Step by step
The question
Question: diagonalize A.
Diagonalizing means writing A as P times D times P inverse.
The columns of P are eigenvectors.
D is diagonal: the eigenvalues down the diagonal, 0s elsewhere.
1 Find the eigenvalues
As in the eigenvalue lesson: take λ off the diagonal of A.
Its determinant, down minus up, must be 0.
So (6 − λ)(1 − λ) − 2 × (−2) = 0.
Multiply out the brackets, and 2 × (−2) = −4.
Like terms together, the highest power first: λ² − 7λ + 10 = 0.
It factors, as in the factoring lesson.
So λ = 2 or λ = 5.
2 Eigenvector for λ = 2
We take 2 off the diagonal: 6 − 2 = 4 and 1 − 2 = −1.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − y/2 = 0.
We want x alone.
To remove − y/2, we add y/2 on both sides.
They cancel on the left.
So x = y/2.
We pick y = 2 to keep whole numbers.
So an eigenvector for λ = 2 is (1, 2).
3 Eigenvector for λ = 5
We take 5 off the diagonal: 6 − 5 = 1 and 1 − 5 = −4.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − 2y = 0.
We want x alone.
To remove − 2y, we add 2y on both sides.
They cancel on the left.
So x = 2y.
We pick y = 1.
So an eigenvector for λ = 5 is (2, 1).
4 Build P and D
P: the eigenvector for λ = 2 is column 1, the one for λ = 5 is column 2.
D: 2 first, then 5, in the same order.
Careful!
The order matters.
The first λ in D goes with the first column of P.
So A = P D P⁻¹, with this P and D.
5 Check
Let's check without the inverse: A = P D P⁻¹ means A P = P D.
We work out both sides.
First A times P.
Now P times D.
Both give rows 2, 10 and 4, 5.
So A P = P D, and the answer is right!
Let's look back.
First: find the eigenvalues.
Then: eigenvector for λ = 2.
Then: eigenvector for λ = 5.
Then: build P and D.
Last: check.