# restriction of an operator to invariant subspace is diagonalizable?

Can you help me out with solving the following problems. I am stuck with theses problems and I really appreciate any help.

Here is the problem 1 that I wanted to prove:

Let $M\colon H\to H$ be diagonalizable ($H$ vector space and $W$ subspace of $H$). $W$ is $M$-invariant. Prove that the restriction of $M$ to $W$ is diagonalizable and that $T\colon M/W\to M/W$ is diagonalizable (note: $M/W$ means $M$ quotient $W$ and not restriction of $M$ to $W$, i.e the quotient space).

Problem 2:

Let $M\colon H\to H$ has $n$ distinct eigenvalues (note: the dimension of $H$ is $n$). Let $U\colon H\to H$ commute with $M$. Show that $U$ is diagonalizable

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For the first question, see here. If you don't know about the minimal polynomial yet, please edit your question to indicate what material you do have on hand.

For the second: Suppose that $\lambda$ is an eigenvalue of $M$, and let $\mathbf{w}$ be an eigenvector of $\mathbf{M}$ corresponding to $\lambda$. Then $\mathrm{span}(\mathbf{w})$ is the eigenspace of $\lambda$ (since $M$ has $n$ distinct eigenvalues, so each eigenvalue has algebraic and geometric dimensions equal to $1$).

Now notice that $$\lambda U(\mathbf{w}) = U\left(\lambda\mathbf{w}\right) =U(M(\mathbf{w})) = M(U(\mathbf{w})).$$ That is, if $\mathbf{z}=U(\mathbf{w})$, then $M(\mathbf{z}) = \lambda\mathbf{z}$. Thus, $U(\mathbf{w}) = \mathbf{z}\in\langle \mathbf{w}\rangle$.

Can you go from here to conclude that $U$ is diagonalizable?

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I can't figure out how I prove that U is diagonalizable? – M.Krov Oct 8 '11 at 22:54
Can you elaborate more? Thanks – M.Krov Oct 8 '11 at 22:55
@Zi2018Alpha: Since $U(\mathbf{w})\in\langle \mathbf{w}\rangle$, that means that $U(\mathbf{w})=\mu\mathbf{w}$ for some scalar $\mu$ (since that is what every vector in $\langle\mathbf{w}\rangle$ looks like); so $U(\mathbf{w})$ is a scalar multiple of $\mathbf{w}$. That is, $\mathbf{w}$ is an eigenvector of $U$. So, if $\mathbf{w}$ is an eigenvector of $M$, then it is also an eigenvector of $U$. Is that sufficient? – Arturo Magidin Oct 8 '11 at 23:31
Definitely. Now I can see it. Thanks a lot for your help. I appreciate it. – M.Krov Oct 9 '11 at 0:18

While the current answer gives the necessary indications I'll try to give a concise and self contained answer. For diagonalizability, the following criterion is often useful.

Proposition. If a linear operator $T$ on a vector space over a field $K$ satisfies a polynomial equation $P[T]=0$ where $P$ factors over $K$ into distinct monic factors of degree$~1$, then $T$ is diagonalisable.

Proof. Write $P=(X-\lambda_1)\ldots(X-\lambda_k)$ where $\lambda_1,\ldots,\lambda_k$ are its distinct roots (in$~K$). The vector space becomes a module over the ring $K[X]/(P)$ by having $X$ act as $T$. By the Chinese remainder theorem for $K[X]$, one has $K[X]/(P)=(K[X]/(X-\lambda_1))\times\cdots\times(K[X]/(X-\lambda_k))$; this is a product of $k$ fields isomorphic to$~K$, with $X$ mapped to $\lambda_i$ in factor$~i$. Any module over such a product decomposes canonically as a direct sum, where the unit of factor$~i$ acts as projection onto summand$~i$. For our vector space, summand$~i$ (if nonzero) is the eigenspace of $T$ for $\lambda_i$, whence the result. QED.

Now if $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues of$~M$, then for $P=(X-\lambda_1)\ldots(X-\lambda_k)$ one has $P[M]=0$, and obviously substitution of the restriction of $M$ to the invariant subspace $W$ into $P$ is also zero, as well as that of the operator $M_{/W}$ induced by$~W$ in the quotient space $V/W$. By the proposition these linear operators are therefore diagonalisable.

For the second problem, note that (by a simple calculation) any linear operator commuting with $M$ stabilises each kernel (and also each image) of a polynomial in$~M$, and in particular each eigenspace of$~M$ (which is the kernel of $(X-\lambda_i)[M]$). Then $U$ stabilises each of the eigenspaces of $M$, and as these are all $1$-dimensional, any eigenvector for$~M$ is also eigenvector for$~U$. In particular any basis of diagonalisation for$~M$ is also one for$~U$.

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