For the matrix A below, find a value of k so that A has two basic eigenvectors associated with the eigenvalue λ = 3.
A = [−3 −18 54 204
0 3 −18 −54
0 0 −3 k
0 0 0 3]
k = ?
I'm specifically having troubles reducing this and knowing what it'll look when there are 2 parameters aka 2 eigenvectors?
k = -27
To find the value of k such that matrix A has two basic eigenvectors associated with the eigenvalue λ = 3, we need to follow these steps:
Step 1: Write down the characteristic equation
The characteristic equation is formed by setting the determinant of (A - λI) equal to zero, where A is the matrix and λ is the eigenvalue we are interested in.
In this case, our eigenvalue is λ = 3. We represent the identity matrix as I.
So, (A - 3I) is:
[-6 -18 54 204
0 0 -18 -54
0 0 -6 k
0 0 0 0]
Step 2: Calculate the determinant of (A - 3I)
To find the determinant, we can use the row reduction method. Let's row reduce the matrix to bring it to row-echelon form.
R1 = R1/(-6)
R2 = R2 - R1 * (0)
R3 = R3/(-6)
R4 = R4 - R1 * (0)
Now the matrix is:
[1 3 -9 -34
0 0 -18 -54
0 0 1 -k/6
0 0 0 0]
Step 3: Continue row reduction
We want to make the second column to have zeros below the leading entry (pivot) of the first column. We can multiply the third row by 18 and add it to the second row.
R2 = R2 + R3 * (-18)
Now the matrix is:
[1 3 -9 -34
0 0 0 0
0 0 1 -k/6
0 0 0 0]
Step 4: Identify the basic eigenvectors
Since the second row consists entirely of zeros, it does not contribute to the basic eigenvectors. Hence, we focus on the remaining rows:
Row 1: [1 3 -9 -34] represents the equation x + 3y - 9z - 34w = 0.
Row 3: [0 0 1 -k/6] represents the equation z - (k/6)w = 0.
Step 5: Set up a system of equations
We have two equations based on the rows we selected in Step 4:
x + 3y - 9z - 34w = 0
z - (k/6)w = 0
Step 6: Solve the system of equations
We can solve this system of equations to find the values of k.
From the second equation, we have: z = (k/6)w.
Substituting this value of z in the first equation:
x + 3y - 9(k/6)w - 34w = 0.
x + 3y - (9k/6 + 34)w = 0.
For this system of equations to have a non-trivial solution (the eigenvector), the determinant of the coefficient matrix must equal zero.
So, the determinant is:
|1 3 -(9k/6 + 34)|
|0 0 1 |
The determinant 1 * 1 - 0 * (9k/6 + 34) = 1 - 0 = 1 must equal zero.
Therefore, we have the equation:
1 - 0 = 1.
1 = 0 (Not possible).
Since the determinant is non-zero, there is no value of k that will result in two basic eigenvectors for the eigenvalue λ = 3.
To find a value of k such that matrix A has two basic eigenvectors associated with eigenvalue λ = 3, we can follow these steps:
Step 1: Set up the eigenvalue equation
Start by setting up the eigenvalue equation: (A - λI)x = 0, where λ is the eigenvalue we are interested in, I is the identity matrix, and x is the eigenvector.
Step 2: Substitute the values into the equation
Substitute the values from matrix A into the equation:
[A - 3I]x = 0
⇒ [−3 −18 54 204
0 3 −18 −54
0 0 −3 k
0 0 0 3 - 3]x = 0
⇒ [−3 −18 54 204
0 3 −18 −54
0 0 −3 k
0 0 0 0]x = 0
Step 3: Reduce the matrix
Reduce the matrix to row-echelon form using row operations:
R2 = R2 + 6R1 ⇒ [−3 −18 54 204
0 0 0 0
0 0 −3 k
0 0 0 0]x = 0
R1 = -R1 ⇒ [3 18 -54 -204
0 0 0 0
0 0 −3 k
0 0 0 0]x = 0
Step 4: Identify the basic eigenvectors
To have two basic eigenvectors associated with λ = 3, we should have at least two rows of zeros below the diagonal elements of the third column. If there are any nonzero elements in the last two rows of the third column, both rows will become the basic eigenvectors.
From the reduced matrix above, we can see that there are nonzero elements in the last two rows of the third column. Therefore, the basic eigenvectors associated with λ = 3 will be the last two rows of the reduced matrix.
Step 5: Determine the value of k
The value of k can be any real number since it does not affect the basic eigenvectors associated with λ = 3. Therefore, k can be any real number.
In conclusion, the value of k can be any real number for matrix A to have two basic eigenvectors associated with eigenvalue λ = 3.