# Why does a diagonalization of a matrix B with the basis of a commuting matrix A give a block diagonal matrix?

I am trying to understand a proof concerning commuting matrices and simultaneous diagonalization of these.

It seems to be a well known result that when you take the eigenvectors of $A$ as a basis and diagonalize $B$ with it then you get a block diagonal matrix:

$$B= \begin{pmatrix} B_{1} & 0 & \cdots & 0 \\ 0 & B_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & B_{m} \end{pmatrix},$$

where each $B_{i}$ is an $m_{g}(\lambda_{i}) \times m_{g}(\lambda_{i})$ block ($m_{g}(\lambda_{i})$ being the geometric multiplicity of $\lambda_{i}$).

My question
Why is this so? I calculated an example and, lo and behold, it really works :-) But I don't understand how it works out so neatly.

Can you please explain this result to me in an intuitive and step-by-step manner - Thank you!

• "diagonalize B with it" doesn't seem to be what you want to say, does it? – Rasmus Jun 20 '11 at 20:49
• So, we are assuming that $A$ and $B$ are both diagonalizable, and that $AB=BA$. Then you take a basis $\beta$ of eigenvectors of $A$ (listed with all the vectors corresponding to the same eigenvalue together), and you look at the coordinate matrix of $B$ relative to $\beta$, and you wonder why this matrix is a block-diagonal matrix. Is this correct? – Arturo Magidin Jun 20 '11 at 21:04
• @Arturo: now I am back - yes, this is correct! – vonjd Jun 21 '11 at 6:36

Suppose that $$A$$ and $$B$$ are matrices that commute. Let $$\lambda$$ be an eigenvalue for $$A$$, and let $$E_{\lambda}$$ be the eigenspace of $$A$$ corresponding to $$\lambda$$. Let $$\mathbf{v}_1,\ldots,\mathbf{v}_k$$ be a basis for $$E_{\lambda}$$.

I claim that $$B$$ maps $$E_{\lambda}$$ to itself; in particular, $$B\mathbf{v}_i$$ can be expressed as a linear combination of $$\mathbf{v}_1,\ldots,\mathbf{v}_k$$, for $$i=1,\ldots,k$$.

To show that $$B$$ maps $$E_{\lambda}$$ to itself, it is enough to show that $$B\mathbf{v}_i$$ lies in $$E_{\lambda}$$; that is, that if we apply $$A$$ to $$B\mathbf{v}_i$$, the result ill be $$\lambda(B\mathbf{v}_i)$$. This is where the fact that $$A$$ and $$B$$ commute comes in. We have: $$A\Bigl(B\mathbf{v}_i\Bigr) = (AB)\mathbf{v}_i = (BA)\mathbf{v}_i = B\Bigl(A\mathbf{v}_i\Bigr) = B(\lambda\mathbf{v}_i) = \lambda(B\mathbf{v}_i).$$ Therefore, $$B\mathbf{v}_i\in E_{\lambda}$$, as claimed.

So, now take the basis $$\mathbf{v}_1,\ldots,\mathbf{v}_k$$, and extend it to a basis for $$\mathbf{V}$$, $$\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_k,\mathbf{v}_{k+1},\ldots,\mathbf{v}_n]$$. To find the coordinate matrix of $$B$$ relative to $$\beta$$, we compute $$B\mathbf{v}_i$$ for each $$i$$, write $$B\mathbf{v}_i$$ as a linear combination of the vectors in $$\beta$$, and then place the corresponding coefficients in the $$i$$th column of the matrix.

When we compute $$B\mathbf{v}_1,\ldots,B\mathbf{v}_k$$, each of these will lie in $$E_{\lambda}$$. Therefore, each of these can be expressed as a linear combination of $$\mathbf{v}_1,\ldots,\mathbf{v}_k$$ (since they form a basis for $$E_{\lambda}$$. So, to express them as linear combinations of $$\beta$$, we just add $$0$$s; we will have: \begin{align*} B\mathbf{v}_1 &= b_{11}\mathbf{v}_1 + b_{21}\mathbf{v}_2+\cdots+b_{k1}\mathbf{v}_k + 0\mathbf{v}_{k+1}+\cdots + 0\mathbf{v}_n\\ B\mathbf{v}_2 &= b_{12}\mathbf{v}_1 + b_{22}\mathbf{v}_2 + \cdots +b_{k2}\mathbf{v}_k + 0\mathbf{v}_{k+1}+\cdots + 0\mathbf{v}_n\\ &\vdots\\ B\mathbf{v}_k &= b_{1k}\mathbf{v}_1 + b_{2k}\mathbf{v}_2 + \cdots + b_{kk}\mathbf{v}_k + 0\mathbf{v}_{k+1}+\cdots + 0\mathbf{v}_n \end{align*} where $$b_{ij}$$ are some scalars (some possibly equal to $$0$$). So the matrix of $$B$$ relative to $$\beta$$ would start off something like: $$\left(\begin{array}{ccccccc} b_{11} & b_{12} & \cdots & b_{1k} & * & \cdots & *\\ b_{21} & b_{22} & \cdots & b_{2k} & * & \cdots & *\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ b_{k1} & b_{k2} & \cdots & b_{kk} & * & \cdots & *\\ 0 & 0 & \cdots & 0 & * & \cdots & *\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & * & \cdots & * \end{array}\right).$$

So, now suppose that you have a basis for $$\mathbf{V}$$ that consists entirely of eigenvectors of $$A$$; let $$\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$$ be this basis, with $$\mathbf{v}_1,\ldots,\mathbf{v}_{m_1}$$ corresponding to $$\lambda_1$$ (with $$m_1$$ the algebraic multiplicity of $$\lambda_1$$, which equals the geometric multiplicity of $$\lambda_1$$); $$\mathbf{v}_{m_1+1},\ldots,\mathbf{v}_{m_1+m_2}$$ the eigenvectors corresponding to $$\lambda_2$$, and so on until we get to $$\mathbf{v}_{m_1+\cdots+m_{k-1}+1},\ldots,\mathbf{v}_{m_1+\cdots+m_k}$$ corresponding to $$\lambda_k$$. Note that $$\mathbf{v}_{1},\ldots,\mathbf{v}_{m_1}$$ are a basis for $$E_{\lambda_1}$$; that $$\mathbf{v}_{m_1+1},\ldots,\mathbf{v}_{m_1+m_2}$$ are a basis for $$E_{\lambda_2}$$, etc.

By what we just saw, each of $$B\mathbf{v}_1,\ldots,B\mathbf{v}_{m_1}$$ lies in $$E_{\lambda_1}$$, and so when we express it as a linear combination of vectors in $$\beta$$, the only vectors with nonzero coefficients are $$\mathbf{v}_1,\ldots,\mathbf{v}_{m_1}$$, because they are a basis for $$E_{\lambda_1}$$. So in the first $$m_1$$ columns of $$[B]_{\beta}^{\beta}$$ (the coordinate matrix of $$B$$ relative to $$\beta$$), the only nonzero entries in the first $$m_1$$ columns occur in the first $$m_1$$ rows.

Likewise, each of $$B\mathbf{v}_{m_1+1},\ldots,B\mathbf{v}_{m_1+m_2}$$ lies in $$E_{\lambda_2}$$, so when we express them as linear combinations of $$\beta$$, the only places where you can have nonzero coefficients are in the coefficients of $$\mathbf{v}_{m_1+1},\ldots,\mathbf{v}_{m_1+m_2}$$. So the $$(m_1+1)$$st through $$(m_1+m_2)$$st column of $$[B]_{\beta}^{\beta}$$ can only have nonzero entries in the $$(m_1+1)$$st through $$(m_1+m_2)$$st rows. And so on.

That means that $$[B]_{\beta}^{\beta}$$ is in fact block-diagonal, with the blocks corresponding to the eigenspaces $$E_{\lambda_i}$$ of $$A$$, exactly as described.

I will write $k_i=m_g(\lambda_i)$.

You are looking for the general form of a matrix $B$ that commutes with $$A= \begin{pmatrix} \lambda_1 I_{k_1} & 0 & \cdots & 0 \\ 0 & \lambda_2 I_{k_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_m I_{k_m} \end{pmatrix}.$$

If you put $B$ in the same block structure, you have

$$B= \begin{pmatrix} B_{11} & B_{12} & \cdots & B_{1m} \\ B_{21} & B_{22} & \cdots & B_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ B_{m1} & B_{m2} & \cdots & B_{mm} \end{pmatrix},$$ where $B_{ij}$ is a $k_i$-by-$k_j$ matrix.

Then $$AB= \begin{pmatrix} \lambda_1 B_{11} & \lambda_1 B_{12} & \cdots & \lambda_1 B_{1m} \\ \lambda_2 B_{21} & \lambda_2 B_{22} & \cdots & \lambda_2 B_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_m B_{m1} & \lambda_m B_{m2} & \cdots & \lambda_m B_{mm} \end{pmatrix},$$ while $$BA= \begin{pmatrix} \lambda_1 B_{11} & \lambda_2 B_{12} & \cdots & \lambda_m B_{1m} \\ \lambda_1 B_{21} & \lambda_2 B_{22} & \cdots & \lambda_m B_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_1 B_{m1} & \lambda_2 B_{m2} & \cdots & \lambda_m B_{mm} \end{pmatrix}.$$

You can compare off-diagonal blocks to see that $B$ must have the desired form if $BA=AB$, because $\lambda_i\neq \lambda j$ if $i\neq j$.

Getting a block diagonal matrix expressing$~B$ just means that each eigenspace for$~A$ (whose direct sum fills the entire space since $A$ is supposed diagonalisable) is $B$-stable. And a useful fact that applies here is that when two linear operators commute, then every subspace that is the kernel or image of a polynomial in one of the operators is automatically stable for the other operator. A polynomial in the first operator is just another operator$~\psi$ that commutes with the second operator$~\phi$, so it suffices to show that for the kernel and the image of $\psi$ are $\phi$-stable when $\psi$ and $\phi$ commute:

• Kernel: if $v\in\ker\psi$ then $\psi(\phi(v))=\phi(\psi(v))=\phi(0)=0$, so indeed $\phi(v)\in\ker\psi$.
• Image: if $v=\psi(w)$ then $\phi(v)=\phi(\psi(w))=\psi(\phi(w))$, which indeed is in the image of $\psi$.

The eigenspace of$~A$ for $\lambda$ is of course just the special case of the kernel of the polynomial $A-\lambda I$ in$~A$.