From Eigenspaces to a Diagonalization Test
A real matrix is diagonalizable when there is a basis of \(\mathbb{R}^n\) made entirely of eigenvectors. Finding a few eigenvectors is not enough: the question is whether the eigenspaces contain enough independent vectors to form a basis of the whole space. The dimension of each eigenspace records how many basis vectors it can contribute. The criterion in this tutorial makes that idea exact, and its proof explains why dimensions—not merely the number of eigenvalues—are what matter.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), where \(n\geq 1\). A real number \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^n\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The corresponding real eigenspace is the set of all such eigenvectors together with the zero vector. In particular, it is the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\). A matrix is diagonalizable over \(\mathbb{R}\) if \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for some invertible real matrix \(\mathbf{P}\) and real diagonal matrix \(\mathbf{D}\).
Only distinct eigenvalues appear in this sum: each eigenspace is included once. Each \(E_\lambda\) is a subspace of the finite-dimensional space \(\mathbb{R}^n\), so it has a basis and a well-defined dimension. Also, there can be at most \(n\) distinct real eigenvalues: choosing one nonzero eigenvector for each would give an independent list, by the result on eigenvectors for distinct eigenvalues, and an independent list in \(\mathbb{R}^n\) has at most \(n\) members. Thus the displayed sum has finitely many terms.
Why Distinct Eigenspaces Do Not Overlap in Their Directions
To combine bases from different eigenspaces, we need to know that their vectors do not introduce hidden dependence. The following statement establishes this directly. It is the independence of eigenvectors for distinct eigenvalues, expressed for whole eigenspaces rather than for just one chosen eigenvector from each.
Proof. We prove the first statement by induction on \(k\). For \(k=1\), the equation is \(\mathbf{v}_1=\mathbf{0}\), as required. Suppose the statement holds for \(k-1\) distinct eigenvalues, and suppose \(\mathbf{v}_1+\cdots+\mathbf{v}_k=\mathbf{0}\). Apply the linear map \(\mathbf{A}-\lambda_k\mathbf{I}_n\) to both sides. Since \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\), the result is \[ \sum_{i=1}^{k-1}(\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}. \] For \(1\leq i<k\), \(\lambda_i-\lambda_k\neq 0\), and \((\lambda_i-\lambda_k)\mathbf{v}_i\in E_{\lambda_i}\). The induction hypothesis therefore gives \((\lambda_i-\lambda_k)\mathbf{v}_i=\mathbf{0}\) for every \(i<k\). A nonzero real scalar times a vector is zero only if the vector is zero, so \(\mathbf{v}_i=\mathbf{0}\) for every \(i<k\). The original equation now gives \(\mathbf{v}_k=\mathbf{0}\). This proves the first statement. For the consequence, take any linear relation among vectors in the union of the eigenspace bases, and group its terms according to their eigenvalue. Each group is a vector in the associated eigenspace, and the first statement forces every group to be zero. Since each chosen list is a basis, all coefficients in each group are zero. Thus the union is independent. \(\square\)
The lemma also covers the case of a single eigenspace. When there are no real eigenvalues, there are no eigenspace bases to combine; the union is the empty list, which is independent. These conventions let the dimension criterion include matrices with no real eigenvalues without treating them as a special exception.
The Eigenspace Dimension Criterion
Proof. First suppose \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\). Equivalently, it has a basis of \(\mathbb{R}^n\) consisting of real eigenvectors. Group those \(n\) basis vectors according to their eigenvalues. Every eigenvalue in this grouping is real, and the vectors in the group for \(\lambda\) are an independent list in \(E_\lambda\). If there are \(k_\lambda\) vectors in that group, then \(k_\lambda\leq\dim E_\lambda\). The groups together contain the whole basis, so \(\sum_\lambda k_\lambda=n\); hence \(\sum_\lambda\dim E_\lambda\geq n\). On the other hand, choose a basis of every real eigenspace. By the lemma, their union is independent in \(\mathbb{R}^n\), so it has at most \(n\) vectors. Its number of vectors is exactly \(\sum_\lambda\dim E_\lambda\), giving \(\sum_\lambda\dim E_\lambda\leq n\). The two inequalities imply equality.
Conversely, suppose \(\sum_{\lambda\in\Lambda_{\mathbb{R}}(\mathbf{A})}\dim E_\lambda=n\). Choose a basis for each real eigenspace and join these lists. The lemma says the resulting list is independent, and the assumed sum says that it has \(n\) vectors. An independent list of \(n\) vectors in \(\mathbb{R}^n\) is a basis. Put these vectors, in their chosen order, into the columns of a matrix \(\mathbf{P}\). Then \(\mathbf{P}\) is invertible. Each column is an eigenvector, so multiplying each column by \(\mathbf{A}\) multiplies it by its eigenvalue. If \(\mathbf{D}\) is the diagonal matrix listing those eigenvalues in the corresponding order, this column-by-column statement is \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), so \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\). \(\square\)
The proof gives both directions for a reason. If the matrix is diagonalizable, an eigenbasis cannot contain more vectors from any one eigenspace than that space’s dimension; meanwhile the full eigenspace bases together cannot exceed \(n\) independent vectors. In the reverse direction, when the dimensions add to \(n\), the lemma guarantees that joining the bases really does produce \(n\) independent eigenvectors. The sum is therefore a test for whether the eigenspaces supply a complete real eigenbasis.
Applying the Criterion
Worked Example: Eigenspace Dimensions Add to the Full Dimension
Consider \(\mathbf{A}=\begin{bmatrix}2&0&1\\0&2&0\\0&0&-1\end{bmatrix}\). Determine whether it is diagonalizable over \(\mathbb{R}\) by finding its real eigenspaces.
Since \(\mathbf{A}-\lambda\mathbf{I}_3\) is upper triangular, its determinant is the product of its diagonal entries: \[ \det(\mathbf{A}-\lambda\mathbf{I}_3)=(2-\lambda)^2(-1-\lambda). \] Thus the real eigenvalues are \(2\) and \(-1\). For \(\lambda=2\), the equation \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\), with \(\mathbf{x}=(x,y,z)^\top\), is \[ \begin{bmatrix}0&0&1\\0&0&0\\0&0&-3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}z\\0\\-3z\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. \] The condition is \(z=0\), while \(x\) and \(y\) are free. Therefore \(E_2=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\), with dimension \(2\).
For \(\lambda=-1\), the equation is \[ \begin{bmatrix}3&0&1\\0&3&0\\0&0&0\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}3x+z\\3y\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. \] Thus \(z=-3x\), \(y=0\), and \(x\) is free. Hence \(E_{-1}=\operatorname{span}\{(-1,0,3)^\top\}\), with dimension \(1\). The sum is \(2+1=3\), so the criterion proves that \(\mathbf{A}\) is diagonalizable.
Check: Let \(\mathbf{P}=\begin{bmatrix}1&0&-1\\0&1&0\\0&0&3\end{bmatrix}\), whose determinant is \(3\neq0\), and let \(\mathbf{D}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&-1\end{bmatrix}\). The columns of \(\mathbf{P}\) are the eigenvectors found above. Direct multiplication gives \[ \mathbf{A}\mathbf{P} =\begin{bmatrix}2&0&1\\0&2&0\\0&0&-1\end{bmatrix} \begin{bmatrix}1&0&-1\\0&1&0\\0&0&3\end{bmatrix} =\begin{bmatrix}2&0&1\\0&2&0\\0&0&-3\end{bmatrix} =\mathbf{P}\mathbf{D}. \] Since \(\mathbf{P}\) is invertible, \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), as claimed.
Worked Example: A Sum Smaller Than the Dimension
Consider \(\mathbf{B}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&-1\end{bmatrix}\). Find its real eigenspaces and apply the criterion.
The diagonal entries give \(\det(\mathbf{B}-\lambda\mathbf{I}_3)=(2-\lambda)^2(-1-\lambda)\), so the real eigenvalues are \(2\) and \(-1\). For \(\lambda=2\), \[ \begin{bmatrix}0&1&0\\0&0&0\\0&0&-3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}y\\0\\-3z\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. \] Thus \(y=z=0\), with \(x\) free, and \(E_2=\operatorname{span}\{(1,0,0)^\top\}\), of dimension \(1\).
For \(\lambda=-1\), \[ \begin{bmatrix}3&1&0\\0&3&0\\0&0&0\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}3x+y\\3y\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. \] The second equation gives \(y=0\), and the first then gives \(x=0\); \(z\) is free. So \(E_{-1}=\operatorname{span}\{(0,0,1)^\top\}\), of dimension \(1\). The sum is \(1+1=2<3\). The criterion shows that \(\mathbf{B}\) is not diagonalizable over \(\mathbb{R}\).
Check: \(\mathbf{B}(1,0,0)^\top=(2,0,0)^\top=2(1,0,0)^\top\), and \(\mathbf{B}(0,0,1)^\top=(0,0,-1)^\top=-(0,0,1)^\top\). The computed vectors are indeed eigenvectors, and the two one-dimensional eigenspaces do not provide three basis vectors.
Worked Example: No Real Eigenspaces
Consider the real matrix \(\mathbf{C}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). Determine whether it is diagonalizable over \(\mathbb{R}\).
For real \(\lambda\), \[ \det(\mathbf{C}-\lambda\mathbf{I}_2) =\det\begin{bmatrix}-\lambda&-1\\1&-\lambda\end{bmatrix} =\lambda^2+1. \] Since \(\lambda^2+1>0\) for every real \(\lambda\), there is no real eigenvalue and \(\Lambda_{\mathbb{R}}(\mathbf{C})\) is empty. The sum of the real eigenspace dimensions is therefore the empty sum \(0\), which is less than \(2\). By the criterion, \(\mathbf{C}\) is not diagonalizable over \(\mathbb{R}\).
Check: A real diagonal matrix has real diagonal entries as eigenvalues. If \(\mathbf{C}\) were similar over \(\mathbb{R}\) to such a matrix, it would have real eigenvalues, contradicting the determinant computation. This confirms the conclusion without treating complex eigenvalues as real eigenspaces.
A Common Counting Pitfall
Count each distinct eigenspace once, not once for every basis vector it contains. In the first worked example, \(E_2\) has dimension \(2\), so its basis contributes two vectors, but there is still just one eigenspace associated with the eigenvalue \(2\). A mistaken count might write \(2+2+1=5\), counting the dimension of \(E_2\) twice because its basis has two vectors. The correct sum is \(\dim E_2+\dim E_{-1}=2+1=3\). The list of basis vectors has three members; the sum of eigenspace dimensions is also three, not five.
Another important distinction is the field. This criterion concerns diagonalization over \(\mathbb{R}\) and real eigenspaces. A real matrix with no real eigenvalues cannot have a real eigenbasis, even if it has eigenvalues when considered over \(\mathbb{C}\). In practice, identify the distinct real eigenvalues, find the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\) for each, calculate each dimension, and then compare the total with \(n\).
Check Your Understanding
Use the theorem and its proof to answer each question. When finding an eigenspace, show the null-space equations that determine its dimension.
- For a real \(4\times4\) matrix, the dimensions of its distinct real eigenspaces are \(2\) and \(2\). What does the criterion imply?
- Why does the union of bases from distinct real eigenspaces form an independent list?
- A real \(3\times3\) matrix has one real eigenspace of dimension \(1\) and another of dimension \(1\), and no other real eigenvalues. Can it be diagonalizable over \(\mathbb{R}\)? Explain.
- What is the sum in the criterion when a real matrix has no real eigenvalues, and what does that imply for a positive-sized real square matrix?
- Explain why an eigenbasis for a real matrix can be grouped by real eigenvalues.
