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Suppose $V$ is a vector space over $\mathbb{C}$, and $A$ is a linear transformation on $V$ which is diagonalizable. I.e. there is a basis of $V$ consisting of eigenvectors of $A$. If $W\subseteq V$ is an invariant subspace of $A$ (so $A(W)\subseteq W$), show that $A|_W$ is also diagonalizable.

I tried supposing $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_m$, with $V_i=\{v\in V: Av=\lambda_i v\}$. Then we can write $V=V_1\oplus\cdots\oplus V_m,$ but I'm not sure whether it is true that

$$W=(W\cap V_1)\oplus\cdots\oplus (W\cap V_m),.$$

If it is true, then we're done, but it may be wrong.

azif00
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NGY
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  • @NGY: Since $W$ is in fact diagonalizable, and any eigenvector of $A|_W$ is necessarily an eigenvector of $A$, yes, it's true: you know $W=W_1\oplus \cdots \oplus W_m$ (some of them possibly trivial). Find a basis of eigenvectors for each $W_i$, then extend to a basis of eigenvectors of the corresponding $V_i$. But I think proving it ex nihilo will be difficult. – Arturo Magidin Sep 06 '11 at 17:10
  • The desired result can also be phrased as: let $A$ be a linear operator on a finite dimensional vector space (over any field). If $A$ is diagonalizable, then every invariant subspace is a direct sum of one-dimensional invariant subspaces. As Arturo says though, it doesn't seem straightforward to prove this directly. His nice answer using canonical forms looks like the best way to go. – Pete L. Clark Sep 06 '11 at 17:55
  • I'm curious about whether this theorem can also be generalized to infinite-dimensional vector spaces? – alan23273850 Nov 26 '21 at 05:07
  • @alan23273850 It was proved below in full generality by Zorn and Pierre-Yves Gaillard. – Anne Bauval Oct 19 '24 at 12:47

4 Answers4

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This theorem is true for arbitrary $V$ (over an arbitrary field $\mathbb{F}$).

We can prove the following

Lemma. If $v_1 + v_2 + \cdots + v_k \in W$ and each of the $v_i$ are eigenvectors of $A$ corresponding to distinct eigenvalues, then each of the $v_i$ lie in $W$.

Proof. Proceed by induction. If $k = 1$ there is nothing to prove. Otherwise, let $w = v_1 + \cdots + v_k$, and $\lambda_i$ be the eigenvalue corresponding to $v_i$. Then:

$$Aw - \lambda_1w = (\lambda_2 - \lambda_1)v_2 + \cdots + (\lambda_k - \lambda_1)v_k \in W.$$

By induction hypothesis, $(\lambda_i - \lambda_1)v_i \in W$, and since the eigenvalues $\lambda_i$ are distinct, $v_i \in W$ for $2 \leq i \leq k$, then we also have $v_1 \in W$. $\quad \square$

Now each $w \in W$ can be written as a finite sum of nonzero eigenvectors of $A$ with distinct eigenvalues, and by the Lemma these eigenvectors lie in $W$. Then we have $W = \bigoplus_{\lambda \in F}(W \cap V_{\lambda})$ as desired (where $V_{\lambda} = \{v \in V\mid Av = \lambda v\}$).

user26857
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Zorn
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  • Another view on this is the following: One typically proves that eigenvectors corresponding to distinct eigenvalues are linearly independent. Now apply this to the quotient space $V/W$ and we immediately get from this that if $v_1+\dots+v_k\in W$, then $v_1,\dots, v_k\in W$. Of course, if one expands the proof of that fact above, it becomes the same proof as in this answer. But I think it is helpful to note that this is really a consequence of this other perhaps more well-known fact. – leoli1 May 08 '25 at 01:37
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Theorem. A linear transformation is diagonalizable if and only if its minimal polynomial splits and has no repeated factors.

Proof. This follows by examining the Jordan canonical form, since the largest power of $(x-\lambda)$ that divides the minimal polynomial is equal to the size of the largest block of corresponding to $\lambda$ of the Jordan canonical form of the linear transformation. (Use the fact that every irreducible factor of the characteristic polynomial divides the minimal polynomial, and that the characteristic polynomial must split for the linear transformation to be diagonalizable to argue that you can restrict yourself to linear transformations with Jordan canonical forms). QED

Theorem. Let $A$ be a linear transformation on $V$, and let $W\subseteq V$ be an $A$-invariant subspace. Then the minimal polynomial of the restriction of $A$ to $W$, $A|_{W}$, divides the minimal polynomial of $A$.

Proof. Let $B=A|_{W}$, and let $\mu(x)$ be the minimal polynomial of $A$. Since $\mu(A)=0$ on all of $V$, the restriction of $\mu(A)$ to $W$ is $0$; but $\mu(A)|_{W} = \mu(A|_{W}) = \mu(B)$. Since $\mu(B)=0$, then the minimal polynomial of $B$ divides $\mu(x)$. QED

Corollary. If $A$ is diagonalizable, and $W$ is $A$-invariant, then the restriction of $A$ to $W$ is diagonalizable.

Proof. The minimal polynomial of $A$ splits and has no repeated factors; since the minimal polynomial of $A|_W$ divides a polynomial that splits and has no repeated factors, it follows that it itself has no repeated factors and splits. Thus, the restriction of $A$ to $W%$ is also diagonalizable. QED

Willie Wong
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Arturo Magidin
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    How do I see that $\mu(A)\restriction_W=\mu(A\restriction_W)$? Also, if $\nu(x)$ is the minimal polynomial of $B$, then the second theorem claims that $\mu(x)=\nu(x)f(x)$ for some polynomial $f(x)$. But how does it follow from the proof? – user557 Aug 07 '18 at 16:55
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    @user437309: You are replying to a post that is almost 7 years old... the chances of quickly remembering what one was thinking at the time are dim. The reason that $\mu(x) = \nu(x)f(x)$ follows because the minimal polynomial of $B$ divides any polynomial that vanishes at $B$ (using the division algorithm). This is a standard property of minimal polynomials. The proof shows that $\mu(B)=0$, hence $\mu(x)$ vanishes at $B$, hence must be a multiple of the minimal polynomial. What is the problem? – Arturo Magidin Aug 07 '18 at 17:20
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    @user437309: $\mu(A) = A^k + a_{k-1}A^{k-1} + \cdots + a_1A + a_0I$. Restricting the sum to $W$ is the same thing as adding the restrictions; and restricting $A^k$ is the same thing as restricting $A$ and raising the restriction to the $k$th power. – Arturo Magidin Aug 07 '18 at 17:22
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Here is a minor variation of Zorn's very nice argument. I'll use Zorn's notation:

Let $w=v_1 + v_2 + \cdots + v_k$ be in $W$, each $v_i$ being a $\lambda_i$-eigenvector of $A$, and the $\lambda_i$ being distinct.

It suffices to check that each $v_i$ is in $W$.

But this is clear since

$$v_i=\left(\prod_{j\neq i}\ \frac{A-\lambda_j\,I}{\lambda_i-\lambda_j}\right)(w)\quad.$$

Marc van Leeuwen
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Pierre-Yves Gaillard
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    Very nice. This is the the explicit formula for writing the projection from the sum of all eigenspaces onto one of them as a polynomial in $A$, using the formula for Lagrange interpolation that gives the solution for the particular Chinese remainder theorem instance needed here (a polynomial whose evaluation at $\lambda_i$ is $1$, and at all other eigenvalues is$~0$). I'll add the application of the operator to $w$, which you forgot. – Marc van Leeuwen Jan 12 '15 at 09:23
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    Since $W$ is $A$-invariant, it is also $p(A)$-invariant where $p$ is any polynomial with coefficients from the underlying field of $V$. In particular, here, $p(x)=\prod_{j\neq i} \frac{x-\lambda_j}{\lambda_i-\lambda_j}$ so $p(A)w = v_i\in W$. Am I getting it right? – Nothing special Oct 03 '24 at 12:40
  • @Nothingspecial - Yes! I find your wording outstanding! – Pierre-Yves Gaillard Oct 03 '24 at 21:05
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Claim: There is an $A$-invariant subspace $W'$ such that $V=W\oplus W'$.

Assuming this, let $P:V\to W$ be the projection with kernel $W'$. Since $W'$ is also $A$-invariant, we have $AP=PA$. So if $(v_i)_{i\in I}$ is an eigenbasis for $V$, then $(Pv_i)_{i\in I}$ are eigenvectors (or $0$) and span $W$, hence $A\rvert_W$ is diagonalizable.

Proof of claim: Pick a maximal subset $S\subseteq (v_i)_i$ such that $W\cap \operatorname{Span S}=0$. Then $W'=\operatorname{Span} S$ works.

leoli1
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