From Eigenvalues to Eigenspace Dimensions
Finding the eigenvalues of a matrix is only part of a diagonalizability test. For each eigenvalue, we must also determine how many linearly independent eigenvectors belong to it. Row reduction gives this information directly: the eigenvectors for \(\lambda\), together with the zero vector, are the solutions of a homogeneous system with coefficient matrix \( \mathbf{A}-\lambda\mathbf{I}\).
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), unless another field is stated. An eigenvalue \(\lambda\in\mathbb{R}\) is a scalar for which there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^n\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Rearranging gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Thus the eigenvectors for \(\lambda\), along with \(\mathbf{0}\), form the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\).
To find this dimension, row reduce \(\mathbf{A}-\lambda\mathbf{I}_n\) and count its free-variable columns. The row operations preserve the solutions of the homogeneous system, so they preserve the eigenspace as a solution set. They do not, in general, preserve the original eigenvectors as columns: use the reduced equations to describe the solutions, and substitute any proposed eigenvector into \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) to check it.
Proof. Row operations preserve the solutions of \(\mathbf{M}\mathbf{x}=\mathbf{0}\), so we may use the RREF. Let it have \(q\) free-variable columns. Each choice of values for those \(q\) variables determines the pivot variables uniquely from the reduced equations. In particular, setting the free variables gives a linear one-to-one correspondence between \(\mathbb{R}^q\) and the solution space: every solution has one free-variable assignment, and each assignment gives exactly one solution. Therefore the solution space has dimension \(q\). Taking \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_n\) gives the eigenspace statement. \(\square\)
Why the Total Dimension Is the Test
Suppose the distinct eigenvalues of \(\mathbf{A}\) are \(\lambda_1,\dots,\lambda_s\), and all lie in the scalar field being used. The eigenspaces for distinct eigenvalues fit together without overlap in their nonzero directions: a basis for each eigenspace, combined into one list, is linearly independent. Therefore, the sum of their dimensions counts the number of independent eigenvectors available from all the eigenspaces.
Proof. First, eigenspaces for distinct eigenvalues have a direct sum. To verify this, suppose \(\mathbf{v}_1+\cdots+\mathbf{v}_s=\mathbf{0}\), where \(\mathbf{v}_i\in E_{\lambda_i}\). Fix an index \(k\). Apply the matrix \(\prod_{j\ne k}(\mathbf{A}-\lambda_j\mathbf{I}_n)\) to the equation. Each term \(\mathbf{v}_i\) with \(i\ne k\) is sent to zero, because one factor is \(\mathbf{A}-\lambda_i\mathbf{I}_n\). The remaining term is \(\bigl(\prod_{j\ne k}(\lambda_k-\lambda_j)\bigr)\mathbf{v}_k=\mathbf{0}\). The scalar product is nonzero because the eigenvalues are distinct, so \(\mathbf{v}_k=\mathbf{0}\). This holds for each \(k\), proving that the sum is direct.
Choose a basis for each eigenspace. By the direct-sum argument, the combined list is linearly independent, and it spans the sum of the eigenspaces. If the dimensions sum to \(n\), this list has \(n\) independent vectors in \(\mathbb{R}^n\), so it is a basis consisting of eigenvectors. Thus \(\mathbf{A}\) is diagonalizable. Conversely, if \(\mathbf{A}\) is diagonalizable, it has a basis of eigenvectors. Group those basis vectors by their eigenvalues. For each \(\lambda_i\), the group lies in \(E_{\lambda_i}\); moreover, it spans that eigenspace, since a vector in \(E_{\lambda_i}\) expanded in the eigenbasis cannot have a nonzero component from a different eigenspace, by the direct-sum property just proved. Hence the grouped vectors form bases for the eigenspaces, and their total number is \(n\). Therefore the eigenspace dimensions sum to \(n\). \(\square\)
This test concerns dimensions, not the sizes of the eigenvalues’ repetitions in the characteristic polynomial. The algebraic multiplicity \(\operatorname{am}(\lambda)\) is the multiplicity of \(\lambda\) as a root of \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\); the geometric multiplicity is found by row reduction. They are different quantities.
Worked Row-Reduction Tests
Worked Example: A Repeated Eigenvalue with Too Few Eigenvectors
Let \(\mathbf{A}=\begin{bmatrix}3&1\\0&3\end{bmatrix}\). Its characteristic polynomial is \((3-\lambda)^2\), so its only eigenvalue is \(\lambda=3\), with algebraic multiplicity \(2\). Find its geometric multiplicity and test diagonalizability.
Subtract \(3\mathbf{I}_2\) and row reduce:
The reduced equations give \(x_2=0\), while \(x_1\) is free. Thus \(E_3=\{(x_1,0)^{\top}:x_1\in\mathbb{R}\}\), with basis \(\{(1,0)^{\top}\}\), and \(\operatorname{gm}(3)=1\). Check the basis vector: \(\mathbf{A}(1,0)^{\top}=(3,0)^{\top}=3(1,0)^{\top}\). The total eigenspace dimension is \(1<2\), so \(\mathbf{A}\) is not diagonalizable over \(\mathbb{R}\).
Worked Example: A Repeated Eigenvalue with a Full Eigenspace
Let \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\1&0&5\end{bmatrix}\). Since this matrix is triangular, its eigenvalues are \(2\) and \(5\); its characteristic polynomial is \((2-\lambda)^2(5-\lambda)\). Find both geometric multiplicities.
For \(\lambda=2\), row reduce:
The equation is \(x_1+3x_3=0\); \(x_2\) and \(x_3\) are free. A basis for \(E_2\) is \(\{(0,1,0)^{\top},(-3,0,1)^{\top}\}\), so \(\operatorname{gm}(2)=2\). The checks are \(\mathbf{A}(0,1,0)^{\top}=(0,2,0)^{\top}=2(0,1,0)^{\top}\) and \(\mathbf{A}(-3,0,1)^{\top}=(-6,0,2)^{\top}=2(-3,0,1)^{\top}\).
For \(\lambda=5\), reduce separately:
The first two variables are zero and \(x_3\) is free, so \(E_5\) has basis \(\{(0,0,1)^{\top}\}\) and \(\operatorname{gm}(5)=1\). Indeed, \(\mathbf{A}(0,0,1)^{\top}=(0,0,5)^{\top}=5(0,0,1)^{\top}\). The total is \(2+1=3\), the matrix size, so \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\).
Worked Example: Three Distinct Eigenvalues
Let \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&2&1\\0&0&3\end{bmatrix}\). Its characteristic polynomial is \((1-\lambda)(2-\lambda)(3-\lambda)\). Reduce \(\mathbf{A}-\lambda\mathbf{I}_3\) for each eigenvalue and compare the dimensions.
For \(\lambda=1\):
There are pivots in columns 2 and 3, leaving \(x_1\) free. Hence \(E_1\) has basis \(\{(1,0,0)^{\top}\}\). The check is \(\mathbf{A}(1,0,0)^{\top}=(1,0,0)^{\top}\).
For \(\lambda=2\):
The equations are \(x_1-x_2=0\) and \(x_3=0\), so \(E_2\) has basis \(\{(1,1,0)^{\top}\}\). Indeed, \(\mathbf{A}(1,1,0)^{\top}=(2,2,0)^{\top}=2(1,1,0)^{\top}\).
For \(\lambda=3\), scale the first two rows and clear the entry above the second pivot:
Taking \(x_3=2\) gives \(x_2=2\) and \(x_1=1\). Thus \(E_3\) has basis \(\{(1,2,2)^{\top}\}\); the check is \(\mathbf{A}(1,2,2)^{\top}=(3,6,6)^{\top}=3(1,2,2)^{\top}\). Each eigenspace has dimension \(1\), so the total is \(1+1+1=3\), and \(\mathbf{A}\) is diagonalizable.
Common Pitfall: Counting Zero Rows Instead of Free Variables
For the first example, a mistaken calculation might look at \(\begin{bmatrix}0&1\\0&0\end{bmatrix}\), see one zero row, and conclude that there are two free variables because there are two rows minus one nonzero row. That is incorrect: the number of free variables is determined by the columns. Column 2 contains a pivot, and column 1 does not, so there is exactly one free variable. The corrected conclusion is \(\operatorname{gm}(3)=1\), not \(2\).
A related error is to set the geometric multiplicity equal to the algebraic multiplicity. In that same example, \(\operatorname{am}(3)=2\), but row reduction shows \(\operatorname{gm}(3)=1\). The diagonalizability test uses the dimensions found from the null spaces. For a real matrix with non-real eigenvalues, the real-eigenvalue version of the test above is not sufficient to decide diagonalizability over \(\mathbb{R}\); the field and its eigenvalues must match. Over \(\mathbb{C}\), apply the same method using complex scalars and all complex eigenvalues.
Check Your Understanding
For each question, distinguish the number of pivots from the number of free variables and check proposed eigenvectors by substitution.
- For \(\mathbf{A}=\begin{bmatrix}4&1\\0&4\end{bmatrix}\), find \(\operatorname{gm}(4)\) and decide whether \(\mathbf{A}\) is diagonalizable.
- If \(\operatorname{rref}(\mathbf{A}-\lambda\mathbf{I}_4)\) has three pivot columns, what is \(\operatorname{gm}(\lambda)\)?
- For \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&7\end{bmatrix}\), find the geometric multiplicity of each eigenvalue and compare their sum with the matrix size.
- Explain why a zero row in the reduced matrix does not itself count as a free variable.
- A \(3\times3\) real matrix has only real eigenvalues, and the dimensions of its distinct eigenspaces sum to \(2\). Can it be diagonalizable over \(\mathbb{R}\)?
