A Dimension Count That Decides Diagonalizability
To diagonalize a real matrix, one needs enough linearly independent real eigenvectors to form a basis of the whole space. When an eigenvalue is repeated, counting its repetitions in the characteristic polynomial does not tell us how many independent eigenvectors it supplies. The diagnostic is instead to find a basis for each eigenspace, add those dimensions over the distinct real eigenvalues, and compare the total with the size of the matrix.
Throughout, \(\mathbf{A}\) is an \(n\times n\) real matrix, and diagonalizability means diagonalizability over \(\mathbb{R}\). An eigenvalue is a real scalar \(\lambda\) for which there is a nonzero vector \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Its eigenspace is the set of all solutions of the corresponding homogeneous equation, including the zero vector.
The columns of \(\mathbf{P}\) in a diagonalization are eigenvectors: from \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), we get \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Column by column, this says that each column of \(\mathbf{P}\) is an eigenvector, with its eigenvalue on the matching diagonal entry of \(\mathbf{D}\). Thus the practical issue is whether the eigenspaces together contain a basis of \(\mathbb{R}^n\).
Proof. First, eigenspaces for distinct eigenvalues are independent in the following sense: if \(\mathbf{v}_i\in E_{\lambda_i}\) and \(\mathbf{v}_1+\cdots+\mathbf{v}_s=\mathbf{0}\), then every \(\mathbf{v}_i=\mathbf{0}\). We prove this by induction on the number of distinct eigenvalues. For one eigenvalue the claim follows directly from \(\mathbf{v}_1=\mathbf{0}\). Suppose it holds for \(s-1\) eigenvalues. Apply \(\mathbf{A}-\lambda_s\mathbf{I}_n\) to the displayed relation. The term from \(\mathbf{v}_s\) becomes zero, while \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\) gives $$ \sum_{i=1}^{s-1}(\lambda_i-\lambda_s)\mathbf{v}_i=\mathbf{0}. $$ By the induction hypothesis, \((\lambda_i-\lambda_s)\mathbf{v}_i=\mathbf{0}\) for each \(i<s\). Since the eigenvalues are distinct, each scalar \(\lambda_i-\lambda_s\ne0\), so \(\mathbf{v}_i=\mathbf{0}\) for \(i<s\). The original relation then gives \(\mathbf{v}_s=\mathbf{0}\). This proves the independence claim.
Choose a basis for each \(E_{\lambda_i}\). By the claim, putting these bases together gives an independent list, so the sum of their lengths is at most \(n\). If the sum of the eigenspace dimensions equals \(n\), this list is \(n\) independent vectors in \(\mathbb{R}^n\), hence a basis. Put them as the columns of \(\mathbf{P}\), and let \(\mathbf{D}\) have the corresponding eigenvalues on its diagonal. The eigenvector equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), and because \(\mathbf{P}\) is invertible, \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Conversely, if \(\mathbf{A}\) is diagonalizable, the columns of its invertible eigenvector matrix give a basis of \(n\) eigenvectors. Group those columns by their eigenvalues. Each group lies in the matching eigenspace, so each eigenspace dimension is at least the size of its group. The groups contain \(n\) columns in total, so the sum of dimensions is at least \(n\); the independence claim gives that it is at most \(n\). Therefore it equals \(n\). \(\square\)
The Diagnostic Procedure
For each distinct real eigenvalue, row-reduce \(\mathbf{A}-\lambda\mathbf{I}_n\) and solve the resulting homogeneous system. The free variables describe the eigenspace; a basis is obtained by setting one free variable at a time to \(1\) and the others to \(0\). The number of vectors in that basis is the eigenspace dimension. Count each distinct eigenvalue once, add these dimensions, and compare the result with \(n\).
Use the characteristic polynomial or other available information. Repeated roots are listed once for this dimension count.
For each listed \(\lambda\), solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) by row reduction.
Add the numbers of free variables, equivalently the dimensions of the eigenspaces. The matrix is diagonalizable over \(\mathbb{R}\) exactly when this sum is \(n\).
Worked Diagnostics
Worked Example: A Repeated Eigenvalue With Enough Eigenvectors
Decide whether \(\mathbf{A}=\begin{bmatrix}4&0&0\\3&1&0\\3&0&1\end{bmatrix}\) is diagonalizable over \(\mathbb{R}\), and find bases for its eigenspaces.
The matrix is triangular, so its characteristic polynomial is \((4-\lambda)(1-\lambda)^2\). Its distinct eigenvalues are \(4\) and \(1\). For \(\lambda=4\), row-reduce \(\mathbf{A}-4\mathbf{I}_3\):
The equations are \(x-y=0\) and \(y-z=0\), so \(x=y=z\). Thus \(E_4=\operatorname{span}\{(1,1,1)^{\top}\}\), with dimension \(1\). In particular, the second equation links \(y\) and \(z\); it does not set \(z\) to zero.
For \(\lambda=1\), the reduction is
Here \(x=0\), while \(y\) and \(z\) are free. Hence \(E_1=\operatorname{span}\{(0,1,0)^{\top},(0,0,1)^{\top}\}\), with dimension \(2\). The dimensions sum to \(1+2=3\), so the criterion proves that \(\mathbf{A}\) is diagonalizable.
Check: \(\mathbf{A}(1,1,1)^{\top}=(4,4,4)^{\top}=4(1,1,1)^{\top}\), \(\mathbf{A}(0,1,0)^{\top}=(0,1,0)^{\top}\), and \(\mathbf{A}(0,0,1)^{\top}=(0,0,1)^{\top}\). These three eigenvectors are independent: their columns form a matrix with determinant \(1\).
Worked Example: A Repeated Eigenvalue With Too Few Eigenvectors
Test \(\mathbf{B}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&5\end{bmatrix}\) for diagonalizability over \(\mathbb{R}\).
Its characteristic polynomial is \((2-\lambda)^2(5-\lambda)\), so its distinct eigenvalues are \(2\) and \(5\). For \(\lambda=2\), row reduction gives
The equations are \(y=0\) and \(z=0\), with \(x\) free. Therefore \(E_2=\operatorname{span}\{(1,0,0)^{\top}\}\) and \(\dim E_2=1\). For \(\lambda=5\),
Thus \(x=y=0\), while \(z\) is free. So \(E_5=\operatorname{span}\{(0,0,1)^{\top}\}\) and \(\dim E_5=1\). The dimension sum is \(1+1=2<3\); \(\mathbf{B}\) is not diagonalizable over \(\mathbb{R}\).
Check: \(\mathbf{B}(1,0,0)^{\top}=(2,0,0)^{\top}=2(1,0,0)^{\top}\), and \(\mathbf{B}(0,0,1)^{\top}=(0,0,5)^{\top}=5(0,0,1)^{\top}\). For \(\lambda=2\), a vector \((x,y,z)^{\top}\) must satisfy \(\mathbf{B}\mathbf{x}=2\mathbf{x}\), whose second coordinate gives \(y=0\) and third gives \(z=0\). The repeated eigenvalue \(2\) supplies only one independent eigenvector.
Worked Example: Three Distinct Eigenspaces
Classify \(\mathbf{C}=\begin{bmatrix}2&1&0\\1&2&0\\0&0&5\end{bmatrix}\) and find an eigenbasis.
The characteristic polynomial is \(((2-\lambda)^2-1)(5-\lambda)=(1-\lambda)(3-\lambda)(5-\lambda)\). For \(\lambda=3\),
Hence \(x=y\), \(z=0\), and \(E_3=\operatorname{span}\{(1,1,0)^{\top}\}\). For \(\lambda=1\),
Thus \(x=-y\), \(z=0\), and \(E_1=\operatorname{span}\{(1,-1,0)^{\top}\}\). Finally, for \(\lambda=5\),
So \(x=y=0\), \(z\) is free, and \(E_5=\operatorname{span}\{(0,0,1)^{\top}\}\). The dimensions sum to \(1+1+1=3\), giving diagonalizability. The three basis vectors form an eigenbasis.
Check: \(\mathbf{C}(1,1,0)^{\top}=(3,3,0)^{\top}\), \(\mathbf{C}(1,-1,0)^{\top}=(1,-1,0)^{\top}\), and \(\mathbf{C}(0,0,1)^{\top}=(0,0,5)^{\top}\). Each is the claimed eigenvalue times its vector.
Multiplicity Is Not Eigenspace Dimension
The algebraic multiplicity of an eigenvalue is its number of occurrences as a root of the characteristic polynomial. Its geometric multiplicity is the dimension of its eigenspace. For an \(n\times n\) matrix, the dimension criterion concerns the geometric multiplicities, not the algebraic ones. In the second example, \(2\) has algebraic multiplicity \(2\) but geometric multiplicity \(1\). The dimension sum therefore falls short of \(3\), even though the roots of the characteristic polynomial, counted with repetition, total \(3\).
A common incorrect calculation is: “The characteristic polynomial of \(\mathbf{B}\) has roots \(2,2,5\), so there are three eigenvectors and \(\mathbf{B}\) is diagonalizable.” This treats a repeated root as if it guaranteed a second independent eigenvector. The correction is to solve \((\mathbf{B}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\): row reduction gives \(y=z=0\), leaving only the one-dimensional space spanned by \((1,0,0)^{\top}\). The root \(2\) occurs twice, but its eigenspace dimension is \(1\), so the dimension criterion rules out diagonalizability.
Also count dimensions only once per distinct eigenvalue. In the first example, the root \(1\) occurs twice, but its eigenspace contributes \(2\) to the dimension sum—not \(2\) copies of its eigenspace dimension. The row-reduction result already counts all independent eigenvectors associated with that eigenvalue.
Check Your Understanding
For each question, distinguish the eigenvalue from its eigenspace and use the dimension criterion where appropriate.
- For \(\mathbf{A}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\), find \(E_3\), its dimension, and decide whether \(\mathbf{A}\) is diagonalizable.
- A \(4\times4\) real matrix has distinct real eigenvalues \(2\) and \(7\), with eigenspace dimensions \(2\) and \(1\). Is it diagonalizable over \(\mathbb{R}\)? Explain using the criterion.
- For \(\mathbf{B}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), solve \((\mathbf{B}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\). What does the dimension count imply?
- In your own words, why does a repeated root of the characteristic polynomial not by itself guarantee enough eigenvectors?
