Blockwise Changes of Basis for Block Matrices

Changing Coordinates One Block at a Time

A matrix may describe an operator acting separately on several parts of a vector space. If the basis lists the vectors for each part together, the matrix has diagonal blocks and zeros between them. A new basis for each part can then be chosen independently: the full basis change is assembled from those smaller changes, and the transformed operator matrix retains the same block structure. This lets us do a larger change-of-basis calculation through smaller ones without mixing the parts.

We work over \(\mathbb{R}\). Suppose a finite-dimensional vector space is written as a direct sum \(V=V_1\oplus\cdots\oplus V_k\). This means every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=\mathbf{v}_1+\cdots+\mathbf{v}_k\) with \(\mathbf{v}_i\in V_i\). Suppose \(T:V\to V\) maps each \(V_i\) into itself. In bases formed by listing a basis of \(V_1\), then a basis of \(V_2\), and so on, the matrix of \(T\) is block diagonal: each diagonal block records the action on one summand.

For a single operator represented in an old basis and a new basis, the coordinate convention is important. Let \(\mathbf{P}\) have as its columns the new basis vectors written in old coordinates. If \(\mathbf{x}_{\mathrm{new}}\) is a coordinate column in the new basis, then its old-coordinate column is \(\mathbf{P}\mathbf{x}_{\mathrm{new}}\). Thus the old matrix \(\mathbf{A}\) sends it to \(\mathbf{A}\mathbf{P}\mathbf{x}_{\mathrm{new}}\). To express that result in new coordinates, multiply by \(\mathbf{P}^{-1}\). The new matrix is therefore \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\), or equivalently it satisfies \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{B}\).

The Blockwise Change-of-Basis Rule

Proof. Define \(\mathbf{Q}=\operatorname{diag}(\mathbf{P}_1^{-1},\dots,\mathbf{P}_k^{-1})\). In block multiplication, the \((i,j)\) block of \(\mathbf{Q}\mathbf{P}\) is \(\mathbf{P}_i^{-1}\mathbf{P}_i=\mathbf{I}_{n_i}\) when \(i=j\), and is the zero matrix when \(i\ne j\). Hence \(\mathbf{Q}\mathbf{P}=\mathbf{I}\). The same block multiplication gives \(\mathbf{P}\mathbf{Q}=\mathbf{I}\), so \(\mathbf{Q}=\mathbf{P}^{-1}\). Next, \(\mathbf{A}\mathbf{P}\) has diagonal block \(\mathbf{A}_i\mathbf{P}_i\) in position \((i,i)\) and zero blocks elsewhere. Multiplying on the left by \(\mathbf{Q}\) gives diagonal block \(\mathbf{P}_i^{-1}\mathbf{A}_i\mathbf{P}_i\) in position \((i,i)\) and zero blocks elsewhere. This is the stated formula. \(\square\)

The theorem uses the same ordering of summands in the old and new bases. Each \(\mathbf{P}_i\) acts only on coordinates belonging to \(V_i\). Its inverse converts the resulting coordinates back into that summand’s new coordinate system. The calculation within one block is exactly an ordinary similarity transformation; the block-diagonal arrangement ensures that these transformations do not interfere with one another.

Different Block Sizes, Independent Choices

The theorem does not require all blocks to have the same size. A one-dimensional summand can be left unchanged while a larger summand gets a nontrivial basis change. The only requirement is that each \(\mathbf{P}_i\) be square and invertible with the size matching its block.

Why the Assembly Works in Both Directions

The rule is useful not only for computing a new matrix but also for recognizing when blockwise similarity is enough. Here “similar” means that one matrix is \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) for some invertible \(\mathbf{P}\). If each pair of corresponding blocks is similar, the individual change-of-basis matrices can be assembled into one block-diagonal matrix.

Proof. In block position \((i,i)\), the product \(\mathbf{A}\mathbf{P}\) has block \(\mathbf{A}_i\mathbf{P}_i\), while \(\mathbf{P}\mathbf{B}\) has block \(\mathbf{P}_i\mathbf{B}_i\). These are equal by hypothesis. In every off-diagonal position both products have a zero block, since both factors are block diagonal. Therefore \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{B}\). Each \(\mathbf{P}_i\) is invertible, so the block-diagonal matrix \(\mathbf{P}\) is invertible by the theorem. Multiplying the equality on the left by \(\mathbf{P}^{-1}\) gives \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{B}\). \(\square\)

This proposition asserts a sufficient construction: matching blockwise changes produce a whole-matrix similarity. It does not say that every similarity between two block-diagonal matrices must use a block-diagonal change-of-basis matrix. A general change of basis may mix vectors from different summands, so its matrix need not preserve the displayed partition.

Common Pitfall: Reversing the Similarity Direction

A frequent error is to use \(\mathbf{P}\mathbf{A}\mathbf{P}^{-1}\) when the columns of \(\mathbf{P}\) are the new basis vectors written in old coordinates. For the first block in the earlier example, the correct calculation gave \(\mathbf{P}_1^{-1}\mathbf{A}_1\mathbf{P}_1=\begin{bmatrix}2&0\\0&3\end{bmatrix}\). Reversing the factors instead gives

That is not the matrix in the new basis under this coordinate convention. A direct check exposes the mistake: for the correct new matrix \(\mathbf{B}_1\), the equation \(\mathbf{A}_1\mathbf{P}_1=\mathbf{P}_1\mathbf{B}_1\) holds, since both sides equal \(\begin{bmatrix}2&3\\0&3\end{bmatrix}\). Substituting the reversed-factor result for \(\mathbf{B}_1\) does not give that equality. Before multiplying, identify what the columns of \(\mathbf{P}\) mean; then check \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{B}\).

A second limitation is just as important: a block-diagonal change of basis preserves zero off-diagonal blocks, but does not generally turn nonzero off-diagonal blocks into zero. For example, let \(\mathbf{A}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\) be partitioned into two one-by-one blocks and take \(\mathbf{P}=\operatorname{diag}(2,3)\). Then

The off-diagonal entry remains nonzero. Blockwise changes of basis change coordinates within the chosen pieces; they do not by themselves separate pieces that are coupled by off-diagonal blocks.

Using the Rule Efficiently

For a blockwise calculation, first fix the partition and order of the summands. Then identify the old matrix \(\mathbf{A}_i\) and the new-coordinate matrix \(\mathbf{P}_i\) for each block. Compute \(\mathbf{P}_i^{-1}\mathbf{A}_i\mathbf{P}_i\) separately, and assemble the answers in the original block positions. The block products are usually easier to check than one large multiplication.

For additional verification, check each local equation \(\mathbf{A}_i\mathbf{P}_i=\mathbf{P}_i\mathbf{B}_i\). If those equations hold and every \(\mathbf{P}_i\) is invertible, the assembled matrix is a valid change of basis. This approach applies whenever the operator preserves the summands and the new basis is obtained by changing basis within each summand, without reordering or mixing them.