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Let $V$ be a vector space with $\text{dim}(V)=n$ and $T: V \to V$ be a linear transformation. Let $W \subseteq V$ be a $T$-invariant subspace, i.e. $T(W) \subseteq W$, with $\text{dim}(W)=k$.

Show, that if $T$ is diagonalizable, then there exists a basis for $W$ consisting of eigenvectors of $T$. Show also that $T_W: W \to W, T_W(w):=T(w)$ is diagonalizable.

I found only a proof using the minimal polynomial which I haven't learned yet and couldn't find a proof without it. Can somebody check my attempt below?


Since $T$ is diagonalizable, there exists a Basis $B=\{v_1,\ldots,v_n\}$ for $V$ consisting of eigenvectors of $T$. Also the characteristic polynomial splits in linear factors. If we denote with $\lambda_1,\ldots,\lambda_n$ the corresponding eigenvalues to $v_1,\ldots,v_n$, then $p_T(t)= p_{[T]_B}(t)=(t-\lambda_1) \cdots (t-\lambda_n)$.

Construct from some $w_j \in V$ a basis $C=\{w_1,\ldots,w_k\}$ of $W$ and $T_W:W \to W$ with $T_W(w)=T(w)$. Since $T(W)\subseteq W$, we have $T(w_j)=\sum_{i=1}^k c_iw_i$ for some scalars $c_i$, which means

$$[T_W(w_j)]_{C}=\begin{bmatrix} c_1 \\ \vdots \\ c_k \end{bmatrix}$$

And by definition $[T_W]_C=\begin{bmatrix} [T_W(w_1)]_{C} & \ldots & [T_W(w_k)]_{C} \end{bmatrix}$

We can extend $C$ to a basis of $V$ with additional $v_i$'s (since $B$ is a basis of $V$ already). Assume for the sake of simplicity of notation that it's the first $n-k$ basis vectors, i.e. $B':=C\cup\{v_1,\ldots,v_{n-k}\}$ forms a basis for $V$. Then

$$[T]_{B'} = \begin{bmatrix} \vert & & \vert & \vert & & \vert \\ [T(w_1)]_{B'} & \ldots & [T(w_k)]_{B'} & [T(v_1)]_{B'} & \ldots & [T(v_{n-k})]_{B'} \\ \vert & & \vert & \vert & & \vert \end{bmatrix}$$

But since $T(W)\subseteq W$, we have $T(w_j)=\sum_{i=1}^k c_iw_i$ for some scalars $c_i$, which means

$$[T(w_j)]_{B'}=\begin{bmatrix} c_1 \\ \vdots \\ c_k \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} \vert \\ [T_W(w_j)]_{C} \\ \vert \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$

And $T(v_j)=\lambda_j v_j$ and so $[T(v_j)]_{B'}$ has only a $\lambda_j$ in the $k+j$-th entry and everywhere else zeros. So we get

$$[T]_B = \begin{bmatrix} \ulcorner & & \urcorner & & & \\ & [T_W]_C & & & & \\ \llcorner & & \lrcorner & & & \\ & & & \lambda_1 & & \\ & & & & \ddots & \\ & & & & & \lambda_{n-k} \end{bmatrix}$$

If we compute the characteristic polynomial we get

$$\begin{align} p_{[T]_B}(t) &= (t-\lambda_1) \cdots (t-\lambda_n) = p_{[T]_{B'}}(t) = \text{det}\left([T]_B-tI_n\right) \\&= (-1)^{n-k}(t-\lambda_1) \cdots (t-\lambda_{n-k}) \cdot \text{det}\left([T_W]_C-t I_k\right) \\&= (-1)^{n-k}(t-\lambda_1) \cdots (t-\lambda_{n-k}) \cdot p_{[T_W]_C}(t) \end{align}$$

And so $p_{T_W}(t) = p_{[T_W]_C}(t) = (-1)^{n-k}(t-\lambda_{n-k+1})\cdots(t-\lambda_{n})$, which means $T_W$ has eigenvalues $\lambda_{n-k+1},\ldots,\lambda_n$.

Let $u_{n-k+1},\ldots,u_n \in W$ be the corresponding eigenvectors of $T_W$, i.e. $T_W(u_j)=\lambda_j u_j$. Since for all $w \in W$ we have $T_W(w)=T(w)$, we have $\lambda_j u_j = T_W(u_j) = T(u_j)$ for all $n-k+1 \le j \le n$. But $T(v_j)=\lambda_j v_j$ and therefore $u_j = c v_j$ for some scalar $c$.

We have found $n-(n-k+1)+1 = k$ eigenvalues of $T_W$ which are also eigenvalues of $T$ and the corresponding $k$ eigenvectors of $T_W$ are also exactly eigenvectors of $T$. Since $\text{dim}(W)=k$, we can construct a basis for $W$ of eigenvectors of $T_W$ and therefore $T_W$ is diagonalizable.

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  • $\begingroup$ Your argument will only work if the eigenvalues are all distinct. Otherwise, how do you know that $T_W$ has linearly independent eigenvectors corresponding to $\lambda_{n-k+1},\dots,\lambda_n$? You must argue that the geometric multiplicity of each such eigenvalue equals its algebraic multiplicity. Obviously, the fact that $T$ is diagonalizable has to get used here! $\endgroup$ – Ted Shifrin Apr 13 '18 at 16:18
  • $\begingroup$ @TedShifrin Yes, I didn't write it clearly enough. When I wrote in the beginning there are $n$ eigenvalues $\lambda_1,\ldots,\lambda_n$, I meant including multiplicity. So $\lambda_5=\lambda_6=\lambda_7$ is possible, but since $T$ is diagonalizable, there are also linearly independent $v_5,v_6,v_7$. And in the end where we've got $k$ (not necessarily distinct) eigenvalues, there are $k$ linearly independent eigenvectors associated with them. I tried to avoid keeping track of algebraic multiplicities by writing the eigenvalues one by one. $\endgroup$ – philmcole Apr 13 '18 at 17:36
  • $\begingroup$ But you need to prove that there will be sufficiently many eigenvectors for $T_W$. That's the key part of the proof. Having $k$ linearly independent eigenvectors for $T$ in the first place doesn't help you if they're not in $W$. $\endgroup$ – Ted Shifrin Apr 13 '18 at 18:10
  • $\begingroup$ @TedShifrin Isn't this exactly the second last paragraph? We have those $k$ eigenvalues of $T_W$ and say $u_j$ are the corresponding eigenvectors in $W$. Then we show that those $u_j=v_j$ and since the $u_j$ are in $W$ the $v_j$ must be too and those $v_j$ are exactly the $k$ eigenvectors of $T$ we can use to build a basis of $W$. $\endgroup$ – philmcole Apr 13 '18 at 18:42
  • $\begingroup$ Nope. This is a "proof by wishful thinking." Never have you proved that the geometric multiplicities are equal to the algebraic multiplicities. You are forgetting that with repeated eigenvalues, you no longer have "unique" eigenvectors for each eigenvalue. $\endgroup$ – Ted Shifrin Apr 13 '18 at 18:45
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If you start with a basis $\{w_1,\dots,w_k\}$ for $W$ and extend it to a basis $\{w_1,\dots,w_k, v_{k+1},\dots,v_n\}$ for $V$, we get the matrix representation $$[T]=A=\left[\begin{array}{c|c} B & D \\ \hline O & C \end{array}\right]\,.$$ (You are correct that you can actually assume $v_{k+1},\dots,v_n$ are eigenvectors, but that's actually unnecessary.) Fix an eigenvalue $\lambda$ of $B$. Let $d_A$, $d_B$, and $d_C$ be the respective geometric multiplicities for $\lambda$, and let $m_A$, $m_B$, and $m_C$ be the respective algebraic multiplicities for $\lambda.$ We want to see that $d_B = m_B$.

As you pointed out, $p_A(t) = p_B(t)p_C(t)$, so $m_A=m_B+m_C$. Moreover, since $T$ is diagonalizable, $d_A=m_A$. And, as always, $d_B\le m_B$ and $d_C\le m_C$.

Now $\text{rank}(A-\lambda I) \ge \text{rank}(B-\lambda I)+\text{rank}(C-\lambda I)$ (why?). Since $d_A = n-\text{rank}(A-\lambda I)$, $d_B = k-\text{rank}(B-\lambda I)$ and $d_C = (n-k)-\text{rank}(C-\lambda I)$, we have $d_A\le d_B+d_C$. But since $d_A=m_A$, we must have $d_B=m_B$ (why?).

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  • $\begingroup$ Thanks! I think I found where I made a wrong assumption in my proof. When I built $B'$ with some eigenvectors of $T$, I assumed that there are no more eigenvectors corresponding to the eigenvalues of those chosen eigenvectors left, i.e. all would be in this extended basis. This would mean that the remaining eigenvectors not in $B'$ would all have different eigenvalues than those that are in $B'$. But for a repeated eigenvalue some of its eigenvectors could be in $B'$ and some not. I hadn't thought of that. It would have helped if I had written the multiplicities explicitly. Thanks anyways! $\endgroup$ – philmcole Apr 14 '18 at 8:36

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