Eigenvalues and eigenvectors for the moment of inertia matrix

Question

Find the eigenvalues and eigenvectors for the moment of inertia matrix given by $$I={m\over 2}\left(\begin{matrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{matrix}\right)$$

I'm not quite sure whether or not I should include $m\over 2$ when determining the eigenvalues and eigenvectors or if I should let $A=\left(\begin{matrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{matrix}\right)$ and solve for its eigenvalues and eigenvectors.

I determined the eigenvalues to be $\ \lambda=0 \ $ and$\ \lambda=m$. Solving for the eigenvector given $\ \lambda = 0$: $$I-\lambda={m\over 2}\left(\begin{matrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{matrix}\right)$$ and by Gauss-Jordan elimination: $$I-\lambda={m\over 2}\left(\begin{matrix} 0 & 0 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{matrix}\right)$$ The corresponding eigenvector for $\ \lambda = 0 \ $ was determined: $$\left(\begin{matrix} x\\ y\\ z\end{matrix}\right)_{\lambda = 0}=x\left(\begin{matrix} 1\\ 1\\0\end{matrix}\right)$$ Similarly for $\ \lambda = m$: $$I-\lambda={m\over 2}\left(\begin{matrix} -1 & -1 & 0\\ -1 & -1 & 0\\ 0 & 0 & 0\end{matrix}\right)$$ and by Gauss-Jordan elimination: $$I-\lambda={m\over 2}\left(\begin{matrix} 0 & 0 & 0\\ -1 & -1 & 0\\ 0 & 0 & 0\end{matrix}\right)$$ The corresponding eigenvector for $\ \lambda = m \ $ was determined: $$\left(\begin{matrix} x\\ y\\ z\end{matrix}\right)_{\lambda = m}=x\left(\begin{matrix} 1\\ -1\\0\end{matrix}\right)+z\left(\begin{matrix} 0\\ 0\\ 1\end{matrix}\right)$$ Please correct me if I'm wrong!

Answer 1

The eigenvalues of the matrix $$\begin{bmatrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}$$ are not the same as the eigenvalues of the matrix

$$\frac{m}{2}\begin{bmatrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}.$$

You most certainly cannot just ignore the factor $\frac{m}{2}$.

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However, you might notice that for any matrix $A$ and nonzero scalar $\alpha$, the matrix $A$ and $\alpha A$ have the same eigenvectors, and from that, you can also show that $\lambda$ is an eigenvalue of $A$ if and only if $\alpha\lambda$ is an eigenvalue of $\alpha A$.

This means that when looking for eigenvectors and eigenvalues of $\alpha A$, you can calculate them for $A$, and then keep the vectors and multiply the eigenvalues by $\alpha$ to get your final result.

Answer 2

To answer your main question, yes, you do have to include the $m/2$. It affects the eigenvalues, but not the corresponding eigenvectors: If you have $A=cB$ and $B\mathbf v=\lambda\mathbf v$, then $A\mathbf v=cB\mathbf v=c\lambda\mathbf v$, so $\mathbf v$ is an eigenvector of $A$ with eigenvalue $c\lambda$.

That aside, the structure of this matrix allows you to find its eigenvalues and eigenvectors by inspection. Recalling that the columns of a matrix are images of the basis vectors, we have $A(0,0,1)^T=(0,0,2)^T=2(0,0,1)^T$, therefore $(0,0,1)^T$ is an eigenvector of $A$ with eigenvalue $2$. If we add the first two columns together, we get the zero vector, so $A(1,1,0)^T = 0(1,1,0)^T$ and we have another eigenvector-eigenvalue pair. You can get the last eigenvalue “for free” by examining the trace: $(1+1+2)-2-0=2$. To get a corresponding eigenvector, you might notice that subtracting the second column from the first gives $(2,-2,0)^T$, so $(1,-1,0)$ is another eigenvector of $2$. Alternatively, since the matrix is symmetric, you know that it’s diagonalizable and that eigenspaces of distinct eigenvalues are orthogonal, so you can take any other vector that’s orthogonal to $(1,1,0)^T$. Since you’re working with moments of inertia, you’ll probably want an orthogonal set of eigenvectors for the principal axes, so $(1,1,0)^T\times(0,0,1)^T=(1,-1,0)^T$ is a convenient choice, which not entirely coincidentally is the same as the one found by inspection.