic formula, and the next thing we&#x27;re going to do is determine the corresponding Eigen vectors for each Eigen value. So let&#x27;s focus on Lambda one To start out, we can solve the system a minus lambda one, which is two minus two I times X equals zero vector. So let&#x27;s augment and say Our matrix is going to be distributing the negative sign in. In the first calm will have negative two plus two I and the negative eight below. In the second column, we have one. Then take four and at together the result of taking negative two plus two. I now we&#x27;ll have altogether two plus two. I they were augmenting with a zero vector, which I&#x27;ll place here. Now it&#x27;s due to operations as we ro reduce. I want to go to the second row, divide it by negative eight, then make it become the first row will have one. Then negative 0.25 plus 0.25 I and a zero. Then the second row is going to be the old first row, which is negative. Two plus two I one and a zero. Now we have a pivot that&#x27;s here and uncle is to eliminate this entry. So let&#x27;s start by copying the first row. It is a one negative points to five plus 50.25 I and a zero. And what we&#x27;ll be doing is multiplying Row one by two minus two i, and adding the results to the second row will obtain zero. And if we take negative points to five plus 50.25 I and multiply it by Tu minus two, I we obtain exactly negative one, and that combines with this entry producing a zero here. Then we have a zero for the last column where we augmented. So in the row one of our Matrix, which is in row reduced echelon form, we find that if we saw for the very bill, X one will be equal to the opposite of 0.25 plus 0.25 i times x two and X two is a free variable Salt. Just write X two equals X two Oh, it will have a negative here that was neglected to be copied from there. So now we need to describe what our first vector, Aiken vector V one, is that corresponds to Lambda One and we have some choices. If we decide to not deal with these decimal values we see here, we can allow X two to be any real number except zero since zero is never Nike in Vector. So is pick X two to be, say, a four. If we do that, then all together will get a one minus I. That&#x27;s a result of placing for here distributing as well as distributing the negative sign. Now we have to re piece repeat this row reduction to find the second I get vector, but there is a wonderful shortcut. That shortcut says that V two is the congregate of the one and the one conjugated would be one plus I and the contra gave a real number is itself. So what we&#x27;ve done is we found dykan values V one, V two and the corresponding Eigen vectors V one and V two. So these V one V two also form a basis for their corresponding Eigen spaces.

Runpeng Li · Answer

Okay, so for problem for given the matrix K to be I negative too. 123 Now we to find out the wagon values consider is determinant to be zero. Now, this will be everyday and tough Find minus Lunda three miles. Lambda Three minus lambda and negative. Two and one. He&#x27;s ill now. I&#x27;ll skip the calculation. So they&#x27;re The answer is Lambda is, um for us or minus I Okay, now, to find out the faces, we first take lumber to be for us. I then consider this matrix a minus. Some die. So we have, um so in this case, it will be one minus. I negative too. And one and active one minus I. Now I&#x27;ll just noted Here, take take four plus I and by couching indignation, then we find out the issue of warm will be one minus I next to 00 Okay, right now be plugging the unknown factors. So if we plug in the end of a vector that we will have one minus i times x one, which is same as two off x two just to buy our first rodeo. This matrix and we divide both. Excuse me um we divide both sides by one Miss I. So he&#x27;s we have x one. You closed two divided by one minus I time. Sex too. Then we simplify. It will be one. My one. Plus I time sex too. So our first of faces be take x two To be one that rx one would be a one fuzz. I so corresponding second basis will be one minus I and and one

Runpeng Li · Answer

Okay, so for problem, too, Were given the Matrix. I negative five, and I want one. So to find the argon value that is too gay. All right. Um, I&#x27;m just to rewrite this like this to be a and find the Eiken value determinant a money&#x27;s love die, which is zero now. So this turns out to be five minus lambda and 95 11 by the slob zero now expanded way have minus dum dum. One man is standard minus next to five. So it&#x27;s plus five. No, we have five minus five. London minus lambda. First number squared. Pacify, Cyril. Now keep going. Then We have nam the squares and find the six Loved, uh, plus 10. Right? Plus 10 which is zero. So dissolution would be That would be too in the denominator neck, too. Uh, minus negative. Six weeks. Just six asshole. Minus square. Rude. 36 minus four times. Four times, four times 10. So that&#x27;s 40. So six plus or minus. Active square root of negative for. So it&#x27;s just six by plus or minus. To I divided by two. That was three. Plus or minus. I. So he&#x27;s hard to. I confess. So now we need to find out the basis for each again Space. Excuse me. Now here. We need to consider a minus Lambda chi Those first consider three plus I as a as the conductor. So we have a five negative five 11 minus. I love the time size with its three plus I three plus I and 00 So that gives two. Find this I and next 25 and one and elected to mine aside. Okay, so the next thing is, just do the gallery indignation to find out the location of form. So we times we used second row times two minus i to minus I row two and minus the first room. So first entry will be zero now to minus I times active to minus I minus r one. So he&#x27;s minus negative five. So it&#x27;s just a plus five. Hence, we have negative for, uh, minus one plus five. Right? So he said elective. So we expend. Is I negative? I squared minus four. So he&#x27;s just Ah, negative four. That&#x27;s four. So he&#x27;s zero Cyril and we leave, uh, leave the first roll. So it&#x27;s two minus I and negative five. Now that&#x27;s it. So we can see her. The ViaGen vector here will be to minus I time. Sex one equals five of x two. Okay, so which kind of So what kind of excellent x two to take? Well, in this case, we can just take, um x X one to be five divided by two months. I time sex too. So I see we can simplify this. There&#x27;s five times two plus I. So it&#x27;s four minus. I squared, so it it will be fine. And this gives us too to us. Hi, I&#x27;m sex squared. Uh, sorry to plus I time sex to sow the first. The first base is we find will be now take a X two to be one. The next one will be to sigh and correspondingly. That&#x27;s second basis will be to minus I and one so

Problem

In Exercises $7-12$ , use Example 6 to list the e…

Problem 6 Easy Difficulty

Let each matrix in Exercises $1-6$ act on $\mathbb{C}^{2} .$ Find the eigenvalues and a basis for each eigenspace in $\mathbb{C}^{2} .$
$$
\left[\begin{array}{rr}{4} & {3} \\ {-3} & {4}\end{array}\right]
$$

Answer

$\mathbf { v } _ { 2 } = \overline { \mathbf { v } _ { 1 } } = \left[ \begin{array} { l } { i } \\ { 1 } \end{array} \right]$

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Linear Algebra and Its Applications

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$\mathbf { v } _ { 2 } = \overline { \mathbf { v } _ { 1 } } = \left[ \begin{array} { l } { i } \\ { 1 } \end{array} \right]$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Michael J.

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.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications