What Determines the Jordan Canonical Form of a Matrix?

[Ressurection]

Homework Statement

Let A be an nxn matrix (of real or complex components) and

[tex]J=\left(\begin{array}{c} λ & 1 & 0 & 0\\
0 & λ & 1 & 0\\
& & ... & \\
0 & 0 & λ & 1\\
0 & 0 & 0 & λ
&\end{array}\right) \,with\, λ \in ℂ[/tex]


Show that there is
[tex]S = \left(\begin{array}{c} v1 & v2 & ... & v_n &\end{array}\right)
\,with\, v1, v2, ...,v_n \inℂ [/tex]
such that A = SJS^-1 if and only if:
(A - λI)v₁ = 0
(A - λI)v_i+1 = v_i , for i=1,2,...,n-1

Homework Equations

The Attempt at a Solution

My first step was A = SJS^-1 ⇔ AS = SJ
Now, developing the right side I get SJ = [λv₁ , v₁ + λv₂ ... v_n-1 + λv_n ]

So, column by column I get: Av₁ = λv₁ ⇔ (A-λ)v₁ = 0
Av₂ = v₁ + λv₂ ⇔ (A-λ)v₂ = v₁

and extending, I get
(A - λI)v_i+1 = v_i , for i=1,2,...,n-1

My only question is, does this solve the problem? I thought that to prove a ⇔ b , I had to prove a [itex]\Rightarrow[/itex] b and b [itex]\Rightarrow[/itex] a, but it seems to me that this proves both ways.

 

[mighty2000]

Am I missing something?

Thank you for your post. Your approach to the problem is correct and you have indeed proved both directions of the statement. To show that a statement is true if and only if another statement is true, you only need to prove one direction and the other direction automatically follows. So your solution is complete and correct.

One way to think about it is that if we start with A and construct S and J as given, then we will end up with (A-λI)v1 = 0 and (A-λI)vi+1 = vi for i = 1,2,...,n-1. Conversely, if we start with (A-λI)v1 = 0 and (A-λI)vi+1 = vi for i = 1,2,...,n-1 and construct S and J as given, we will end up with A. Therefore, the two statements are equivalent and proving one direction is sufficient.

I hope this helps clarify any confusion. Keep up the good work in your studies!