Diagonalization, eigenvectors, eigenvalues

In summary, the conversation discusses finding a nonsingular matrix P that will result in a diagonal matrix when multiplied with the given matrix A. The conversation covers various approaches, such as finding eigenvalues and eigenvectors, and using row reduction to solve for the matrix. Ultimately, the method of expanding the middle row of the determinant is recommended to solve for the eigenvalues, which are 1, -1, and 4.
  • #1
109
1
[SOLVED] diagonalization, eigenvectors, eigenvalues

Homework Statement



Find a nonsingular matrix P such that (P^-1)*A*P is diagonal

| 1 2 3 |
| 0 1 0 |
| 2 1 2 |


Homework Equations


I am at a loss on how to do this. I've tried finding the eigen values but its getting me nowhere


The Attempt at a Solution



i row reduced to 1 2 3
0 1 0
0 0 -4

and found : (x-1)^2 and (x+4), which gives me eigen values of 1 and -4.

Using A - Ix, when x = -4, the matrix becomes 5 2 3 0
0 5 0 0
2 1 6 0

Using row reduction: 1 2/5 1/5 0
0 5 0 0
2 1 6 0

1 2/5 1/5 0
0 1 0 0
0 1/5 28/5 0

1 2/5 1/5 0
0 1 0 0
0 0 28/5 0

1 0 0 0
0 1 0 0
0 0 1 0
which means X = y = z = 0, but that's wrong


When i use x = 1: 0 2 3 0
0 0 0 0
2 1 1 0
this gives 2y = -3z
2X + y + z = 0 ==> 2X = z/2

If z = 4, then x = 1, y = -6

That's one correct answer, but i can't get the second one, which is suppose to be
X = -3, y = 0, z = 2
 
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  • #2
aznkid310 said:

Homework Statement



Find a nonsingular matrix P such that (P^-1)*A*P is diagonal

| 1 2 3 |
| 0 1 0 |
| 2 1 2 |


Homework Equations


I am at a loss on how to do this. I've tried finding the eigen values but its getting me nowhere
Yes, the "diagonal" elements will be the eigenvalues of the matrix and the matrix P has eigenvectors corresponding to those eigenvalues as columns.


The Attempt at a Solution



i row reduced to 1 2 3
0 1 0
0 0 -4

and found : (x-1)^2 and (x+4), which gives me eigen values of 1 and -4.
Why in the world would you row-reduce? In general, the eigenvalues of the row-reduced matrix are not the eigenvalues of the original matrix. If it were, there wouldn't be such complicated computer routines for finding eigenvalues!

The eigenvalue equation for your matrix is
[tex]\left|\begin{array}{ccc} 1-\lambda & 2 & 3 \\ 0 & 1-\lambda & 0 \\ 2 & 1 & 2-\lambda \end{array}\right|= 0[/tex].

I recommend expanding on the middle row. That gives
[tex](1- \lambda)\left|\begin{array}{cc} 1- \lambda & 3 \\ 2 & 2-\lambda \end{array}\right|= 0[/tex]
which is easily solved for [itex]\lambda[/itex]= 1, -1, and 4.

Using A - Ix, when x = -4, the matrix becomes 5 2 3 0
0 5 0 0
2 1 6 0

Using row reduction: 1 2/5 1/5 0
0 5 0 0
2 1 6 0

1 2/5 1/5 0
0 1 0 0
0 1/5 28/5 0

1 2/5 1/5 0
0 1 0 0
0 0 28/5 0

1 0 0 0
0 1 0 0
0 0 1 0
which means X = y = z = 0, but that's wrong
Yes, it is wrong: -4 is not an eigenvalue!

Saying that 4 (not -4) is an eigenvalue gives
[tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 2 & 1 & -2 & 0\end{array}\right][/tex]
subtracting 2/3 of the first row from the third row gives
[tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & \frac{7}{3} & 0 & 0\end{array}\right][/tex]
and clearly now adding 7/9 of the second row to the third row gives
[tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right][/tex]
(You don't really need to carry that fourth column along- it will always be 0s). That gives y= 0, z= x so any eigenvector corresponding to eigen value 4 is a multiple of < 1, 0, 1>.





When i use x = 1: 0 2 3 0
0 0 0 0
2 1 1 0
this gives 2y = -3z
2X + y + z = 0 ==> 2X = z/2

If z = 4, then x = 1, y = -6

That's one correct answer, but i can't get the second one, which is suppose to be
X = -3, y = 0, z = 2
Yes, 1 is an eigenvalue, but not a double eigenvalue. -1 is the third eigenvalue.
 
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  • #3
HallsofIvy said:
I recommend expanding on the middle row. That gives
[tex](1- \lambda)\left|\begin{array}{cc} 1- \lambda & 3 \\ 2 & 2-\lambda \end{array}\right|= 0[/tex]
which is easily solved for [itex]\lambda[/itex]= 1, -1, and 4.

Could you explain what you did here? What do you mean my expanding the middle row?
 
  • #4


You can use the zeros in a matrix to make your work easier (and less likely to introduce arithmetic mistakes - a worthy skill to develop :smile:).

Remember that if you switch a row/column in the determinant, you change the sign of determinant? Well, imagine switching row1 with row2, then col2 with col1 (for a net sign change of +1).
Now, if you calculate your determinant using expansion of minors, all the zeros help you out, so that you are taking (1-λ) times the determinant of the remaining, smaller matrix, as HallsofIvy showed.
If you calculate your determinant by drawing the diagonal lines & multiplying, only the lines that go through (1-λ) will be non-zero, and you'll get the same result.
 
Last edited:
  • #5


This thread is 3 years old...
 

1. What is diagonalization?

Diagonalization is the process of finding a diagonal matrix that is similar to a given square matrix. This is achieved by finding a set of eigenvectors and eigenvalues for the original matrix and using them to transform the matrix into a diagonal form.

2. What are eigenvectors and eigenvalues?

Eigenvectors are special vectors that do not change direction when multiplied by a matrix. Eigenvalues are the corresponding scalars that represent the amount by which the eigenvectors are scaled. In other words, when a matrix is multiplied by its eigenvector, the resulting vector is a scaled version of the original eigenvector.

3. Why is diagonalization important?

Diagonalization is important because it simplifies many mathematical calculations involving matrices. It also allows us to more easily analyze the behavior of a system described by a matrix, as the diagonal form provides clear information about the system's stability and behavior.

4. How do you find the eigenvectors and eigenvalues of a matrix?

To find the eigenvectors and eigenvalues of a matrix, one can solve the characteristic equation of the matrix, which is obtained by subtracting the identity matrix multiplied by a scalar from the original matrix. The eigenvectors are then found by solving the system of equations formed by setting the characteristic equation equal to zero.

5. Can all matrices be diagonalized?

No, not all matrices can be diagonalized. A square matrix can only be diagonalized if it has a full set of linearly independent eigenvectors. If a matrix does not have enough eigenvectors, it is considered defective and cannot be diagonalized.

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