Why Distinct Eigenvalues Matter
A symmetric matrix may have eigenvectors pointing in quite different directions. The connection between the directions is not arbitrary: eigenvectors belonging to different eigenvalues must be orthogonal. This gives a strong geometric restriction that can simplify calculations and help organize the eigenvectors of a matrix. The hypothesis that the matrix is real and symmetric matters, as does the requirement that the eigenvalues be distinct.
Throughout this tutorial, matrices and vectors are real. For vectors \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\), their dot product is \(\mathbf{u}\cdot\mathbf{v}=\mathbf{u}^{\top}\mathbf{v}\). They are orthogonal when \(\mathbf{u}\cdot\mathbf{v}=0\). A real matrix \(\mathbf{A}\) is symmetric when \(\mathbf{A}^{\top}=\mathbf{A}\). An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{x}\) such that \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\); the vectors satisfying this equation, including \(\mathbf{0}\), form the eigenspace \(E_{\lambda}\).
The key calculation comes from symmetry. For a symmetric matrix, multiplication by the matrix can be shifted from one side of a dot product to the other without changing the value. We establish this identity first; the orthogonality result will follow by applying it to eigenvectors.
Proof. By the definition of the dot product and the transpose rule for a matrix product, \((\mathbf{A}\mathbf{x})\cdot\mathbf{y}=(\mathbf{A}\mathbf{x})^{\top}\mathbf{y}=\mathbf{x}^{\top}\mathbf{A}^{\top}\mathbf{y}\). Since \(\mathbf{A}\) is symmetric, \(\mathbf{A}^{\top}=\mathbf{A}\), so this equals \(\mathbf{x}^{\top}\mathbf{A}\mathbf{y}=\mathbf{x}\cdot(\mathbf{A}\mathbf{y})\). \(\square\)
The Orthogonality Theorem
Proof. Choose any \(\mathbf{u}\in E_{\lambda}\) and \(\mathbf{v}\in E_{\mu}\). By the definition of eigenspace, \(\mathbf{A}\mathbf{u}=\lambda\mathbf{u}\) and \(\mathbf{A}\mathbf{v}=\mu\mathbf{v}\). The lemma and these eigenvector equations give
Subtracting the right side from the left yields \((\lambda-\mu)(\mathbf{u}\cdot\mathbf{v})=0\). The eigenvalues are distinct, so \(\lambda-\mu\ne0\). Dividing by this nonzero scalar gives \(\mathbf{u}\cdot\mathbf{v}=0\). The choices of \(\mathbf{u}\) and \(\mathbf{v}\) were arbitrary, so every vector in \(E_{\lambda}\) is orthogonal to every vector in \(E_{\mu}\). This also covers the case where either chosen vector is \(\mathbf{0}\), since its dot product with every vector is zero. \(\square\)
The proof has two ingredients: symmetry lets us equate the two ways of evaluating the dot product, and distinctness lets us divide by \(\lambda-\mu\). Without the second ingredient, the calculation says only that \(0\cdot(\mathbf{u}\cdot\mathbf{v})=0\), which imposes no restriction on the dot product.
Computing and Checking Orthogonal Eigenspaces
Worked Example: Two Distinct Eigenvalues in R²
Find eigenvectors for the distinct eigenvalues of \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), and check their orthogonality directly.
For \(\mathbf{x}=(x,y)^{\top}\), the equation \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) is
When \(\lambda=3\), the equations reduce to \(y=x\). Choosing \(x=1\) gives \(\mathbf{u}=(1,1)^{\top}\). When \(\lambda=1\), they reduce to \(y=-x\), so choose \(\mathbf{v}=(1,-1)^{\top}\). Verify both eigenvector equations:
The eigenvalues \(3\) and \(1\) are distinct, and \(\mathbf{A}\) is symmetric. The theorem therefore predicts orthogonality; direct substitution checks it:
In fact, the equations show \(E_3=\operatorname{span}\{(1,1)^{\top}\}\) and \(E_1=\operatorname{span}\{(1,-1)^{\top}\}\). Every vector in either eigenspace is a scalar multiple of the displayed vector, so the zero dot product also verifies orthogonality for every pair of vectors in the two eigenspaces.
Worked Example: One Eigenspace Has Dimension Two
Let \(\mathbf{A}=\begin{bmatrix}4&0&0\\0&4&0\\0&0&-2\end{bmatrix}\). Find the eigenspaces for its distinct eigenvalues and check that the spaces, not just one pair of eigenvectors, are orthogonal.
For \(\mathbf{x}=(x,y,z)^{\top}\), the equation \(\mathbf{A}\mathbf{x}=4\mathbf{x}\) gives \((4x,4y,-2z)^{\top}=(4x,4y,4z)^{\top}\), so \(-2z=4z\), hence \(z=0\). Thus \(E_4=\{(x,y,0)^{\top}:x,y\in\mathbb{R}\}\). For eigenvalue \(-2\), the first two equations are \(4x=-2x\) and \(4y=-2y\), so \(x=y=0\). Hence \(E_{-2}=\{(0,0,z)^{\top}:z\in\mathbb{R}\}\).
Take arbitrary vectors \(\mathbf{u}=(a,b,0)^{\top}\in E_4\) and \(\mathbf{v}=(0,0,c)^{\top}\in E_{-2}\). Their dot product is
For a nonzero check using exact vectors, choose \(\mathbf{u}=(2,-1,0)^{\top}\) and \(\mathbf{v}=(0,0,3)^{\top}\). Then \(\mathbf{A}\mathbf{u}=(8,-4,0)^{\top}=4\mathbf{u}\), and \(\mathbf{A}\mathbf{v}=(0,0,-6)^{\top}=-2\mathbf{v}\). Also \(\mathbf{u}\cdot\mathbf{v}=2(0)+(-1)(0)+0(3)=0\). The arbitrary-vector calculation establishes the full claim, while this substitution confirms it on a concrete pair.
This example illustrates why the theorem is about eigenspaces. The eigenspace \(E_4\) contains many vectors, not just one chosen eigenvector, but each of them is orthogonal to every vector in \(E_{-2}\).
Using the Result with a Repeated Eigenvalue
Worked Example: Orthogonality to a Repeated Eigenspace
Consider the symmetric matrix \(\mathbf{A}=\begin{bmatrix}3&1&0\\1&3&0\\0&0&2\end{bmatrix}\). Find the eigenspaces for \(4\) and \(2\), then use the theorem to check orthogonality for a chosen pair.
For eigenvalue \(4\), the system \((\mathbf{A}-4\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) has coefficient matrix
Its equations are \(-x+y=0\), \(x-y=0\), and \(-2z=0\). Thus \(x=y\) and \(z=0\), so \(E_4=\{(t,t,0)^{\top}:t\in\mathbb{R}\}\). For eigenvalue \(2\), the equations from \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) are \(x+y=0\), \(x+y=0\), and \(0=0\). Hence \(E_2=\{(s,-s,r)^{\top}:s,r\in\mathbb{R}\}\).
Choose \(\mathbf{u}=(1,1,0)^{\top}\in E_4\) and \(\mathbf{v}=(3,-3,2)^{\top}\in E_2\). The eigenvector equations and dot product are verified directly:
More generally, for any \((t,t,0)^{\top}\in E_4\) and \((s,-s,r)^{\top}\in E_2\), the dot product is \(ts-t s+0r=0\). Thus the two eigenspaces are orthogonal even though \(E_2\) has two free parameters. Its dimension reflects the repeated eigenvalue \(2\); the theorem concerns its orthogonality to the eigenspace for the distinct eigenvalue \(4\).
Common Mistakes and the Role of the Hypotheses
Mistake: applying the theorem to a matrix that is not symmetric. Consider \(\mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\). It is not symmetric, since its entry in row 1, column 2 is \(1\), while its entry in row 2, column 1 is \(0\). The vector \(\mathbf{u}=(1,0)^{\top}\) satisfies \(\mathbf{B}\mathbf{u}=(1,0)^{\top}=1\mathbf{u}\), and \(\mathbf{v}=(1,1)^{\top}\) satisfies \(\mathbf{B}\mathbf{v}=(2,2)^{\top}=2\mathbf{v}\). These are eigenvectors for distinct eigenvalues, but \(\mathbf{u}\cdot\mathbf{v}=1\). The conclusion fails because the symmetry hypothesis is absent.
Mistake: assuming eigenvectors for the same eigenvalue must be orthogonal. For the symmetric matrix \(\mathbf{I}_2\), both \((1,0)^{\top}\) and \((1,1)^{\top}\) are eigenvectors with eigenvalue \(1\), since multiplying either vector by \(\mathbf{I}_2\) leaves it unchanged. Their dot product is \(1\), not \(0\). The theorem does not apply: the eigenvalues are equal. The eigenspace for this eigenvalue is all of \(\mathbb{R}^2\), and it contains pairs of vectors that are not orthogonal.
A reliable check is to identify the hypotheses before using the conclusion. The matrix must be real and symmetric, and the eigenvalues associated with the two vectors must be distinct. If those conditions hold, the theorem establishes orthogonality without requiring the vectors to be normalized. If either condition fails, orthogonality may still happen for particular vectors, but this theorem does not guarantee it.
This result is useful because it lets us separate directions associated with different eigenvalues: their dot products vanish automatically, even when the eigenspaces have several independent directions. It also provides a quick consistency check for exact calculations. If a symmetric matrix calculation gives nonorthogonal eigenvectors for distinct eigenvalues, at least one claimed eigenvector, eigenvalue, or matrix entry should be checked. The theorem does not, by itself, select special orthogonal vectors within a single eigenspace; that is a different task.
Check Your Understanding
Use the hypotheses and calculations in this tutorial to answer each question.
- State the dot-product identity that holds for a real symmetric matrix, and show where symmetry enters its proof.
- Let \(\mathbf{A}\) be real and symmetric, and let \(\mathbf{u}\in E_5\) and \(\mathbf{v}\in E_{-1}\). Which equation forces \(\mathbf{u}\cdot\mathbf{v}=0\)?
- For \(\mathbf{A}=\begin{bmatrix}4&0&0\\0&4&0\\0&0&-2\end{bmatrix}\), describe both eigenspaces and verify their orthogonality for arbitrary vectors in them.
- Why does the orthogonality theorem not imply that every pair of vectors in \(E_2\) from the third worked example is orthogonal?
- Give the eigenvector equations and dot product that show why the theorem cannot be applied to \(\mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\).
