We are asked to find the bases for the eigenspaces of the matrices in Exercise 4. We need to repeat the whole process for each matrix (in each part). Part a) First of all, let us look at the matrix [λI-A], where A is the given matrix.
A=
[λI-A] = In question 4, we found using the Characteristic Equation, det(λI-A)=0, that the eigenvalues for A are λ1=1, λ2=2, λ3=3. Next, we will follow the steps outlined below for each eigenvalue: We will construct a matrix by subsituting the value for each λi , use elementary row operations to put it in Row Echelon Form and find the solution vector, which will be the corresponding eigenvector, xi. λ1=1
λ2=2
λ3=3
Now, we have all the eigenvectors. The basis for the eigenspace of A is simply the set of all the eigenvectors we found:
= Part b) First of all, let us look at the matrix [λI-A], where A is the given matrix.
A=
[λI-A] = In question 4, we found using the Characteristic Equation, det(λI-A)=0, that the eigenvalues for A are λ1=0, λ2=√2, λ3=-√2. Next, we will follow the steps outlined below for each eigenvalue: We will construct a matrix by subsituting the value for each λi , use elementary row operations to put it in Row Echelon Form and find the solution vector, which will be the corresponding eigenvector, xi. λ1=1
λ2=√2
λ3=-√2
Now, we have all the eigenvectors. The basis for the eigenspace of A is simply the set of all the eigenvectors we found:
=