Linear Algebra Jordan Canonical Form- 3, guaranteed and verified 100% Pass

1


Jordan Canonical Form


Recall that earlier we saw that if 𝑇: 𝑉 → 𝑉 was a linear operator on an
𝑛-dimensional vector space represented in an ordered basis by a matrix 𝐴, then 𝑇
(or 𝐴) was diagonalizable if
1. The characteristic polynomial splits over ℝ, ie
𝑝(𝜆) = det(𝐴 − 𝜆𝐼) = 𝑐(𝜆1 − 𝜆) ⋯ (𝜆𝑛 − 𝜆); 𝑐∈ℝ
2. For each eigenvalue 𝜆𝑖 , the multiplicity of 𝜆𝑖 equals the dim(𝑁(𝑇 − 𝜆𝑖 𝐼)).


However, we also saw that if the characteristic polynomial of 𝑇 splits over ℝ that
1 1
𝑇 might not be diagonalizable (eg, 𝐴 = [ ]). Given that the characteristic
0 1
polynomial of 𝑇 splits over ℝ, we want to find an ordered basis for 𝑉 so that 𝑇 is
as close to being diagonal as possible. We will see that we can find an ordered
basis 𝐵 for 𝑉 such that:
𝐴1 0 0 ⋯ 0
0 𝐴2 0 ⋯ 0
[𝑇]𝐵 = 0 0 𝐴3 ⋯ 0
0 0 0 ⋱ ⋮
[0 0 0 ⋯ 𝐴𝑘 ]
where 0 is a zero matrix and
𝜆𝑖 1 0 ⋯ ⋯ 0
0 𝜆𝑖 1 ⋯ ⋯ 0
0 0 𝜆𝑖 ⋱ ⋯ 0
𝐴𝑖 = .
⋮ ⋮ ⋮ ⋱ ⋱ ⋮
0 0 0 ⋯ ⋱ 1
[0 0 0 ⋯ 0 𝜆𝑖 ]


That is, each 𝐴𝑖 will have 𝜆𝑖 , the 𝑖 𝑡ℎ eigenvalue, along the diagonal, ones along the
“superdiagonal” of 𝐴𝑖 , and zeros everywhere else. The matrix [𝑇]𝐵 is called the
Jordan canonical form of 𝑻.

, 2


Ex. Let 𝐵 = {𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 } be an ordered basis for 𝑉 and 𝑇: 𝑉 → 𝑉 a linear
operator with
2 1 0 0
0 2 1 0
𝐴 = [𝑇]𝐵 = [ ].
0 0 2 0
0 0 0 3
Identify 𝑁(𝑇 − 𝜆𝑖 𝐼) for each eigenvalue of 𝑇.
Notice that in this case:

𝐴 0 2 1 0
𝐴=[ 1 ], where 𝐴1 = [0 2 1] and 𝐴2 = [3].
0 𝐴2
0 0 2


The characteristic polynomial for 𝑇 is
2−𝜆 1 0 0
0 2−𝜆 1 0
det(𝐴 − 𝜆𝐼) = 𝑑𝑒𝑡 [ ]
0 0 2−𝜆 0
0 0 0 3−𝜆


= (2 − 𝜆)3 (3 − 𝜆).
Thus 𝑇 has 𝜆 = 2 as an eigenvalue of multiplicity 3 and 𝜆 = 3 as an eigenvalue of
multiplicity 1. Let’s find the eigenvectors of 𝑇.


For 𝜆 = 2 we have to find vectors that span the null space of 𝐴 − 2𝐼:
2 1 0 0 2 0 0 0 0 1 0 0
0 2 1 0 0 2 0 0 0 0 1 0
𝐴 − 2𝐼 = [ ]−[ ]=[ ].
0 0 2 0 0 0 2 0 0 0 0 0
0 0 0 3 0 0 0 2 0 0 0 1

, 3


0 1 0 0 𝑥1 0
0 0 1 0 𝑥2 0
(𝐴 − 2𝐼)𝑣 = [ ] [𝑥 ] = [ ]
0 0 0 0 3 0
0 0 0 1 𝑥4 0


𝑥2 0
𝑥3 0
or [ 0 ] = [ ].
0
𝑥4 0
So 𝑥2 = 𝑥3 = 𝑥4 = 0 and 𝑥1 can be any real number.
Thus the null space of 𝐴 − 2𝐼 is given by {< 𝑎, 0,0,0 >| 𝑎 ∈ ℝ} and is spanned
by < 1,0,0,0 >. Since the basis for 𝑉 is {𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 }, 𝑣1 =< 1,0,0,0 > is an
eigenvector associated with 𝜆 = 2 for 𝑇. We can check this by:
2 1 0 0 1 2 1
0 2 1 0 0 0 0
𝐴𝑣1 = [ ] [ ] = [ ] = 2 [ ] = 2𝑣1 .
0 0 2 0 0 0 0
0 0 0 3 0 0 0


For 𝜆 = 3 we need to find the null space of
2 1 0 0 3 0 0 0 −1 1 0 0
0 2 1 0 0 3 0 0 0 −1 1 0
𝐴 − 3𝐼 = [ ]−[ ]=[ ].
0 0 2 0 0 0 3 0 0 0 −1 0
0 0 0 3 0 0 0 3 0 0 0 0

−1 1 0 0 𝑥1 0
0 −1 1 0 𝑥2 0
(𝐴 − 3𝐼)𝑣 = [ ] [𝑥 ] = [ ]
0 0 −1 0 3 0
0 0 0 0 𝑥4 0
−𝑥1 + 𝑥2 0
−𝑥 + 𝑥3 0
or [ 2 ] = [ ].
−𝑥3 0
0 0