Similarity Transformation of the Inertia Tensor
SimilarityTransformationOfTheInertiaTensor
2026-09-19 18:37:10
2026-09-19 19:03:51
Definition
new test if preview is fixed on embedded videos
inertia tensor
similarity transformation
direction cosine matrix
coordinate transformation
principal axes
principal moments
eigenvalues
products of inertia
rigid-body dynamics
kinetic energy
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\section*{Rigid-Body Mechanics: Similarity Transformation of the Inertia Tensor}
The inertia tensor is a physical property of a rigid body about a specified point, but the \emph{matrix used to represent that tensor depends on the coordinate frame}. When the coordinate axes are rotated, the physical mass distribution has not changed; only its matrix representation has changed. The correct coordinate transformation is an orthogonal similarity transformation.
This article derives that result carefully and connects it to angular momentum, rotational kinetic energy, the mass-integral definition of inertia, and principal-axis diagonalization. The key result is
\begin{equation}
\boxed{
\mathbf I^{B}
=\mathbf C_{BA}\,\mathbf I^{A}\,\mathbf C_{AB}
=\mathbf C_{BA}\,\mathbf I^{A}\,\mathbf C_{BA}^{T}
}
\label{eq:main_transform}
\end{equation}
when $\mathbf C_{BA}$ maps vector components from frame $A$ into frame $B$.
\section{Video companion at the start of the derivation}
\PMyoutube{https://youtu.be/6oGjAlrHjtE}{Companion video for the inertia-tensor similarity-transformation derivation.}
The video is placed here intentionally, immediately before the mathematical derivation, so that the PhysicsLibrary video-embedding macro can be tested in the same location where a learner would naturally use it.
\section{What the inertia tensor represents}
Consider a rigid body and a reference point $O$. Let $\mathbf r$ denote the position of a mass element $dm$ relative to $O$. In an orthonormal Cartesian frame, the inertia tensor is represented by the symmetric matrix
\begin{equation}
\boxed{
\mathbf I_O
=\int
\left[
(\mathbf r^{T}\mathbf r)\mathbf 1_3
-\mathbf r\mathbf r^{T}
\right]dm
}
\label{eq:inertia_integral}
\end{equation}
where $\mathbf 1_3$ is the $3\times3$ identity matrix \cite{Goldstein2002,Kane1985,MooreLMD}.
Written component-by-component,
\begin{equation}
\mathbf I
=
\begin{bmatrix}
I_{xx} & I_{xy} & I_{xz}\\
I_{xy} & I_{yy} & I_{yz}\\
I_{xz} & I_{yz} & I_{zz}
\end{bmatrix}.
\label{eq:inertia_matrix}
\end{equation}
Depending on the convention used for products of inertia, some engineering texts write the off-diagonal entries as $-I_{xy}$, $-I_{xz}$, and $-I_{yz}$. The transformation law derived below is unchanged as long as the matrix convention is used consistently.
The tensor appears directly in three important physical relations. Angular momentum about $O$ is
\begin{equation}
\boxed{\mathbf H_O=\mathbf I_O\boldsymbol\omega,}
\label{eq:H_Iw}
\end{equation}
rotational kinetic energy is
\begin{equation}
\boxed{T=\frac{1}{2}\boldsymbol\omega^{T}\mathbf I_O\boldsymbol\omega,}
\label{eq:rot_ke}
\end{equation}
and the scalar moment of inertia about a unit axis $\widehat{\mathbf n}$ through $O$ is
\begin{equation}
\boxed{I_{\widehat n}=\widehat{\mathbf n}^{T}\mathbf I_O\widehat{\mathbf n}.}
\label{eq:axis_moi}
\end{equation}
These relations describe physical quantities and therefore cannot depend on an arbitrary choice of coordinate axes.
\section{Coordinate-frame convention}
Let frames $A$ and $B$ be orthonormal frames with the same origin $O$. Define the direction cosine matrix $\mathbf C_{BA}$ by
\begin{equation}
\boxed{
\mathbf v^{B}=\mathbf C_{BA}\mathbf v^{A}
}
\label{eq:dcm_convention}
\end{equation}
for every geometric vector $\mathbf v$.
Because $\mathbf C_{BA}$ represents a proper orthogonal rotation,
\begin{equation}
\mathbf C_{BA}^{T}\mathbf C_{BA}=\mathbf 1_3,
\qquad
\det\mathbf C_{BA}=+1,
\label{eq:orthogonality}
\end{equation}
and therefore
\begin{equation}
\boxed{
\mathbf C_{AB}=\mathbf C_{BA}^{-1}=\mathbf C_{BA}^{T}.
}
\label{eq:inverse_dcm}
\end{equation}
\begin{center}
\includegraphics{InertiaSimilarity_fig01_rotated_frames.png}
\vspace{0.45em}
\textbf{Figure.} The same rigid body and the same physical inertia tensor are described using two rotated coordinate frames. Only the matrix components change.
\end{center}
\section{Derivation from angular momentum}
This is the shortest physical derivation of the similarity transformation.
In frame $A$,
\begin{equation}
\mathbf H^{A}=\mathbf I^{A}\boldsymbol\omega^{A}.
\label{eq:H_A}
\end{equation}
The geometric vectors $\mathbf H$ and $\boldsymbol\omega$ transform according to the ordinary vector rule,
\begin{equation}
\mathbf H^{B}=\mathbf C_{BA}\mathbf H^{A},
\label{eq:H_transform}
\end{equation}
\begin{equation}
\boldsymbol\omega^{B}=\mathbf C_{BA}\boldsymbol\omega^{A}.
\label{eq:omega_transform}
\end{equation}
From Equation \eqref{eq:inverse_dcm},
\begin{equation}
\boldsymbol\omega^{A}
=\mathbf C_{AB}\boldsymbol\omega^{B}.
\label{eq:omega_inverse}
\end{equation}
Substitute Equation \eqref{eq:H_A} into Equation \eqref{eq:H_transform}:
\begin{equation}
\mathbf H^{B}
=\mathbf C_{BA}\mathbf I^{A}\boldsymbol\omega^{A}.
\label{eq:step1}
\end{equation}
Now substitute Equation \eqref{eq:omega_inverse}:
\begin{equation}
\mathbf H^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{AB}\boldsymbol\omega^{B}.
\label{eq:step2}
\end{equation}
But in frame $B$ the same physical angular-momentum law must be
\begin{equation}
\mathbf H^{B}=\mathbf I^{B}\boldsymbol\omega^{B}.
\label{eq:H_B}
\end{equation}
Since Equations \eqref{eq:step2} and \eqref{eq:H_B} must agree for arbitrary $\boldsymbol\omega^{B}$,
\begin{equation}
\boxed{
\mathbf I^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{AB}.
}
\label{eq:similarity_general}
\end{equation}
Finally, orthogonality gives
\begin{equation}
\boxed{
\mathbf I^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}.
}
\label{eq:similarity_orthogonal}
\end{equation}
The matrix has to appear on \emph{both sides} of $\mathbf I^{A}$. A second-order tensor has two coordinate indices, so both indices must be transformed.
\section{Why this is called a similarity transformation}
In linear algebra, a similarity transformation has the form
\begin{equation}
\mathbf A' = \mathbf S\mathbf A\mathbf S^{-1}.
\label{eq:generic_similarity}
\end{equation}
Equation \eqref{eq:similarity_general} has exactly this form with
\begin{equation}
\mathbf S=\mathbf C_{BA}.
\end{equation}
For an orthogonal rotation,
\begin{equation}
\mathbf S^{-1}=\mathbf S^{T},
\end{equation}
so the transformation becomes
\begin{equation}
\mathbf I^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}.
\label{eq:similarity_congruence}
\end{equation}
This expression is also called an \emph{orthogonal congruence transformation}. For a rotation matrix, the distinction disappears algebraically because inverse and transpose are identical. Conceptually, however, the similarity viewpoint is especially useful because it immediately explains why the eigenvalues are preserved \cite{Strang2006}.
\section{Independent derivation from the mass integral}
The transformation law can also be derived directly from the definition of inertia.
Let
\begin{equation}
\mathbf r^{B}=\mathbf C_{BA}\mathbf r^{A}.
\label{eq:r_transform}
\end{equation}
The Euclidean norm is unchanged by rotation:
\begin{align}
(\mathbf r^{B})^{T}\mathbf r^{B}
&=
(\mathbf r^{A})^{T}
\mathbf C_{BA}^{T}\mathbf C_{BA}
\mathbf r^{A}\\
&=
(\mathbf r^{A})^{T}\mathbf r^{A}.
\label{eq:norm_invariant}
\end{align}
The outer product transforms as
\begin{align}
\mathbf r^{B}(\mathbf r^{B})^{T}
&=
\mathbf C_{BA}\mathbf r^{A}
(\mathbf r^{A})^{T}\mathbf C_{BA}^{T}.
\label{eq:outer_transform}
\end{align}
Insert these expressions into the inertia integral:
\begin{align}
\mathbf I^{B}
&=\int
\left[
(\mathbf r^{B})^{T}\mathbf r^{B}\,\mathbf 1_3
-\mathbf r^{B}(\mathbf r^{B})^{T}
\right]dm\\
&=\int
\left[
(\mathbf r^{A})^{T}\mathbf r^{A}\,\mathbf 1_3
-\mathbf C_{BA}\mathbf r^{A}(\mathbf r^{A})^{T}\mathbf C_{BA}^{T}
\right]dm.
\label{eq:mass_step}
\end{align}
Because
\begin{equation}
\mathbf 1_3
=\mathbf C_{BA}\mathbf 1_3\mathbf C_{BA}^{T},
\end{equation}
we may factor the rotation matrices outside the integral:
\begin{align}
\mathbf I^{B}
&=
\mathbf C_{BA}
\left\{
\int
\left[
(\mathbf r^{A})^{T}\mathbf r^{A}\,\mathbf 1_3
-\mathbf r^{A}(\mathbf r^{A})^{T}
\right]dm
\right\}
\mathbf C_{BA}^{T}\\
&=\boxed{\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}}.
\label{eq:mass_final}
\end{align}
Thus the same transformation follows directly from the mass distribution itself.
\section{Rotational kinetic energy is invariant}
The same physical rigid body must have the same rotational kinetic energy no matter which frame is used to write the components.
Starting in frame $B$,
\begin{equation}
T
=\frac{1}{2}(\boldsymbol\omega^{B})^{T}
\mathbf I^{B}
\boldsymbol\omega^{B}.
\end{equation}
Substitute
\begin{equation}
\boldsymbol\omega^{B}=\mathbf C_{BA}\boldsymbol\omega^{A}
\end{equation}
and Equation \eqref{eq:similarity_orthogonal}:
\begin{align}
T
&=\frac{1}{2}(\boldsymbol\omega^{A})^{T}
\mathbf C_{BA}^{T}
\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}
\mathbf C_{BA}
\boldsymbol\omega^{A}\\
&=\frac{1}{2}(\boldsymbol\omega^{A})^{T}
\mathbf I^{A}
\boldsymbol\omega^{A}.
\end{align}
Hence
\begin{equation}
\boxed{T^{A}=T^{B}.}
\end{equation}
The matrix entries may change, but the physical scalar does not.
\section{Principal axes are the eigenvectors of the inertia tensor}
Because the inertia tensor is real and symmetric, it possesses three mutually orthogonal eigenvectors and real eigenvalues. Let
\begin{equation}
\mathbf I^{A}\mathbf V
=\mathbf V\boldsymbol\Lambda,
\label{eq:eigendecomp_step}
\end{equation}
where the columns of $\mathbf V$ are orthonormal eigenvectors expressed in frame $A$, and
\begin{equation}
\boldsymbol\Lambda
=
\begin{bmatrix}
I_1&0&0\\
0&I_2&0\\
0&0&I_3
\end{bmatrix}
\label{eq:principal_matrix}
\end{equation}
contains the principal moments of inertia.
Since $\mathbf V$ is orthogonal,
\begin{equation}
\mathbf V^{-1}=\mathbf V^{T}.
\end{equation}
Multiplying Equation \eqref{eq:eigendecomp_step} on the left by $\mathbf V^{T}$ gives
\begin{equation}
\boxed{
\mathbf V^{T}\mathbf I^{A}\mathbf V
=\boldsymbol\Lambda.
}
\label{eq:diagonalization}
\end{equation}
This is precisely the inertia-tensor similarity transformation. If frame $P$ is chosen to have its axes along the principal axes, then
\begin{equation}
\mathbf C_{PA}=\mathbf V^{T},
\end{equation}
and
\begin{equation}
\boxed{
\mathbf I^{P}
=\mathbf C_{PA}\mathbf I^{A}\mathbf C_{AP}
=\boldsymbol\Lambda.
}
\label{eq:principal_transform}
\end{equation}
The products of inertia vanish in the principal-axis frame.
\begin{center}
\includegraphics{InertiaSimilarity_fig02_principal_axes.png}
\vspace{0.45em}
\textbf{Figure.} Diagonalization of the inertia tensor is an orthogonal similarity transformation. The eigenvectors define the principal frame and the eigenvalues are the principal moments.
\end{center}
\section{What the similarity transformation preserves}
Similar matrices have the same characteristic polynomial. Therefore rotating the coordinate frame preserves
\begin{itemize}
\item the three eigenvalues, which are the principal moments of inertia;
\item the trace;
\item the determinant;
\item the characteristic polynomial;
\item positive-definite or positive-semidefinite character;
\item rotational kinetic energy when the angular-velocity components are transformed consistently.
\end{itemize}
For example,
\begin{equation}
\operatorname{tr}(\mathbf I^{B})
=\operatorname{tr}(\mathbf C_{BA}\mathbf I^{A}\mathbf C_{AB}).
\end{equation}
Using the cyclic property of the trace,
\begin{align}
\operatorname{tr}(\mathbf I^{B})
&=\operatorname{tr}(\mathbf I^{A}\mathbf C_{AB}\mathbf C_{BA})\\
&=\operatorname{tr}(\mathbf I^{A}).
\end{align}
Likewise,
\begin{align}
\det(\mathbf I^{B})
&=\det(\mathbf C_{BA})
\det(\mathbf I^{A})
\det(\mathbf C_{AB})\\
&=\det(\mathbf I^{A}).
\end{align}
These invariants are useful software checks after transforming an inertia matrix.
\section{Worked numerical example: rotate a principal inertia tensor}
Suppose the principal-axis inertia matrix is
\begin{equation}
\mathbf I^{P}
=
\begin{bmatrix}
2&0&0\\
0&5&0\\
0&0&8
\end{bmatrix}
\ \text{kg m}^{2}.
\label{eq:numeric_principal}
\end{equation}
Let frame $A$ be rotated relative to the principal frame by $\theta=30^{\circ}$ about the common $z$ axis. Let
\begin{equation}
\mathbf C_{AP}
=
\begin{bmatrix}
\cos\theta&-\sin\theta&0\\
\sin\theta& \cos\theta&0\\
0&0&1
\end{bmatrix}.
\label{eq:CAP}
\end{equation}
The inertia matrix expressed in frame $A$ is
\begin{equation}
\mathbf I^{A}
=\mathbf C_{AP}\mathbf I^{P}\mathbf C_{PA}.
\label{eq:numeric_transform}
\end{equation}
Writing $c=\cos\theta$ and $s=\sin\theta$,
\begin{equation}
\mathbf I^{A}
=
\begin{bmatrix}
2c^2+5s^2 & (2-5)cs & 0\\
(2-5)cs & 2s^2+5c^2 & 0\\
0&0&8
\end{bmatrix}.
\label{eq:numeric_symbolic}
\end{equation}
At $\theta=30^{\circ}$,
\begin{equation}
c=\frac{\sqrt{3}}{2},
\qquad
s=\frac{1}{2},
\end{equation}
so
\begin{equation}
\boxed{
\mathbf I^{A}
=
\begin{bmatrix}
2.7500&-1.2990&0\\
-1.2990&4.2500&0\\
0&0&8.0000
\end{bmatrix}
\ \text{kg m}^{2}.
}
\label{eq:numeric_result}
\end{equation}
Nothing physical happened to the body. The off-diagonal entries appeared only because frame $A$ is not aligned with the principal axes.
To recover the principal frame,
\begin{equation}
\mathbf I^{P}
=\mathbf C_{PA}\mathbf I^{A}\mathbf C_{AP},
\end{equation}
which returns Equation \eqref{eq:numeric_principal}.
The invariants provide immediate checks:
\begin{equation}
\operatorname{tr}(\mathbf I^{P})=2+5+8=15,
\end{equation}
while
\begin{equation}
\operatorname{tr}(\mathbf I^{A})=2.75+4.25+8=15.
\end{equation}
The eigenvalues of $\mathbf I^{A}$ are likewise $2$, $5$, and $8\ \text{kg m}^{2}$.
\section{Moment of inertia about a physical axis is also invariant}
Let the physical axis be the $x_A$ direction,
\begin{equation}
\widehat{\mathbf n}^{A}
=
\begin{bmatrix}1\\0\\0\end{bmatrix}.
\end{equation}
From Equation \eqref{eq:numeric_result},
\begin{equation}
I_{\widehat n}
=(\widehat{\mathbf n}^{A})^{T}
\mathbf I^{A}
\widehat{\mathbf n}^{A}
=2.75\ \text{kg m}^{2}.
\end{equation}
In the principal frame,
\begin{equation}
\widehat{\mathbf n}^{P}
=\mathbf C_{PA}\widehat{\mathbf n}^{A}
=
\begin{bmatrix}
\cos\theta\\
-\sin\theta\\
0
\end{bmatrix}.
\end{equation}
Then
\begin{align}
I_{\widehat n}
&=(\widehat{\mathbf n}^{P})^{T}
\mathbf I^{P}
\widehat{\mathbf n}^{P}\\
&=2\cos^2\theta+5\sin^2\theta\\
&=2.75\ \text{kg m}^{2}.
\end{align}
The components changed, but the physical moment of inertia about the chosen physical axis did not.
\section{Similarity transformation versus the parallel-axis theorem}
These two operations are often confused, but they solve different problems.
A similarity transformation changes the \emph{orientation of the coordinate basis} while keeping the same reference point and the same physical inertia tensor:
\begin{equation}
\boxed{
\mathbf I^{B}=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}.
}
\end{equation}
The parallel-axis theorem changes the \emph{reference point} about which the inertia is computed. If $G$ is the center of mass and $O$ is displaced from $G$ by $\mathbf d$, then
\begin{equation}
\boxed{
\mathbf I_O
=\mathbf I_G
+m\left[
(\mathbf d^{T}\mathbf d)\mathbf 1_3
-\mathbf d\mathbf d^{T}
\right].
}
\label{eq:parallel_axis}
\end{equation}
Equation \eqref{eq:parallel_axis} changes the tensor itself because the moment reference point has changed. Equation \eqref{eq:similarity_orthogonal} changes only its coordinate representation.
If both the point and frame change, apply both operations, keeping the geometry and frame convention explicit.
\section{Passive frame change versus active body rotation}
There are two geometrically different situations that can produce the same matrix pattern.
In a \emph{passive} frame change, the body remains fixed and the coordinate axes are changed. With the convention of Equation \eqref{eq:dcm_convention},
\begin{equation}
\mathbf I^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}.
\end{equation}
In an \emph{active} rotation, the coordinate frame is held fixed while the physical body is rotated. If $\mathbf Q$ maps the old physical directions into the new physical directions in that fixed frame, then the newly oriented tensor is
\begin{equation}
\mathbf I_{\text{new}}
=\mathbf Q\mathbf I_{\text{old}}\mathbf Q^{T}.
\end{equation}
The algebra looks the same, but the interpretation is different. Confusing active and passive meanings is a common source of transposes and sign errors.
\section{Connection with covariance transformations}
The same two-sided transformation appears in estimation theory. If a random vector transforms as
\begin{equation}
\delta\mathbf x^{B}
=\mathbf C_{BA}\delta\mathbf x^{A},
\end{equation}
then its covariance transforms as
\begin{equation}
\boxed{
\mathbf P^{B}
=\mathbf C_{BA}\mathbf P^{A}\mathbf C_{BA}^{T}.
}
\end{equation}
The reason is structurally identical: a covariance matrix, like the inertia tensor matrix, represents a second-order object with two coordinate indices. This analogy is especially useful in navigation, estimation, and rigid-body dynamics.
\section{Common mistakes}
\begin{itemize}
\item Transforming inertia like a vector, for example using only $\mathbf C\mathbf I$. A second-order tensor requires a two-sided transformation.
\item Using $\mathbf C^{T}\mathbf I\mathbf C$ without first stating what direction $\mathbf C$ maps. Both orders occur in textbooks because DCM conventions differ.
\item Treating the off-diagonal products of inertia as intrinsic properties that must have the same numerical values in every frame. They are coordinate-dependent components.
\item Assuming diagonal inertia means a body is geometrically symmetric. Every real symmetric inertia tensor has orthogonal principal axes even for an asymmetric body.
\item Confusing a coordinate rotation with the parallel-axis theorem.
\item Forgetting to rotate vectors such as $\boldsymbol\omega$, $\mathbf H$, or the axis vector $\widehat{\mathbf n}$ consistently when checking physical invariance.
\item Calling every expression $\mathbf C\mathbf I\mathbf C^{T}$ a similarity transformation without noting that the inverse equals the transpose only for an orthogonal matrix.
\end{itemize}
\section{Compact derivation chain}
The essential logic can be summarized in four lines:
\begin{align}
\mathbf H^{A}&=\mathbf I^{A}\boldsymbol\omega^{A},\\
\mathbf H^{B}&=\mathbf C_{BA}\mathbf H^{A},\\
\boldsymbol\omega^{A}&=\mathbf C_{AB}\boldsymbol\omega^{B},\\
\therefore\qquad
\boxed{\mathbf I^{B}}
&=\boxed{\mathbf C_{BA}\mathbf I^{A}\mathbf C_{AB}}.
\end{align}
For orthonormal Cartesian frames,
\begin{equation}
\boxed{
\mathbf I^{B}
=\mathbf C_{BA}\mathbf I^{A}\mathbf C_{BA}^{T}.
}
\end{equation}
Choosing $B$ to be the eigenvector, or principal-axis, frame gives
\begin{equation}
\boxed{
\mathbf I^{B}
=\operatorname{diag}(I_1,I_2,I_3).
}
\end{equation}
Thus principal-axis diagonalization is not a separate trick. It is the similarity transformation applied with the special rotation that aligns the coordinate axes with the eigenvectors of the inertia tensor.
\begin{thebibliography}{9}
\bibitem{Goldstein2002}
Herbert Goldstein, Charles Poole, and John Safko,
\emph{Classical Mechanics},
3rd ed., Addison Wesley, 2002.
\bibitem{Kane1985}
Thomas R. Kane and David A. Levinson,
\emph{Dynamics: Theory and Applications},
McGraw-Hill, 1985.
\bibitem{MooreLMD}
Jason K. Moore,
\emph{Learn Multibody Dynamics},
chapter on mass distribution, inertia dyadics, and principal axes.
\bibitem{Strang2006}
Gilbert Strang,
\emph{Linear Algebra and Its Applications},
4th ed., Brooks/Cole, 2006.
\end{thebibliography}