# Thread: Diagonalization, soluton check

1. ## Diagonalization, soluton check

Was just wondering if I did this problem correctly

Diagonalize the matrix B
3 0 0
1 2 0
1 −1 4
That is, find matrices P and D so that D is diagonal
and P(inverse)BP = D.

So to find P, I found the eigenvalues, 2,3,4,
then fond the eigenvectors by plugging in 2,3,4 respectively, so
when eigenvalue is 2,
1 0 0
1 0 0
1 -1 2
which gave me the eigen vector [0 .5 1]
when eigenvalue is 3,
0 0 0
1 -1 0
1 -1 1
which gave me the eigen vector [1 1 0]
when eigenvalue is 4,
-1 0 0
1 -2 0
1 -1 2
which gave me the eigen vector [0 0 1]
so P =
0 1 0
.5 1 0
1 0 1
Then P inverse has to be
-2 2 0
1 0 0
2 -2 1
and so D =
2 0 0
0 3 0
0 0 4

Did I do this problem correctly?

2. Originally Posted by daklutz
Was just wondering if I did this problem correctly

Diagonalize the matrix B
3 0 0
1 2 0
1 −1 4
That is, find matrices P and D so that D is diagonal
and P(inverse)BP = D.

So to find P, I found the eigenvalues, 2,3,4,
then fond the eigenvectors by plugging in 2,3,4 respectively, so
when eigenvalue is 2,
1 0 0
1 0 0
1 -1 2
which gave me the eigen vector [0 .5 1]
when eigenvalue is 3,
0 0 0
1 -1 0
1 -1 1
which gave me the eigen vector [1 1 0]
when eigenvalue is 4,
-1 0 0
1 -2 0
1 -1 2
which gave me the eigen vector [0 0 1]
so P =
0 1 0
.5 1 0
1 0 1
Then P inverse has to be
-2 2 0
1 0 0
2 -2 1
and so D =
2 0 0
0 3 0
0 0 4

Did I do this problem correctly?
The eigenvector I got for $\lambda = 2$ is ${\bf e_1} = \left(\begin{array}{c}0\\1\\0.5\end{array}\right)$. This then changes $P$ and it's inverse.