# The property of commutative matrixes and eigen vectors.

$$A,B$$ are $$n\times n$$ diagonalizable matrixes, and satisfy $$AB=BA.$$

Let $$W_\lambda$$ stands for the eigenspace of eigenvalue of $$A$$, $$\lambda.$$

Prove

(i) For each eigenvalue of $$A$$, $$\lambda$$ ; $$v\in W_\lambda \Rightarrow Bv \in W_\lambda.$$

(ii) For each eigenvalue of $$A$$, $$\lambda$$ ; there exists the base of $$W_\lambda$$ which is composed of the eigenvectors of $$B$$.

I did (i) and I'm having difficulty in showing (ii). I cannot find such base.

Fix $$\lambda$$ ; the eigenvalue of $$A$$, and $$\dim W_\lambda=:k.$$

Intuitively, I think the ONB of $$W_\lambda$$ works (I don't know whether this is correct.)

So, let $$\{ u_1,\cdots , u_k \}$$ be ONB of $$W_\lambda$$, then $$W_\lambda=\langle u_1, \cdots , u_k \rangle$$.

I expect each $$u_j$$ is eigenvector of $$B$$.

I want to find $$\square_j \in \mathbb C$$ s.t. $$B u_j=\square_j u_j$$ for each $$j$$.

For each $$j$$, $$u_j \in W_\lambda$$ so from (i), $$Bu_j\in W_\lambda=\langle u_1, \cdots , u_k \rangle$$.

I can write $$Bu_j=\sum_{i=1}^k c^{(j)}_iu_i$$.

If I show $$c^{(j)}_i=0$$ for $$i \neq j$$, I can see $$Bu_j=c^{(j)}_j u_j$$, but I cannot find the reason why $$c^{(j)}_i=0$$ for $$i \neq j$$.

From the orthonormalness of $$\{ u_1, \cdots, u_k \}$$, I also have $$(Bu_j, u_i)=c_i^{(j)}$$.

Am I on right track ? I want you to give me any help.

• Does ONB mean "orthogonal, normal basis"? There are many of those to choose from, and not every one will work. May 22, 2022 at 5:00
• I don't think that it is necessary to work with orthogonal, normal bases. With that you are assuming the existence of an inner product on the vector space - which is not given in the problem and in fact not necessary. In (a) you showed that $B(W_\lambda) \subset W_\lambda$ - that means $W_\lambda$ is a subspace invariant under $B$. Try to show that $B_{|U}$ is also diagonalizable for an invariant subspace if $B$ is diagonalizable. This way you will find a basis of eigenvectors of $B$ for $W_\lambda$. May 22, 2022 at 11:11
• What $B_{|U}$ means ? $U$ is not defined in my question. @Lukas
– daㅤ
May 22, 2022 at 11:31
• @SABAR Hint: What are the eigenvectors of $B$ and how do they relate to those of $A$. This follows from (i).
– KBS
May 22, 2022 at 13:07
• @KBS I don't know what the eigenvectors of $B$ and those of $A$ are. Could you explain a little more detail ?
– daㅤ
May 23, 2022 at 0:23

Since $$A,B$$ are diagonalizable, we can find a eigenbasis for $$A$$, $$V=[v_1,v_2,...v_n]$$, where $$\lambda_1,...\lambda_n$$. $$\Lambda=diag(\lambda_1,...\lambda_n)$$ $$AV=V\Lambda$$ Then given a comutable matrix $$B$$ $$ABV=BAV=BV\Lambda$$ Let $$U=BV$$, $$AU=U\Lambda$$ Let $$U=[u_1,u_2...u_n]$$, $$u_i=Bv_i$$, it's obvious that $$i$$ th column of $$BV$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda_i$$.

Thus for each $$\lambda$$ $$\forall v\in W^{\lambda},Bv\in W^{\lambda}$$

Now consider an eigenspace $$W^{\lambda}$$ spanned by the eigenvectors of $$A$$, $$W^{\lambda}=Span(v_1,...v_p)$$. From 1), $$W^{\lambda}$$ is an invariant subspace for linear operator $$B$$

$$B[v_1,...v_p]=[v_1,...v_p]M$$

$$M$$ is a $$p\times p$$ matrix, which is the effect of $$B$$ as it's restricted to the basis set $$[v_1,...v_p]$$.

Since $$B$$ is diagonalizable, we can actually use this lemma: diagonalizable linear operator is also diagonalizable when restricted to the invariant subspaces (Diagonalizable transformation restricted to an invariant subspace is diagonalizable). Then diagonalizing $$M=Q\Sigma Q^{-1}$$ we get $$B[v_1,...v_p]=[v_1,...v_p]Q\Sigma Q^{-1}\\ B[v_1,...v_p]Q=[v_1,...v_p]Q\Sigma$$ Thus, $$[v_1,...v_p]Q$$ is a eigen basis for $$Span(v_1,...v_p)$$, each column is an eigen vector of $$B$$.

This brief answer proved the lemma well.

Here is a sketch, for any $$w\in W^{\lambda}$$, if we decompose it into eigenvectors with distinct eigenvalues of $$B$$. $$w=\sum_{i=1}^k v'_i\;\; Bv'_i=\sigma_i v'i$$ Since $$Bw\in W^{\lambda}, w\in W^{\lambda}$$, $$Bw-\lambda_1 w\in W^{\lambda}$$ $$Bw-\lambda_1 w = \sum_{i=1}^k (\lambda_i - \lambda_1) v'_i\\ = \sum_{i=2}^k (\lambda_i - \lambda_1) v'_i$$ Then $$Bw-\lambda_1 w$$ can be decomposed into sum of $$k-1$$ eigenvectors. Repeating this procedure for $$Bw-\lambda_1 w$$ with the remaining $$k-1$$ eigenvectors, until we are left with one eigenvector, we have $$v_k=\prod_{j\neq k}\frac{(B-\lambda_j I)}{\lambda_k-\lambda_j}w \in W^{\lambda}$$ This argument is true for any $$k$$, thus we can find a eigenbasis for vectors in $$W^{\lambda}$$