4. Find all eigenvalues of A = < b m a t r i x >, where...

4. Find all eigenvalues of $A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}$, where $a$ and $b$ are arbitrary constants. 5. Consider the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ k & 3 & 0 \end{bmatrix}$, where $k$ is an arbitrary constant. For which values of $k$ does $A$ have three distinct eigenvalues? What happens in the other cases?

Transcript

00:33 Okay, so for problem 2, we're given three eigen values with corresponding eigenvectors, and we need to find the solution of the equation xk plus 1 equals a xk.

00:48 So the first thing is to find x0, and by the definition in our textbook, we have x0 is equal to c1, which is a scalar, times v1 plus c2, times v2 plus c3 times v3.

01:38 Okay, so for problem 2, we're given because we only have three eigen vectors in this problem, so we only write down three terms.

01:50 Okay, so it turns out to be x0, x0 is given, which is negative 2, negative 5, and 3, and we have the first eigenvector, which is 1 -0, negative 3, and 3 eigenvalues with corresponding eigenvectors.

02:19 And we need to find the solution of the equation xk plus 1 equals a xk.

02:26 So the first thing is to find x0.

02:29 And by the definition in our textbook, we have x0 is equal to c1, which is a scalar, times v1 plus c2 times v2 plus c3 times v3 second eigenvector which is 2 1 negative 5 excuse me now the last argument vector which is negative 3 3 negative 3 at 7 so that is equivalent to find out the solution of this linear system.

03:20 We only have three eigenvectors in this problem, so we only write down three terms.

03:28 Okay, so it turns out to be x0, x0 is given, which is negative 2, negative 5, and 3, and we have the first eigenvector, which is 1 ,0, negative 3.

03:52 And set up 1, 2, negative 3, 0, 1, negative 3, negative 3, negative 5, 7 times a column vector with c1, c2, c3, which is equal to negative 2, negative 5, 3.

04:22 Okay, backhand eigenvector, which is 2 -1, negative 5.

04:40 Excuse me.

04:41 Now the last argument, which is negative 3, negative 3 at 7.

04:52 So that is equivalent to find out the solution of this linear system.

04:58 I solving this system, i will skip the process.

05:02 So the solution is c1, c2, c3 is 2, 12, 2, 2, so that we found, we have already found our coefficient, which is a scalar of the, for each, excuse me, for each eigenvector.

05:26 So it turns out to be twice of 1, 1, 2, negative 3, 0, 0 ,000, 2, 0...

Question

Please give Ace some feedback