What a Scalar Shift Changes
For a real symmetric matrix, adding \(t\mathbf{I}\) changes the quadratic form in a particularly controlled way. The added term contributes the same amount, scaled by the squared length of the input vector, in every direction. Correspondingly, every eigenvalue moves by the same amount \(t\), while the eigenvectors remain unchanged. This makes it possible to locate exactly which values of \(t\) produce positive, negative, semidefinite, or indefinite forms.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be symmetric, let \(t\in\mathbb{R}\), and define \(\mathbf{A}_t=\mathbf{A}+t\mathbf{I}_n\). The quadratic form associated with a symmetric matrix \(\mathbf{B}\) is the function \(Q_{\mathbf{B}}(\mathbf{x})=\mathbf{x}^{\top}\mathbf{B}\mathbf{x}\), for \(\mathbf{x}\in\mathbb{R}^n\). A form is positive definite if \(Q_{\mathbf{B}}(\mathbf{x})>0\) for every nonzero \(\mathbf{x}\); positive semidefinite if \(Q_{\mathbf{B}}(\mathbf{x})\geq 0\) for every \(\mathbf{x}\). Negative definite and negative semidefinite are defined by reversing these inequalities. A form is indefinite if it takes both positive and negative values.
The shift is symmetric because both \(\mathbf{A}\) and \(\mathbf{I}_n\) are symmetric. Also, for every \(\mathbf{x}\), direct multiplication gives
Here \(\|\mathbf{x}\|^2=\mathbf{x}^{\top}\mathbf{x}\), which is positive whenever \(\mathbf{x}\ne\mathbf{0}\). Thus a positive \(t\) raises the value of the form in every nonzero direction; a negative \(t\) lowers it. The eigenvalue description sharpens this observation into exact thresholds.
The Eigenvalues Move Together
Proof. By the real Spectral Theorem, there is an orthogonal matrix \(\mathbf{U}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{\top}\), where the diagonal entries of \(\mathbf{D}\) are the eigenvalues \(\lambda_1,\dots,\lambda_n\). Orthogonality means \(\mathbf{U}\mathbf{U}^{\top}=\mathbf{I}_n\). Therefore
The diagonal entries of \(\mathbf{D}+t\mathbf{I}_n\) are \(\lambda_i+t\), so this diagonalization gives the asserted eigenvalues, including their multiplicities. Alternatively, if \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for a nonzero \(\mathbf{v}\), then \((\mathbf{A}+t\mathbf{I}_n)\mathbf{v}=\lambda\mathbf{v}+t\mathbf{v}=(\lambda+t)\mathbf{v}\). Thus each original eigenvector is retained, with its eigenvalue shifted by \(t\). \(\square\)
Because the matrix is real symmetric, its eigenvalues are real and it has an orthonormal eigenbasis. Write its least and greatest eigenvalues as \(\lambda_{\min}\) and \(\lambda_{\max}\), respectively. The smallest shifted eigenvalue is \(\lambda_{\min}+t\), and the largest is \(\lambda_{\max}+t\). The definiteness thresholds follow by asking when these extreme values are positive, nonnegative, negative, or nonpositive.
- It is positive definite exactly when \(t>-\lambda_{\min}\), and positive semidefinite exactly when \(t\geq-\lambda_{\min}\).
- It is negative definite exactly when \(t<-\lambda_{\max}\), and negative semidefinite exactly when \(t\leq-\lambda_{\max}\).
- It is indefinite exactly when \(-\lambda_{\max}<t<-\lambda_{\min}\). This interval is empty when \(\lambda_{\min}=\lambda_{\max}\).
Proof. In an orthonormal eigenbasis, write \(\mathbf{x}=\mathbf{U}\mathbf{y}\), where \(\mathbf{y}=(y_1,\dots,y_n)^{\top}\). The diagonalization above gives
The form is positive definite if every coefficient \(\lambda_i+t\) is positive: then every nonzero \(\mathbf{y}\) makes the sum positive. Conversely, if some coefficient \(\lambda_j+t\leq 0\), taking \(\mathbf{y}=\mathbf{e}_j\) gives a nonpositive value, so the form is not positive definite. Since the least coefficient is \(\lambda_{\min}+t\), positive definiteness is equivalent to \(t>-\lambda_{\min}\). Replacing “positive” by “nonnegative” gives positive semidefiniteness: all terms are nonnegative exactly when \(\lambda_{\min}+t\geq0\), and a negative coefficient gives a negative value on its coordinate vector.
For negative definiteness, all coefficients must be negative; this is equivalent to the greatest coefficient satisfying \(\lambda_{\max}+t<0\), or \(t<-\lambda_{\max}\). For negative semidefiniteness, all coefficients must be nonpositive, equivalent to \(t\leq-\lambda_{\max}\). Finally, the form takes a negative value if and only if at least one coefficient is negative, which is equivalent to \(t<-\lambda_{\min}\). It takes a positive value if and only if at least one coefficient is positive, equivalent to \(t>-\lambda_{\max}\). It is indefinite exactly when both conditions hold, namely \(-\lambda_{\max}<t<-\lambda_{\min}\). If the two extreme eigenvalues agree, these strict inequalities cannot both hold. \(\square\)
At a threshold, the distinction between strict and non-strict inequalities matters. When \(t=-\lambda_{\min}\), the least shifted eigenvalue is zero, so the form is positive semidefinite but not positive definite. When \(t=-\lambda_{\max}\), the greatest shifted eigenvalue is zero, so it is negative semidefinite but not negative definite. If every eigenvalue is the same, both thresholds coincide: at that one value the shifted matrix is the zero matrix, which is both positive semidefinite and negative semidefinite.
Finding Thresholds in Examples
Worked Example: A Diagonal Matrix with Three Eigenvalues
Let \(\mathbf{A}=\begin{bmatrix}-2&0&0\\0&1&0\\0&0&4\end{bmatrix}\). Determine the definiteness of \(\mathbf{A}+t\mathbf{I}_3\) for all real \(t\).
The eigenvalues are the diagonal entries \(-2,1,4\), with corresponding standard basis eigenvectors. After shifting, the matrix and eigenvalues are
The thresholds are \(-\lambda_{\max}=-4\) and \(-\lambda_{\min}=2\). Therefore the form is negative definite for \(t<-4\), negative semidefinite at \(t=-4\), indefinite for \(-4<t<2\), positive semidefinite at \(t=2\), and positive definite for \(t>2\). For example, at \(t=-4\) the shifted eigenvalues are \(-6,-3,0\); at \(t=2\) they are \(0,3,6\). These boundary cases are semidefinite, not definite.
Check: At \(t=-4\), the shifted matrix is \(\operatorname{diag}(-6,-3,0)\); its quadratic form is \(-6x_1^2-3x_2^2\), which is always nonpositive and vanishes for \(\mathbf{x}=\mathbf{e}_3\). At \(t=2\), the form is \(3x_2^2+6x_3^2\), which is always nonnegative and vanishes for \(\mathbf{x}=\mathbf{e}_1\). The displayed shifted eigenvalues and boundary classifications agree.
Worked Example: Shifting a Nondiagonal Matrix
Let \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\). Find the full range of \(t\) for which the shifted form is indefinite, and the ranges for positive and negative definiteness.
For \(\mathbf{v}_1=(1,1)^{\top}\), multiplication gives \(\mathbf{A}\mathbf{v}_1=(3,3)^{\top}=3\mathbf{v}_1\). For \(\mathbf{v}_2=(1,-1)^{\top}\), it gives \(\mathbf{A}\mathbf{v}_2=(1,-1)^{\top}=\mathbf{v}_2\). These independent eigenvectors give eigenvalues \(3\) and \(1\). The shifted eigenvalues are \(3+t\) and \(1+t\); hence the thresholds are \(-3\) and \(-1\). The form is indefinite for \(-3<t<-1\), positive definite for \(t>-1\), and negative definite for \(t<-3\). It is negative semidefinite at \(t=-3\) and positive semidefinite at \(t=-1\).
Take \(t=-2\), inside the indefinite interval. The shifted matrix is \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\). Applying it to the two eigenvectors verifies the shifted eigenvalues directly:
Check: The quadratic form at this shift is \(2x_1x_2\). At \(\mathbf{x}=(1,1)^{\top}\) its value is \(2>0\), while at \(\mathbf{x}=(1,-1)^{\top}\) its value is \(-2<0\). This confirms indefiniteness and agrees with the shifted eigenvalues \(1\) and \(-1\).
Worked Example: A Semidefinite Boundary and Its Null Direction
Let \(\mathbf{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\). Find the shift at which the form first becomes positive semidefinite as \(t\) increases, and check the boundary case.
Multiplication gives \(\mathbf{A}(1,1)^{\top}=(3,3)^{\top}\), so \(3\) is an eigenvalue. It also gives \(\mathbf{A}(1,-1)^{\top}=(-1,1)^{\top}=-(1,-1)^{\top}\), so \(-1\) is an eigenvalue. Thus \(\lambda_{\min}=-1\), and the positive semidefinite threshold is \(t=-\lambda_{\min}=1\). The shifted eigenvalues are \(t+3\) and \(t-1\). The form is positive semidefinite for \(t\geq1\), and positive definite for \(t>1\).
At the boundary \(t=1\),
This expression is nonnegative for every \(\mathbf{x}\), but it is zero for nonzero vectors with \(x_1+x_2=0\). In particular, the eigenvector \((1,-1)^{\top}\) has shifted eigenvalue \(0\).
Check: Multiplying the boundary matrix by that eigenvector gives \(\begin{bmatrix}2&2\\2&2\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). Thus the zero eigenvalue and the vanishing quadratic-form value agree, confirming positive semidefiniteness without positive definiteness.
A Reliable Threshold Procedure
For a symmetric matrix, the shift calculation can be organized into a short sequence. The essential data are the least and greatest eigenvalues; if the eigenvalues are already known, there is no need to diagonalize the shifted matrix again.
Find \(\lambda_{\min}\) and \(\lambda_{\max}\) of the original real symmetric matrix \(\mathbf{A}\).
The negative-definiteness threshold is \(-\lambda_{\max}\); the positive-definiteness threshold is \(-\lambda_{\min}\).
Strict inequalities give definiteness. Equality gives the corresponding semidefinite boundary, because an eigenvalue is then zero.
The shift is indefinite between the two thresholds only when \(\lambda_{\min}<\lambda_{\max}\). If they are equal, there is no indefinite interval.
A common mistake is to reverse the sign of the shift when updating eigenvalues. For instance, for \(\mathbf{A}=\operatorname{diag}(2,5)\) and \(t=-3\), an incorrect calculation might subtract \(t\) and claim shifted eigenvalues \(5,8\). But the shifted matrix is \(\operatorname{diag}(2-3,5-3)=\operatorname{diag}(-1,2)\), so the correct eigenvalues are \(2+t=-1\) and \(5+t=2\). The shift adds \(t\), including when \(t\) is negative.
Another error is to call a boundary positive definite just because none of the shifted eigenvalues is negative. At \(t=-\lambda_{\min}\), the least shifted eigenvalue is exactly zero. Its eigenvector is nonzero and makes the quadratic form equal to zero, so positive definiteness fails even though positive semidefiniteness holds. The same reasoning, with signs reversed, applies at \(t=-\lambda_{\max}\).
| Condition on \(t\) | Classification | Boundary interpretation |
|---|---|---|
| \(t<-\lambda_{\max}\) | Negative definite | Every shifted eigenvalue is negative |
| \(t=-\lambda_{\max}\) | Negative semidefinite | The greatest shifted eigenvalue is zero |
| \(-\lambda_{\max}<t<-\lambda_{\min}\) | Indefinite | Requires distinct extreme eigenvalues |
| \(t=-\lambda_{\min}\) | Positive semidefinite | The least shifted eigenvalue is zero |
| \(t>-\lambda_{\min}\) | Positive definite | Every shifted eigenvalue is positive |
Why the Thresholds Are Useful
The scalar-shift rule turns a family of quadratic forms into a one-dimensional threshold problem. Once the extreme eigenvalues of \(\mathbf{A}\) are known, the entire definiteness classification of \(\mathbf{A}+t\mathbf{I}_n\) follows from two numbers. The same eigenvectors identify the directions that reach zero at the boundaries, so the thresholds also say where strict definiteness is lost or gained.
Keep the sign convention visible: the matrix is shifted by \(+t\mathbf{I}_n\), so its eigenvalues are shifted by \(+t\). To make the form positive definite, \(t\) must exceed the negative of the least eigenvalue. To make it negative definite, \(t\) must be less than the negative of the greatest eigenvalue. Equality is the semidefinite boundary, not a strict-definiteness case.
Check Your Understanding
For each question, use the shifted eigenvalues and distinguish strict definiteness from semidefiniteness at equality.
- A symmetric matrix has eigenvalues \(-5,0,2\). For which values of \(t\) is \(\mathbf{A}+t\mathbf{I}\) positive definite?
- For the same matrix, determine the negative-semidefinite range and the indefinite range.
- If every eigenvalue of a symmetric matrix equals \(4\), can any scalar shift make its quadratic form indefinite? Explain using the shifted eigenvalues.
- For a symmetric matrix with least eigenvalue \(-2\), what happens at \(t=2\), and why is the form not positive definite there?
- Suppose \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) with \(\mathbf{v}\ne\mathbf{0}\). Verify directly that \(\mathbf{v}\) is an eigenvector of \(\mathbf{A}+t\mathbf{I}\), and state its new eigenvalue.
