Let $A\in M_{1\times3}(\mathbb{R})$ be a arbitrary matrix. Find the eigenvalues and eigenvectors of matrix $A^TA$. 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?

$\DeclareMathOperator{\tr}{tr}$The columns of $A^TA$ are all scalar multiples of $A^T$, so for $A\ne0$, this matrix has rank 1: its column space is spanned by $A^T$ and two of its eigenvalues are $0$. The last eigenvalue you get “for free” since the trace is equal to the sum of the eigenvalues, so it is $\tr A^TA-0-0=AA^T$, with eigenvector $A^T$. You can verify this by observing that $(A^TA)A^T=A^T(AA^T)$. The eigenspace of $0$ is just the null space of $A^TA$. Since each row of this matrix is a scalar multiple of $A$, this amounts to solving $AX=0$, which describes the set of vectors orthogonal to $A^T$. If $A=[a,b,c]\ne0$, then at least two of $[-c,0,a]$, $[0,-c,b]$ and $[b,-a,0]$ are non-zero and are obviously linearly independent, so will form a basis for this eigenspace.
