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I am trying to solve the following exercise (not to be handed in):

Let $T : V \rightarrow V$ be a $\mathbb{F}$-linear map with $\dim V < \infty$. Suppose that $p(T)=0$ for some monic polynomial $p(x) \in \mathbb{F}[x]$ which factorizes into distinct linear factors. Show that $T$ is diagonalisable.

Since $p(T) = 0$ and $p(x)$ factorizes into some, say $n$ distinct linear factors, I can decompose $V$ into a direct sum of kernels in the following way $$V = \ker a_1 \oplus \ker a_2 \oplus \ldots \oplus \ker a_n$$

where $p(x) = a_1 \cdot a_2 \cdots a_n$ is the factorization of $p(x)$. But how do I use this to show that $T$ is diagonalisable?

Since this is meant for exam study and is not really homework, I would like the solution to be more than a hint.

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How do you know it has at least $n$ factors? – anonymous Sep 16 '12 at 13:01
I probably don't know that, and I have removed it from the question. – utdiscant Sep 16 '12 at 13:38
up vote 1 down vote accepted

As stated in my lecture note the theorem say that if we have a decomposition of the minimal polynomial of $T$ into irreducible elements in $\mathbb{F}[\lambda]$


Then there are $r$ invariant subspaces $$W_{1},...W_{r}$$ s.t $$V=W_{1}\oplus...\oplus W_{r}$$ and $T|_{W_{i}}=\varphi_{i}(\lambda))^{c_{i}}$

In your case you have some polynomial $p$ s.t $p(T)=0$ and $p$ splits into distinct linear factors, since $m_{T}\mid p$ we have it that $m_{T}$ splits into distinct linear factors.

Let $\alpha_{1},...,\alpha_{r}\in\mathbb{F}$ s.t $m_{T}(x)=(x-\alpha_{1})...(x-\alpha_{r})$ (and recall $i\neq j\implies\alpha_{i}\neq\alpha_{j})$.

Since each $W_{i}$ is invariant and by the stated above we have that this means $T(W_{i})\subset W_{i}$ and $\forall v\in W_{i}:\, T|_{W_{i}}(v)=\alpha_{i}v$

Denote $k_{i}:=dim(W_{i})$, then by $V=W_{1}\oplus...\oplus W_{r}$ we have it that $k_{1}+...+k_{r}=n$. Take a basis $B_{i}$ for $W_{i}$ and from $\forall v\in W_{i}:\, T|_{W_{i}}(v)=\alpha_{i}v$ deduce that $[T|_{W_{i}}]_{B_{i}}$ is of the form $$diag(\alpha_{i},...,\alpha_{i})$$

Since this is a direct sum we have a basis $B=\cup B_{i}$ and $[T]_{B}$is diagonal as a direct sum of diagonal matrices.

Note: You can also use Jordan normal form to deduce that the minimal polynomial splits into distinct linear factor iff the matrix is diagonalizable but I chose to do it by the theorem as asked (the existence of the Jordan normal form in this case should not worry you since the eigenvalues are on the field in this is sufficient, in any case there is a field containing $\mathbb{F}$ in which the Jordan normal form exist).

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The point here is the word "linear". If the factors are linear, it means that are of the form $t - \lambda$. Then, the invariant subspaces in your decomposition are just eigenvector subspaces:

$$ \mathrm{Ker} (T - \lambda \mathrm{Id}) \ . $$

So, you pick a basis of each one -which will be formed by eigenvectors-, and put all these bases together, obtaining a basis of $V$ made out of eigenvectors. Thus, $T$ is diagonalisable.

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I would argue that the point is "distinct" and not "linear" since there exist a field extension where the polynomial is always splits into linear factor but is diagonalizable iff they are distinct – Belgi Sep 24 '12 at 15:13
They are both important; if $(x-1)^{2}$ is the minimal polynomial, then $x$ can be non-diagonalizable. "distinct linear" makes sense to me. – Bombyx mori Sep 24 '12 at 15:32
Yes, you're right: linear + distinct makes the thing. – a.r. Sep 24 '12 at 16:11
@user32240: $(x-1)^2=(x-1)(x-1)$ so it does splits into linear factors. both are of course important but since the splitting is of no issue in an extension field the real key point is the factors being distinct – Belgi Sep 24 '12 at 18:00
@Belgi: Yes, but in our case the minimal polynomial would be degree 2, so not "linear" anymore. But I agree. – Bombyx mori Sep 24 '12 at 18:09

Simply realize that the minimal polynomial consists of different linear factors. If this is the case T is diagonalisable (To see this use the primary decomposition, every characteristic subspace is of dimension 1).

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What do you mean by "every characteristic subspace is of dimension 1" ? if I understand you correctly then this is wrong. take for exmaple $T=Id$ then $m_T=x-1$ but the $W=W_1$ in the primary decomposition is of dimension $n$ = the dimension of the eigenspace corresponding to the eigenvalue $1$ . In general the dimension is not $1$ but rather the geometric multiplicity of the eigenvalue – Belgi Sep 25 '12 at 1:01

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