Math - Trace, Determinant, Frobenius Norm

Determinant

• \[\det(AB) = \det(A)\,\det(B)\]
• For an orthogonal matrix \(Q\) (i.e., \(Q^\top Q = I\)), we have \(\det(Q) = \pm 1\):

In particular, if \(Q \in SO(n)\) (special orthogonal group), then \(\det Q = 1\).

Trace Properties

The trace of a square matrix is just the sum of its diagonal entries: \(\operatorname{tr}(A) = \sum_i A_{ii}\). It looks simple, but a handful of its properties show up all over robotics, so let’s prove them one by one.

Linearity and Transpose

These all follow directly from the definition, because the trace only ever looks at the diagonal:

Cyclic Shifting

\(\operatorname{tr}(AB) = \operatorname{tr}(BA)\). This holds even when \(AB\) and \(BA\) have different sizes, say \(A\) is \(m \times n\) and \(B\) is \(n \times m\). To see why, write out the diagonal of \(AB\):

The two double sums contain exactly the same terms, just added in a different order, so they are equal. Applying this repeatedly, we can rotate a longer product around:

Note that only cyclic rotations are allowed. In general, \(\operatorname{tr}(ABC) \ne \operatorname{tr}(ACB)\).

Invariance Under Change of Basis

For any invertible \(P\), cyclic shifting gives:

So the trace doesn’t care which coordinate system we use. That’s what allows it to equal something basis-independent, like the sum of eigenvalues below.

Trace Equals the Sum of Eigenvalues

Quick proof for a diagonalizable matrix (this includes every symmetric matrix). Write \(A = V \Lambda V^{-1}\). Then by the change-of-basis invariance above,

General proof, using the characteristic polynomial. The eigenvalues are its roots, so we can factor it as

Now let’s expand the same determinant directly. A determinant is a sum of products, where each product picks one entry from every row and every column. Only the product of the diagonal entries, \(\prod_i (\lambda - A_{ii})\), can reach \(\lambda^{n-1}\). Any other product has to skip at least two diagonal entries, so it has at most \(n-2\) factors of \(\lambda\). Expanding the diagonal product:

Matching the \(\lambda^{n-1}\) coefficients of the two expansions gives \(\sum_i A_{ii} = \sum_i \lambda_i\). This works for every square matrix, as long as we count eigenvalues with multiplicity (and allow them to be complex). As a bonus, comparing the constant terms the same way gives \(\det(A) = \prod_i \lambda_i\).

Trace of an Outer Product

The diagonal entries of \(\mathbf{x}\mathbf{y}^\top\) are \(x_i y_i\), and summing them gives the dot product. Another way to see it: by cyclic shifting, \(\operatorname{tr}(\mathbf{x}\mathbf{y}^\top) = \operatorname{tr}(\mathbf{y}^\top\mathbf{x})\), and \(\mathbf{y}^\top\mathbf{x}\) is a \(1 \times 1\) matrix, which is its own trace.

Trace of a Squared Skew-Symmetric Matrix

Let \(\mathbf{v} \in \mathbb{R}^3\), and let \([\mathbf{v}]_\times\) (also written \(\mathbf{v}^\wedge\)) be its skew-symmetric matrix, so that \([\mathbf{v}]_\times \mathbf{w} = \mathbf{v} \times \mathbf{w}\). Then

First, we find what \([\mathbf{v}]_\times^\top [\mathbf{v}]_\times\) actually is. Since \([\mathbf{v}]_\times^\top = -[\mathbf{v}]_\times\), applying it to any vector \(\mathbf{w}\) and using the triple product expansion \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})\) gives

Then, taking the trace with linearity and the outer-product result above:

Since \([\mathbf{v}]_\times^2 = -[\mathbf{v}]_\times^\top[\mathbf{v}]_\times\), the second form just flips the sign. This identity shows up in the quaternion-to-angle derivation and in the rotation block of an ICP Hessian.

Trace of a Rotation Matrix

A rotation by angle \(\theta\) about a unit axis \(\mathbf{a}\) has trace

To see this, start from Rodrigues’ formula, \(R = \cos\theta\, I + (1 - \cos\theta)\,\mathbf{a}\mathbf{a}^\top + \sin\theta\,[\mathbf{a}]_\times\), and take the trace term by term. We have \(\operatorname{tr}(I) = 3\) and \(\operatorname{tr}(\mathbf{a}\mathbf{a}^\top) = \|\mathbf{a}\|^2 = 1\). The last term vanishes, because a skew-symmetric matrix has zeros on its diagonal, so \(\operatorname{tr}([\mathbf{a}]_\times) = 0\). Putting it together:

This is exactly how we recover the rotation angle from a rotation matrix.

Frobenius Norm as a Trace

The \(i\)-th diagonal entry of \(A^\top A\) is the squared length of the \(i\)-th column of \(A\). Summing over all columns adds up every squared entry of \(A\):

Von-Neumann’s Trace Inequality

In 1937, Von-Neumann proved that if \(A, B\) are complex \(n \times n\) matrices with singular values

then

The bound is reached, for example, when \(A\) and \(B\) are both diagonal with their singular values sorted the same way:

Singular Values of a Rotation Matrix Are All 1

Let \(R \in SO(3)\), i.e., \(R^\top R = I\) and \(\det R = 1\). The singular values \(\{\sigma_i\}_{i=1}^3\) of \(R\) are the square roots of the eigenvalues of \(R^\top R\):

Since \(R^\top R = I\), all eigenvalues of \(R^\top R\) are \(1\). Therefore,

Equivalently, in the SVD \(R = U\Sigma V^\top\) with orthogonal \(U\) and \(V\), we must have \(\Sigma = I\), so all singular values are \(1\).

Frobenius Norm

The Frobenius norm is the square root of the sum of all squared entries: \(\|A\|_F = \sqrt{\sum_i \sum_j a_{ij}^2}\).

E.g., a common task in lidar is: we have an estimate \(R\) of a rotation matrix that has drifted away from \(SO(3)\), and we want the closest true rotation \(X \in SO(3)\), i.e. the one with the lowest \(\|X - R\|_F\). Here’s how to find it.

Step 1: turn the norm into a trace. Using \(\|A\|_F^2 = \operatorname{tr}(A^\top A)\) and linearity,

where the fourth line uses \(\operatorname{tr}(R^\top X) = \operatorname{tr}((X^\top R)^\top) = \operatorname{tr}(X^\top R)\).

Step 2: minimizing the norm means maximizing the trace. \(\|R\|_F^2\) is fixed, and \(\|X\|_F^2 = \operatorname{tr}(X^\top X) = \operatorname{tr}(I) = 3\) for any rotation \(X\). So both are constants, and

Step 3: simplify with the SVD. Take the SVD \(R = U\,\Sigma\,V^\top\), with \(U, V \in \mathrm{O}(3)\) and \(\Sigma = \operatorname{diag}(\sigma_1,\sigma_2,\sigma_3)\), \(\sigma_1 \ge \sigma_2 \ge \sigma_3 \ge 0\). Then define an intermediate variable

\(Y\) is a product of orthogonal matrices, so it is orthogonal too (\(Y \in \mathrm{O}(3)\)). Substituting, and then using cyclic shifting to move \(V^\top\) to the front:

Step 4: watch the determinant. \(X\) must be a proper rotation, \(\det(X) = 1\), which pins down the determinant of \(Y\):

Step 5: pick the best \(Y\). By Von-Neumann’s trace inequality, \(\operatorname{tr}(Y^\top \Sigma)\) is largest when \(Y\) is diagonal, since \(\Sigma\) already is. A diagonal orthogonal matrix has \(\pm 1\) on its diagonal, so we maximize over those, subject to \(\det(Y) = \det(UV^\top)\):

• If \(\det(UV^\top) = 1\), the maximizer is \(Y = I\), giving \(\operatorname{tr}(Y^\top \Sigma) = \sigma_1 + \sigma_2 + \sigma_3\).
• If \(\det(UV^\top) = -1\), we need an odd number of \(-1\)’s, and the cheapest place to put one is on the smallest singular value: \(Y = \operatorname{diag}(1,1,-1)\), giving \(\operatorname{tr}(Y^\top \Sigma) = \sigma_1 + \sigma_2 - \sigma_3\).

Step 6: map back. Since \(X = U Y V^\top\), the closest rotation is

In code:

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Eigen::Matrix3d nearestRotation(const Eigen::Matrix3d& R) {
    Eigen::JacobiSVD<Eigen::Matrix3d> svd(R, Eigen::ComputeFullU | Eigen::ComputeFullV);
    Eigen::Matrix3d U = svd.matrixU();
    Eigen::Matrix3d Vt = svd.matrixV().transpose();
    Eigen::Matrix3d R_ortho = U * Vt;
    if (R_ortho.determinant() < 0.0) {
        Eigen::Matrix3d S = Eigen::Matrix3d::Identity();
        S(2,2) = -1.0;
        R_ortho = U * S * Vt;
    }
    return R_ortho;
}

Trace is the Sum of Eigenvalues for Symmetric Matrices

By definition, \(\operatorname{tr}(H) = \sum_i H_{ii}\). For a diagonal matrix, the trace is obviously the sum of the diagonal:

Now if \(H\) is symmetric, it has an eigendecomposition \(H = V\Lambda V^T\), where \(V\) is orthonormal and

Using the cyclic property of trace, \(\operatorname{tr}(ABC) = \operatorname{tr}(BCA)\), we can move \(V\) around. Because \(V\) is orthonormal, \(V^TV = I\), so

And therefore

So the trace is invariant under this change of basis.

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