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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} .$$$\left[\begin{array}{rr}{1} & {-2} \\ {1} & {3}\end{array}\right]$$

Eigenvalues are $[ 2 - i ]$ and $[ 2 + i ]$ Corresponding eigenvectors are $\left[ \begin{array} { c } { 1 + i } \\ { - 1 } \end{array} \right]$ and $\left[ \begin{array} { c } { 1 - i } \\ { - 1 } \end{array} \right]$ respectively.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 5

Complex Eigenvalues

Vectors

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in this example, we have a two by two matrix, and we're going to start out by finding what it'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'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're going to look for the Eigen vectors. Let's start with, say, Lambda One, and find its associate ID Aiken Vector. We'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're going to solve. And the augmented matrix for this equation is going to be first go to this entry. It's attract too. And I we'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'll get altogether one minus I. And then we're augmenting with the zero vector. So placed that here. Let's begin row reduction. Well, for the row reduction, if we'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'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's one when my sigh and zero, Then we'll have zero. And next we're multiplying one my Asai with one Plus I well, one my side times one plus I. If you foil us out, you'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'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'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'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.

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