Reading Eigenvalue Information from the Blocks
A block diagonal matrix can be large while each of its diagonal blocks is small. Its eigenvalue equation respects that structure: each block acts on its own part of the vector, without mixing coordinates from the other blocks. This lets us find eigenvalues and eigenspaces by working block by block, then assembling the results in the original coordinate order.
We work over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), with all matrices and vectors over the same field. Let \(\mathbf{A}_1,\dots,\mathbf{A}_k\) be square matrices of sizes \(n_1,\dots,n_k\), and let \(n=n_1+\cdots+n_k\). Their block diagonal matrix is the \(n\times n\) matrix with these blocks along the diagonal and zero entries in every off-diagonal block. A vector \(\mathbf{x}\in\mathbb{F}^n\) is partitioned accordingly as \(\mathbf{x}=(\mathbf{x}_1,\dots,\mathbf{x}_k)^{\top}\), where \(\mathbf{x}_i\in\mathbb{F}^{n_i}\).
The eigenspace is a subspace: if \(\mathbf{u}\) and \(\mathbf{v}\) satisfy \(\mathbf{M}\mathbf{u}=\lambda\mathbf{u}\) and \(\mathbf{M}\mathbf{v}=\lambda\mathbf{v}\), then for scalars \(a,b\), \(\mathbf{M}(a\mathbf{u}+b\mathbf{v})=a\lambda\mathbf{u}+b\lambda\mathbf{v}=\lambda(a\mathbf{u}+b\mathbf{v})\). The zero vector belongs to every eigenspace, but it is not itself an eigenvector.
The Eigenspace Rule
For a block diagonal matrix \(\mathbf{A}\), multiplication is componentwise: \(\mathbf{A}\mathbf{x}=(\mathbf{A}_1\mathbf{x}_1,\dots,\mathbf{A}_k\mathbf{x}_k)^{\top}\). Consequently, the equation \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) is equivalent to the \(k\) separate equations \(\mathbf{A}_i\mathbf{x}_i=\lambda\mathbf{x}_i\). A component may be zero; the full vector must be nonzero for \(\lambda\) to be an eigenvalue.
Proof. Suppose \(\lambda\) is an eigenvalue of \(\mathbf{A}\). Then there is a nonzero \(\mathbf{x}=(\mathbf{x}_1,\dots,\mathbf{x}_k)^{\top}\) such that \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Equality of the corresponding components gives \(\mathbf{A}_i\mathbf{x}_i=\lambda\mathbf{x}_i\) for each \(i\). Since \(\mathbf{x}\ne\mathbf{0}\), at least one component \(\mathbf{x}_j\ne\mathbf{0}\). The equation for that component says \(\mathbf{A}_j\mathbf{x}_j=\lambda\mathbf{x}_j\), so \(\lambda\) is an eigenvalue of \(\mathbf{A}_j\). Conversely, if \(\lambda\) is an eigenvalue of some \(\mathbf{A}_j\), choose a nonzero \(\mathbf{x}_j\in E_{\lambda}(\mathbf{A}_j)\) and set every other component to zero. The resulting full vector is nonzero, and each block equation holds, so it is an eigenvector of \(\mathbf{A}\). Finally, a vector satisfies the full eigenvalue equation exactly when every component satisfies its block equation. A block eigenspace is \(\{\mathbf{0}\}\) when \(\lambda\) is not an eigenvalue of that block, which proves the component claim as well. \(\square\)
This gives a useful set-level rule: the eigenvalues of \(\mathbf{A}\) are the union of the eigenvalue sets of its blocks. A value shared by several blocks is still one scalar eigenvalue of the full matrix. Its eigenspace, however, can receive independent contributions from each block in which that value occurs.
Worked Example: A Shared Eigenvalue in Two Blocks
Find the eigenspace for \(\lambda=2\) of the block diagonal matrix with blocks \(\mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\) and \(\mathbf{C}=\begin{bmatrix}2&0\\1&3\end{bmatrix}\).
For the first block, write \(\mathbf{x}_1=(x_1,x_2)^{\top}\). Its eigenvalue equation becomes
Thus \(x_1=x_2\), and \(E_2(\mathbf{B})=\operatorname{span}\{(1,1)^{\top}\}\). For the second block, write \(\mathbf{x}_2=(x_3,x_4)^{\top}\). Then
So \(E_2(\mathbf{C})=\operatorname{span}\{(1,-1)^{\top}\}\). Since both blocks have eigenvalue \(2\), their components can be chosen independently. In the full four-dimensional coordinate order,
Check: The full matrix is \(\mathbf{A}=\begin{bmatrix}1&1&0&0\\0&2&0&0\\0&0&2&0\\0&0&1&3\end{bmatrix}\). Multiplying gives \(\mathbf{A}(1,1,0,0)^{\top}=(2,2,0,0)^{\top}\) and \(\mathbf{A}(0,0,1,-1)^{\top}=(0,0,2,-2)^{\top}\). Each basis vector therefore satisfies \(\mathbf{A}\mathbf{v}=2\mathbf{v}\); every linear combination does too.
When Only One Block Contributes
If \(\lambda\) is an eigenvalue of one block but not of another, the component in the latter block is forced to be zero. It is important to apply the block equation to every component, not just to the block that first reveals the eigenvalue.
Worked Example: A Zero Component in a Noncontributing Block
Find the eigenspace for \(\lambda=2\) when \(\mathbf{A}=\operatorname{diag}(\mathbf{B},\mathbf{C})\), where \(\mathbf{B}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\) and \(\mathbf{C}=\begin{bmatrix}5&0\\0&6\end{bmatrix}\).
For the first block, \(\mathbf{B}-2\mathbf{I}_2=\begin{bmatrix}-2&1\\0&0\end{bmatrix}\). The equation \((\mathbf{B}-2\mathbf{I}_2)\mathbf{x}_1=\mathbf{0}\) gives \(-2x_1+x_2=0\), so \(x_2=2x_1\). Hence the first block eigenspace is spanned by \((1,2)^{\top}\).
For the second block, the equation is
It gives \(3x_3=0\) and \(4x_4=0\), hence \(x_3=x_4=0\). Therefore \(E_2(\mathbf{A})=\operatorname{span}\{(1,2,0,0)^{\top}\}\).
Check: The full matrix is \(\begin{bmatrix}0&1&0&0\\0&2&0&0\\0&0&5&0\\0&0&0&6\end{bmatrix}\), and \(\mathbf{A}(1,2,0,0)^{\top}=(2,4,0,0)^{\top}=2(1,2,0,0)^{\top}\). The vector is nonzero, so \(2\) is indeed an eigenvalue of the full matrix.
How the Eigenspace Dimension Adds
The eigenspace description also determines its dimension. Write \(\operatorname{gm}_{\mathbf{M}}(\lambda)=\dim E_{\lambda}(\mathbf{M})\) for the geometric multiplicity of an eigenvalue \(\lambda\) of \(\mathbf{M}\). For a block where \(\lambda\) is not an eigenvalue, use the zero eigenspace, whose dimension is zero. Because each block occupies its own coordinates, a basis vector from one block cannot be formed by combining vectors supported only in the other blocks.
Proof. For each \(i\), choose a basis of \(E_{\lambda}(\mathbf{A}_i)\); if that space is zero, its basis is the empty list. Embed each basis vector into the full coordinate space by placing it in the \(i\)-th block of coordinates and placing zero in every other block. By the eigenspace rule, every embedded vector lies in \(E_{\lambda}(\mathbf{A})\). These embedded vectors span the full eigenspace: any vector in it has components in the block eigenspaces, and each component is a linear combination of that block’s chosen basis. They are independent as well: in a linear combination of embedded vectors that equals zero, examining the coordinates of block \(i\) gives a zero combination of the basis for \(E_{\lambda}(\mathbf{A}_i)\). Its coefficients must all be zero. This holds for every block, so all coefficients in the full combination are zero. The embedded lists therefore form a basis of the full eigenspace, and its size is the sum of their sizes. \(\square\)
Worked Example: Shared Eigenvalue with Different Block Eigenspaces
Consider three blocks: the scalar block \([3]\), the block \(\mathbf{J}=\begin{bmatrix}3&1\\0&3\end{bmatrix}\), and the scalar block \([1]\). Find \(E_3\) for their block diagonal matrix.
In the first block, every scalar \(x_1\) satisfies \(3x_1=3x_1\), so the eigenspace is \(\mathbb{R}\), of dimension one. In the second block, subtracting \(3\mathbf{I}_2\) gives \(\begin{bmatrix}0&1\\0&0\end{bmatrix}\). Its equation is \(x_2^{(2)}=0\), where \(x_2^{(2)}\) denotes the second coordinate within that block. Thus its eigenspace is spanned by \((1,0)^{\top}\), also of dimension one. In the last block the equation \(1x=3x\) implies \(-2x=0\), so its eigenspace is zero.
With the full coordinate order \((x_1,x_2,x_3,x_4)\), the eigenspace is
Check: Here \(\mathbf{A}=\begin{bmatrix}3&0&0&0\\0&3&1&0\\0&0&3&0\\0&0&0&1\end{bmatrix}\). Multiplication sends \((1,0,0,0)^{\top}\) to \((3,0,0,0)^{\top}\), and sends \((0,1,0,0)^{\top}\) to \((0,3,0,0)^{\top}\). Both are eigenvectors for \(3\). The dimension is \(1+1+0=2\), in agreement with the basis found.
A Reliable Block-by-Block Procedure
For a specified scalar \(\lambda\), the calculation can be organized without building or row reducing the entire matrix at once. Keep the original block order so that the component solutions return to the correct coordinates.
Partition the unknown vector into components \(\mathbf{x}_i\). The full equation \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) becomes \((\mathbf{A}_i-\lambda\mathbf{I}_{n_i})\mathbf{x}_i=\mathbf{0}\) for each block.
Find \(E_{\lambda}(\mathbf{A}_i)\) for every block. If the system has only the zero solution, that block contributes only a zero component.
Embed each block basis vector by filling the other blocks with zeros. The union of these embedded bases gives a basis of the full eigenspace.
The procedure is justified by the componentwise matrix equation: it neither omits a possible eigenvector nor adds a vector that fails the full equation. It also makes the dimension count transparent. If \(\lambda\) occurs in several blocks, each block contributes its own number of independent eigendirections.
Common Mistake: Keeping Only One Contributing Block
In the first worked example, a tempting but incomplete answer is to report only \(\operatorname{span}\{(1,1,0,0)^{\top}\}\), because \((1,1)^{\top}\) is an eigenvector of the first block. This vector is valid, but it does not describe the whole eigenspace: the second block also has eigenvalue \(2\).
The missing vector is \((0,0,1,-1)^{\top}\). It satisfies the second block equation because \(\mathbf{C}(1,-1)^{\top}=(2,-2)^{\top}=2(1,-1)^{\top}\), while its first block component is zero. In fact, the two block components can be selected independently, giving every vector \((a,a,b,-b)^{\top}\). The correction is to solve every block equation, not to stop after finding one eigenvector or one contributing block.
A related distinction is that a shared eigenvalue is not a demand that the component vectors be identical, or even related by a common scalar choice. Each block has its own coordinates and its own eigenspace. The full eigenspace combines those choices by concatenation. Conversely, a block that does not have \(\lambda\) as an eigenvalue cannot supply a nonzero component for an eigenvector with eigenvalue \(\lambda\).
This structure is useful when studying matrices built from independent subsystems: the eigenvalue question can be localized to each subsystem, while the full eigenspace records all ways their eigenvectors can coexist. In calculations, the main safeguards are to preserve block coordinate order, include every block, and check the assembled vectors in the full equation \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\).
Check Your Understanding
For each question, use the block equations and state how the component solutions fit together.
- If \(\mathbf{A}=\operatorname{diag}(\mathbf{A}_1,\mathbf{A}_2)\), what condition on the two blocks is necessary and sufficient for \(\lambda\) to be an eigenvalue of \(\mathbf{A}\)?
- Suppose \(E_{\lambda}(\mathbf{A}_1)=\operatorname{span}\{(1,0)^{\top}\}\) and \(E_{\lambda}(\mathbf{A}_2)=\operatorname{span}\{(1,1)^{\top}\}\). Write a basis for the full eigenspace.
- If \(\lambda\) is not an eigenvalue of \(\mathbf{A}_2\), what must the \(\mathbf{A}_2\)-component of every \(\lambda\)-eigenvector of \(\operatorname{diag}(\mathbf{A}_1,\mathbf{A}_2)\) be?
- For blocks with eigenspace dimensions \(2\), \(0\), and \(1\) at \(\lambda\), what is the dimension of the full eigenspace?
- Explain why a nonzero eigenvector from just one block does not, by itself, describe the entire eigenspace when another block also has that eigenvalue.
