From Eigenvalues to Signs of a Quadratic Form
For a real symmetric matrix \(\mathbf{A}\), the value of its quadratic form at a vector \(\mathbf{x}\in\mathbb{R}^n\) is \(q_{\mathbf{A}}(\mathbf{x})=\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). The definition of definiteness asks about the sign of this value over all nonzero vectors, so checking a few inputs is not enough. The key is that a symmetric matrix has an orthonormal basis of eigenvectors: in coordinates along those eigenvectors, the quadratic form becomes a sum of eigenvalues multiplied by squares.
The distinction between strict and non-strict inequalities matters. Positive definite forms must be strictly positive away from \(\mathbf{0}\); positive semidefinite forms are allowed to take the value zero at nonzero vectors. The corresponding distinction applies to negative definite and negative semidefinite forms.
Why Orthogonal Coordinates Preserve the Sign
The Spectral Theorem for real symmetric matrices states that if \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is symmetric, then there are an orthogonal matrix \(\mathbf{Q}\) and a real diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). Here orthogonal means \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). The diagonal entries of \(\mathbf{D}\) are the eigenvalues of \(\mathbf{A}\), and the columns of \(\mathbf{Q}\) are corresponding unit eigenvectors. This form of the Spectral Theorem is used here as an established result.
Proof. Set \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\). Then \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=\mathbf{x}^{\top}\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\mathbf{x}=(\mathbf{Q}^{\top}\mathbf{x})^{\top}\mathbf{D}(\mathbf{Q}^{\top}\mathbf{x})=\mathbf{y}^{\top}\mathbf{D}\mathbf{y}\). Also, \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\) implies \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\), so \(\mathbf{x}=\mathbf{Q}\mathbf{y}\). Therefore \(\mathbf{x}=\mathbf{0}\) exactly when \(\mathbf{y}=\mathbf{0}\). In particular, changing to these coordinates does not introduce or remove nonzero vectors, and the quadratic-form values are unchanged. \(\square\)
If the eigenvalues are \(\lambda_1,\dots,\lambda_n\), then \(\mathbf{D}\) has diagonal entries \(\lambda_1,\dots,\lambda_n\), and the lemma gives the useful identity
Each squared coordinate is nonnegative. Thus, if every eigenvalue is positive, every nonzero \(\mathbf{y}\) makes this sum positive. If an eigenvalue is negative, its eigenvector gives a direction in which the form is negative. These observations extend to all five definiteness classes.
The Exact Eigenvalue Sign Tests
- \(\mathbf{A}\) is positive definite if and only if every \(\lambda_i>0\).
- \(\mathbf{A}\) is positive semidefinite if and only if every \(\lambda_i\ge 0\).
- \(\mathbf{A}\) is negative definite if and only if every \(\lambda_i<0\).
- \(\mathbf{A}\) is negative semidefinite if and only if every \(\lambda_i\le 0\).
- \(\mathbf{A}\) is indefinite if and only if at least one eigenvalue is positive and at least one is negative.
Proof. By the Spectral Theorem and the lemma, for each \(\mathbf{x}\) the quadratic-form value is \(\sum_{i=1}^n\lambda_i y_i^2\), where \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\), and \(\mathbf{x}\ne\mathbf{0}\) exactly when \(\mathbf{y}\ne\mathbf{0}\).
First suppose every \(\lambda_i>0\). For any nonzero \(\mathbf{y}\), at least one \(y_i\ne0\). Every term \(\lambda_i y_i^2\) is nonnegative and the term for that nonzero coordinate is positive, so the sum is positive. Thus \(\mathbf{A}\) is positive definite. Conversely, if \(\lambda_k\le0\), choose \(\mathbf{y}=\mathbf{e}_k\), the \(k\)th standard basis vector. The corresponding \(\mathbf{x}=\mathbf{Q}\mathbf{e}_k\) is nonzero, and \(q_{\mathbf{A}}(\mathbf{x})=\lambda_k\le0\), contradicting positive definiteness. This proves the positive definite equivalence.
If every \(\lambda_i\ge0\), every term in the sum is nonnegative, so \(q_{\mathbf{A}}(\mathbf{x})\ge0\) for all \(\mathbf{x}\). Conversely, if some \(\lambda_k<0\), the same choice \(\mathbf{y}=\mathbf{e}_k\) gives a nonzero \(\mathbf{x}\) with \(q_{\mathbf{A}}(\mathbf{x})=\lambda_k<0\). Hence the form is not positive semidefinite. This proves the positive semidefinite equivalence.
If every \(\lambda_i<0\), for any nonzero \(\mathbf{y}\) at least one term is strictly negative and all other terms are nonpositive, so the sum is negative. Conversely, if \(\lambda_k\ge0\), choosing \(\mathbf{y}=\mathbf{e}_k\) gives a nonzero \(\mathbf{x}\) with \(q_{\mathbf{A}}(\mathbf{x})=\lambda_k\ge0\), which rules out negative definiteness. Thus the negative definite equivalence holds.
If every \(\lambda_i\le0\), every term is nonpositive, so the quadratic form is nonpositive everywhere. Conversely, if some \(\lambda_k>0\), the choice \(\mathbf{y}=\mathbf{e}_k\) yields a positive value, which rules out negative semidefiniteness. This proves the negative semidefinite equivalence.
Finally, if some eigenvalue \(\lambda_i>0\), choosing \(\mathbf{y}=\mathbf{e}_i\) gives a positive quadratic-form value; if some other eigenvalue \(\lambda_j<0\), choosing \(\mathbf{y}=\mathbf{e}_j\) gives a negative value. So eigenvalues of both signs imply indefiniteness. In the other direction, suppose the form has a positive value. In its sum \(\sum_i\lambda_i y_i^2\), at least one term must be positive, which requires some \(\lambda_i>0\). A negative value requires at least one \(\lambda_j<0\), since a sum of nonnegative terms cannot be negative. Therefore an indefinite form has eigenvalues of both signs. All five equivalences follow. \(\square\)
Worked Classifications
Worked Example: A Positive Definite Matrix
Classify \(\mathbf{A}=\begin{bmatrix}3&1\\1&3\end{bmatrix}\) using its eigenvalues.
For \(\mathbf{v}_1=\begin{bmatrix}1\\1\end{bmatrix}\), multiplication gives \(\mathbf{A}\mathbf{v}_1=\begin{bmatrix}3+1\\1+3\end{bmatrix}=\begin{bmatrix}4\\4\end{bmatrix}=4\mathbf{v}_1\). For \(\mathbf{v}_2=\begin{bmatrix}1\\-1\end{bmatrix}\), it gives \(\mathbf{A}\mathbf{v}_2=\begin{bmatrix}3-1\\1-3\end{bmatrix}=\begin{bmatrix}2\\-2\end{bmatrix}=2\mathbf{v}_2\). These vectors are perpendicular and form an eigenbasis, so the eigenvalues are \(4\) and \(2\), both positive. The theorem says \(\mathbf{A}\) is positive definite.
Check: For \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\), direct multiplication gives \(q_{\mathbf{A}}(\mathbf{x})=3x^2+2xy+3y^2\). In the orthonormal eigenvector coordinates \(s=(x+y)/\sqrt{2}\) and \(t=(x-y)/\sqrt{2}\), this is \(4s^2+2t^2\), which is positive whenever \((x,y)\ne(0,0)\), agreeing with the classification.
Worked Example: A Positive Semidefinite Matrix That Is Not Positive Definite
Classify \(\mathbf{B}=\begin{bmatrix}1&1\\1&1\end{bmatrix}\). The vectors \(\mathbf{w}_1=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\mathbf{w}_2=\begin{bmatrix}1\\-1\end{bmatrix}\) satisfy \(\mathbf{B}\mathbf{w}_1=\begin{bmatrix}2\\2\end{bmatrix}=2\mathbf{w}_1\) and \(\mathbf{B}\mathbf{w}_2=\begin{bmatrix}0\\0\end{bmatrix}=0\mathbf{w}_2\). Thus the eigenvalues are \(2\) and \(0\). They are both nonnegative, so the form is positive semidefinite, but the zero eigenvalue means it is not positive definite.
Check: Directly, \(q_{\mathbf{B}}(\mathbf{x})=x^2+2xy+y^2=(x+y)^2\ge0\). At the nonzero vector \(\mathbf{x}=\begin{bmatrix}1\\-1\end{bmatrix}\), the value is \(0\), verifying that strict positivity fails.
Worked Example: An Indefinite Matrix
Classify \(\mathbf{C}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\). With \(\mathbf{z}_1=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\mathbf{z}_2=\begin{bmatrix}1\\-1\end{bmatrix}\), \(\mathbf{C}\mathbf{z}_1=\begin{bmatrix}1\\1\end{bmatrix}=1\mathbf{z}_1\), while \(\mathbf{C}\mathbf{z}_2=\begin{bmatrix}-1\\1\end{bmatrix}=-1\mathbf{z}_2\). The eigenvalues are \(1\) and \(-1\), so the form is indefinite.
Check: The quadratic form is \(q_{\mathbf{C}}(\mathbf{x})=2xy\). At \(\mathbf{z}_1\) its value is \(2\), and at \(\mathbf{z}_2\) its value is \(-2\). It therefore takes both signs, exactly as the definition requires.
Worked Example: A Negative Definite Matrix
For \(\mathbf{D}=\begin{bmatrix}-2&0\\0&-3\end{bmatrix}\), the standard basis vectors satisfy \(\mathbf{D}\mathbf{e}_1=-2\mathbf{e}_1\) and \(\mathbf{D}\mathbf{e}_2=-3\mathbf{e}_2\). Its eigenvalues are both negative, so \(\mathbf{D}\) is negative definite.
Check: For \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\), \(q_{\mathbf{D}}(\mathbf{x})=-2x^2-3y^2\). If \(\mathbf{x}\ne\mathbf{0}\), at least one of \(x^2,y^2\) is positive, so the value is strictly negative. This confirms the classification directly.
Common Mistake: Treating Zero as a Positive Eigenvalue
A common incorrect step is to conclude that a matrix is positive definite because all its eigenvalues are “positive or zero.” That condition is the test for positive semidefiniteness, not positive definiteness. For the matrix \(\mathbf{B}\) above, the eigenvalues \(2\) and \(0\) are nonnegative, but the calculation \(\mathbf{B}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\) gives \(q_{\mathbf{B}}\bigl(\begin{bmatrix}1\\-1\end{bmatrix}\bigr)=0\) at a nonzero vector. The correction is to keep the inequalities distinct: every eigenvalue must be strictly greater than zero for positive definiteness; allowing zero gives positive semidefiniteness.
| Definiteness class | Required eigenvalue signs | Can zero occur? |
|---|---|---|
| Positive definite | All \(\lambda_i>0\) | No |
| Positive semidefinite | All \(\lambda_i\ge0\) | Yes |
| Negative definite | All \(\lambda_i<0\) | No |
| Negative semidefinite | All \(\lambda_i\le0\) | Yes |
| Indefinite | At least one positive and one negative eigenvalue | Possibly |
The sign tests work because orthogonal diagonalization expresses every quadratic-form value as a sum of eigenvalues weighted by squares, and the eigenvector directions test each eigenvalue individually. The criteria are exact, not merely sufficient: the weighted sums prove one direction, and choosing an eigenvector proves the other.
Check Your Understanding
Use the eigenvalue sign tests and, where requested, verify the result directly from the quadratic form.
- A real symmetric matrix has eigenvalues \(3,1,0\). Which definiteness classes apply, and which do not?
- For \(\mathbf{A}=\begin{bmatrix}2&0\\0&-5\end{bmatrix}\), find vectors at which the quadratic form is positive and negative.
- Explain why a symmetric matrix with all eigenvalues strictly negative is negative definite, rather than merely negative semidefinite.
- Can a matrix be both positive semidefinite and negative semidefinite? Give an eigenvalue condition that allows this.
- Prove from the sum \(\sum_i\lambda_i y_i^2\) that a positive value of the quadratic form requires at least one positive eigenvalue.
