Let $A\in M_{1\times3}(\mathbb{R})$ be a arbitrary matrix. Find the eigenvalues and eigenvectors of matrix $A^TA$.
My approach:
$$ A^TA = \begin{bmatrix} a^2 & ab & ac\\ ab & b^2 & bc\\ ac & bc & c^2 \end{bmatrix}; \\ \lambda_1\lambda_2\lambda_3=\det(A^TA)=0 \qquad (1)\\ \lambda_1+\lambda_2+\lambda_3=\text{tr}(A^TA)=a^2+b^2+c^2 \qquad (2) $$
So from these two properties we know that at least one eigenvalue must be $0$. Solving $A^TA-\lambda I=0$ for $\lambda=0$ we get that $\dim(\text{ker}(A^TA))=2$. Since the algebraic multiplicity has to be equal to or larger than the geometric multiplicity and from $(2)$ we conclude that the algebraic multiplicity had to be equal to the geometric multiplicity. So we can say that $\lambda_1=a^2+b^2+c^2,\lambda_2=\lambda_3=0$. And now we just need to find the eigenvectors for the corresponding eigenvalues.
Is my approach correct?