Finding the Diagonalizing Matrix for A

[bennyska]

Homework Statement

find the diagonalizing matrix P of the matrix A
A=:
1 3 0
0 2 0
0 0 2

Homework Equations

The Attempt at a Solution

so i do the whole A-\lambdaI₃ thing and i find my eigenvalues to be \lambda = 1, 2
when i do \lambda = 1, i get the matrix
0 3 0
0 1 0
0 0 1

which turns into
0 1 0
0 0 1
0 0 0

i guess here's my main question. can i have any eigenvectors from here that satisfy this equation? I'm getting a different solution than the book, but it's been a while since I've taken linear algebra (this is a differential equations class and this parts review), and i seem to recall that this doesn't need to be like the book (i.e. the book presents one possible solution; as long as my solution is linearly independent, it should be valid).
...
okay, so i just plugged it in and found out I'm incorrect.

so how do i find the solution to
0 1 0 | 0
0 0 1 | 0
0 0 0 | 0?

 

[vela]

What equations correspond to that augmented matrix?

 

[Dick]

And the first row of your second matrix should be [-1,0,0].

 

[bennyska]

And the first row of your second matrix should be [-1,0,0].

why? i thought the whole point of doing the eigenvalue was to get that to be zero? (at least in this part, when the particular eigenvalue cancels that out)

 

[Dick]

why? i thought the whole point of doing the eigenvalue was to get that to be zero? (at least in this part, when the particular eigenvalue cancels that out)

And I thought you were subtracting the eigenvalue from the matrix diagonal.

 

[bennyska]

the original matrix:
1 3 0
0 2 0
0 0 2

eigenvalues 1, 2, 2

subtracting 1*identity

0 3 0
0 1 0
0 0 1

isn't that right?

 

[Dick]

Yes, it is. I was thinking one of your matrices was the one where you subtracted 2*identity. Now you need to find the null space of that matrix.

 

[VeeEight]

I believe that was his original problem - finding the null space.

 

[HallsofIvy]

Homework Statement

find the diagonalizing matrix P of the matrix A
A=:
1 3 0
0 2 0
0 0 2

I have never particularly liked that method of finding eigenvectors- I prefer working directly from the definition.

It is obvious that the eigenvalues of A are 1 and 2 with 2 being a double root of the eigenvalue equation.

Any eigevector corresponding to eigenvalue 1 must satify
\begin{bmatrix}1 & 3 & 0 \\ 0 & 2 & 0 \\ 0 & 2 & 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix} x \\ y \\ z\end{bmatrix}

Multiplying that out we get x+ 3y= x, 2y= y, and 2z= z. The only solutions to the last two equations are y= z= 0 and any value of x then satifies the first. An eigenvector corresponding to eigenvalue 1 is <x, 0, 0>= x<1, 0, 0>.

Any eigenvector corresponding to eigenvalue 2 must satisfy
\begin{bmatrix}1 &amp; 3 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 0 &amp; 2 &amp; 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 2x \\ 2y \\ 2z\end{bmatrix}

Multiplying that out, we get the 3 equations x+ 3y= 2x, 2y= 2y, and 2z= 2z. The first equation gives us x= 3y and any values of y and z satisfy the last two. An eigenvector corresponding to eigenvalue 2 must be of the form <3y, y, z> = y<3, 1, 0>+ z<0, 0, 1>. That is, {<1, 0, 0>, <3, 1, 0>, <0, 0, 1>} is a basis for the entire space consisting of eigenvectors of A. If we take P to be the matrix having those vectors as columns:
P= \begin{bmatrix}1 &amp; 3 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{bmatrix}
then P^-1AP is a diagonal matrix.

Homework Equations

The Attempt at a Solution

so i do the whole A-\lambdaI₃ thing and i find my eigenvalues to be \lambda = 1, 2
when i do \lambda = 1, i get the matrix
0 3 0
0 1 0
0 0 1

which turns into
0 1 0
0 0 1
0 0 0

i guess here's my main question. can i have any eigenvectors from here that satisfy this equation? I'm getting a different solution than the book, but it's been a while since I've taken linear algebra (this is a differential equations class and this parts review), and i seem to recall that this doesn't need to be like the book (i.e. the book presents one possible solution; as long as my solution is linearly independent, it should be valid).
...
okay, so i just plugged it in and found out I'm incorrect.

so how do i find the solution to
0 1 0 | 0
0 0 1 | 0
0 0 0 | 0?[/QUOTE]