# [SOLVED] diagonalization, eigenvectors, eigenvalues

by aznkid310
Tags: diagonalization, eigenvalues, eigenvectors, solved
 P: 109 1. The problem statement, all variables and given/known data Find a nonsingular matrix P such that (P^-1)*A*P is diagonal | 1 2 3 | | 0 1 0 | | 2 1 2 | 2. Relevant equations I am at a loss on how to do this. I've tried finding the eigen values but its getting me nowhere 3. 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 thats 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 cant get the second one, which is suppose to be X = -3, y = 0, z = 2
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 Quote by aznkid310 1. The problem statement, all variables and given/known data Find a nonsingular matrix P such that (P^-1)*A*P is diagonal | 1 2 3 | | 0 1 0 | | 2 1 2 | 2. Relevant 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.

 3. 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
$$\left|\begin{array}{ccc} 1-\lambda & 2 & 3 \\ 0 & 1-\lambda & 0 \\ 2 & 1 & 2-\lambda \end{array}\right|= 0$$.

I recommend expanding on the middle row. That gives
$$(1- \lambda)\left|\begin{array}{cc} 1- \lambda & 3 \\ 2 & 2-\lambda \end{array}\right|= 0$$
which is easily solved for $\lambda$= 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 thats wrong
Yes, it is wrong: -4 is not an eigenvalue!

Saying that 4 (not -4) is an eigenvalue gives
$$\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 2 & 1 & -2 & 0\end{array}\right]$$
subtracting 2/3 of the first row from the third row gives
$$\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & \frac{7}{3} & 0 & 0\end{array}\right]$$
and clearly now adding 7/9 of the second row to the third row gives
$$\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right]$$
(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 cant 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.
P: 109
 Quote by HallsofIvy I recommend expanding on the middle row. That gives $$(1- \lambda)\left|\begin{array}{cc} 1- \lambda & 3 \\ 2 & 2-\lambda \end{array}\right|= 0$$ which is easily solved for $\lambda$= 1, -1, and 4.
Could you explain what you did here? What do you mean my expanding the middle row?

P: 2

## [SOLVED] diagonalization, eigenvectors, eigenvalues

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 ).

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.
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