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Given that: $$A \in M_{n\times n} (\mathbb F), \; A \; \text{has only one eigenvalue} \; \lambda$$

We have to prove/disprove: $$A \; \text{is diagonaizable if and only if} \; \exists \lambda : A = \lambda I$$

To be honest I've been thinking about this question for an hour now and have no clue on how to even start! Any hints appreciated!

PLEASE take a look at the 3rd comment, that's what I got so far.

TheNotMe
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    Well. For one direction, if there is a $\lambda$ so that $A = \lambda I$, then... In the other direction, if $A$ is diagonalizable, what does that mean? Write it out in terms of your definition. Then, if you know there's only one eigenvalue, what do the diagonal entries look like? – Zach L. May 02 '13 at 17:17
  • Will do. Working on it. – TheNotMe May 02 '13 at 17:26
  • The first side is easy apparently, it is already in diagonal form, and its only eigenvalue is $n$ times $\lambda$ which is $\lambda$. Now in the other direction. I know that $A = P D P^{-1}$ when $D = \begin{pmatrix} \lambda & ... & 0 \ .. & .. & .. \ 0 & .. & \lambda \end{pmatrix}$ and $p = [[v_1][v_2]...[v_n]]$ Now I am asking myself, does one eigenvalue mean one eigenvector? No... so what is the concept? – TheNotMe May 02 '13 at 17:38
  • In the question, they never mentioned if it is algebraically closed or arbitrary.. – TheNotMe May 03 '13 at 08:47

3 Answers3

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You only need the basic definitions for this question. An operator is diagonalisable if the (direct) sum of its eigenspaces fill the whole space (equivalently, every vector can be written as a sum of eigenvectors for distinct eigenvalues). If in addition there is only one eigenvalue$~\lambda$, then the eigenspace for$~\lambda$ must therefore fill the whole space: the operator multiplies every vector by$~\lambda$ and therefore equals$~\lambda I$ (or: every vector is the sum of at most one eigenvector for$~\lambda$, so it either is such an eigenvector or it is$~0$).

Of course the opposite direction that $\lambda I$ is diagonalisable is trivial.

Marc van Leeuwen
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You've almost got it, in the third comment above. You just need to use that $A = PDP^{-1} = P(\lambda I)P^{-1} = \lambda PIP^{-1} = \lambda I$.

Alistair Savage
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An easier approach...but knowing some slightly more advanced stuff:

== $\,A\,$ has one single eigenvalue $\,\lambda\,$ iff its characteristic polynomial is $\,(x-\lambda)^n\,$

== A matrix is diagonalizable iff its minimal polynomial is a product of different linear factors, from which it follows that

$$A\;\text{is diagonalizable}\iff\;(x-\lambda)\;\text{is its minimal polynomial}\iff A-\lambda I=0\iff A=\lambda I$$

DonAntonio
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  • can you please explain for me the difference between a *characteristic polynomial of a matrix* and a *minimal polynomial of a matrix* ? – TheNotMe May 03 '13 at 08:49
  • The characteristic pol. of a matrix square $,A,$ is the polynomial $,p_A(t):=\det(tI-A),$ . The minimal pol. of $,A,$ is the monic non-zero pol. $,m_A(t),$ of minimal degree s.t. $,m_A(A)=0,$ . It is always true that $,m_A(t),\mid,p_A(t),$ (all the above: square matrices and polynomials over a field) – DonAntonio May 03 '13 at 23:54
  • @DonAntonio Can you help me find a link to the proof "A has one single eigenvalue λ iff its characteristic polynomial is (x−λ)n"? I know it is obvious but struggle to prove it formally. I am reading Linear Algebra by Friedberg and stumbled by this claim that was thrown in the middle of theorem 7.2. Thank you. – Lawerance Aug 23 '18 at 12:27
  • @Lawerance Exactly:it is obvious according to theorem/definition, but follow Friedberg: $;\lambda\in F;$ a field is an eigenvalue of operator $;T;$ iff there exists non-zero vector $;v;$ s.t. $$;Tv=\lambda v\iff (\lambda I-T)v=0\iff \text{ the operator};\lambda I-T$$ is singular, and thus $;\lambda;$ is en eigenvalue iff it is a root of the polynomial $;\det(\lambda I-T);$ . – DonAntonio Aug 23 '18 at 16:16