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 (1 − λ)(−1 − λ) − 2 × 4 = 0.
Multiply out the brackets, and 2 × 4 = 8.
Like terms together, the highest power first: λ² − 9 = 0.
It factors, as in the factoring lesson.
So λ = −3 or λ = 3.
2 Eigenvector for λ = −3
We take −3 off the diagonal: 1 − (−3) = 4 and −1 − (−3) = 2.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x + y = 0.
We want x alone.
To remove + y, we take away y on both sides.
They cancel on the left.
So x = −y.
We pick y = 1.
So an eigenvector for λ = −3 is (−1, 1).
3 Eigenvector for λ = 3
We take 3 off the diagonal: 1 − 3 = −2 and −1 − 3 = −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 = 3 is (2, 1).
4 Build P and D
P: the eigenvector for λ = −3 is column 1, the one for λ = 3 is column 2.
D: −3 first, then 3, 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 3, 6 and −3, 3.
So A P = P D, and the answer is right!
Let's look back.
First: find the eigenvalues.
Then: eigenvector for λ = −3.
Then: eigenvector for λ = 3.
Then: build P and D.
Last: check.