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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} .$$$\left[\begin{array}{rr}{1} & {5} \\ {-2} & {3}\end{array}\right]$$

$\mathrm { v } _ { 2 } = \mathrm { v } _ { 1 } = \left[ \begin{array} { c } { 1 + 3 i } \\ { 2 } \end{array} \right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 5

Complex Eigenvalues

Vectors

Campbell University

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

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in this example, we have a two by two matrix say that's provided medical. Here'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'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's constant to with a plus two I. So we confined that using the quadratic formula, and the next thing we're going to do is determine the corresponding Eigen vectors for each Eigen value. So let'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'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'll have altogether two plus two. I they were augmenting with a zero vector, which I'll place here. Now it'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's here and uncle is to eliminate this entry. So let'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'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'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'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.

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