1 compute eigenvalues and eigenvectors of the following...

(1) Compute eigenvalues and eigenvectors of the following matrices. Indicate the geometric and algebraic multiplicities of each eigenvalue. $\begin{pmatrix} 0 & 1 & 0 \\ -4 & 4 & 0 \\ -2 & 1 & 2 \end{pmatrix}$, $\begin{pmatrix} 7 & -12 & 6 \\ 10 & -19 & 10 \\ 12 & -24 & 13 \end{pmatrix}$, $\begin{pmatrix} 3 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 3 & 0 & 5 & -3 \\ 4 & -1 & 3 & -1 \end{pmatrix}$.

Transcript

00:01 Hello students, to find the eigenvalue and eigenvector of a matrix, we need to solve the characteristics equation that is determinant of a -lambda i equal to 0 where a is the matrix and lambda is the eigenvalue and i is the identity matrix.

00:13 After finding the eigenvalue, we can substitute them back into a -lambda i and solve for the equation a -lambda i into v equal to 0 to find the corresponding eigenvector.

00:23 First, matrix a that is equal to 1 -3 3 3 -5 3 6 -6 4.

00:36 Now to find the eigenvalue, we will solve the characteristics equation that is determinant of a -lambda i equal to 0.

00:44 So lambda cube, this lead to the equation lambda cube minus 0 into lambda square minus 7 lambda minus 18 equal to 0.

00:53 This is equal to lambda minus 2 into lambda plus 3 into lambda minus 3 equal to 0.

01:02 So the eigenvalues are lambda equal to lambda 1 equal to 2, lambda 2 equal to 3 minus 3 and lambda 4 equal to, this is lambda plus 3 lambda plus 3.

01:18 This is lambda plus 3 lambda plus 3.

01:21 So we will get lambda 1 equal to 2, lambda 2 equal to minus 3 and lambda 3 is also minus 3.

01:28 To find eigenvector we solve a minus lambda i into b equal to 0.

01:33 For lambda equal to 2, minus 1, 3, 3, 3 minus 7, 3, 6 minus 6, 2, b equal to 0.

01:45 So solving we will get v1 equal to 1, 1, 2.

01:48 Now solving for lambda equal to minus 3, lambda equal to minus 3, solving the system we will get v2 equal to 1, 0, 1 and for, in this case we have two linearly independent eigenvector v1 and v2 which span the space.

02:08 Therefore matrix a is diagonalizable.

02:23 Now moving forward to the second, matrix b.

02:27 So matrix b is given as b equal to 3, minus 1, 1, 7, minus 5, 1, 6, minus 6, 2...

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