# 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?

Your approach is correct, but here's a way to arrive at the same result with less computation. Considering $A^T A$ as a linear transformation, we have the following factorization. $$\newcommand{\R}{\mathbb{R}} \R^3 \overset{A}{\longrightarrow} \R \overset{A^T}{\longrightarrow} \R^3$$ This shows that the map factors through a $1$-dimensional space, so its rank is $\leq 1$. Assuming $A$ is not the zero matrix, then the rank of $A^T A$ is $1$, so its kernel has dimension $3-1 = 2$. Thus two of the eigenvalues must be $0$, and from your trace formula $\DeclareMathOperator{\tr}{tr} \sum_i \lambda_i = \tr(A^T A) = a^2 + b^2 + c^2$, we see that the third eigenvalue must be $a^2 + b^2 + c^2$. Note that this approach generalizes and can be used for $A \in M_{1 \times n}(\R)$ for any $n$.

The factorization above also indicates how to find the eigenvectors. The kernel of $A$ is $2$-dimensional, and these will be the eigenvectors with eigenvalue $0$. (Since $\DeclareMathOperator{\img}{img} \ker(A) = \img(A^T)^\perp$, this agrees with Daniel's answer.) Since $\ker(A)^\perp = \img(A^T)$, the remaining eigenvector is any nonzero multiple of $A^T$.

• Great explanation. Thank you. Aug 8, 2017 at 13:13

Try to argue this way. Suppose that the column vector $X$ is an eigenvector associated to some eigenvalue $\lambda$, then $A^TA(X) = \lambda X$, which we can also write as $A^T(AX) = \lambda X$. Now, notice that $AX$ is an scalar! (recall that $A$ is a row vector), hence, unless $\lambda =0$ and $AX = 0$, the vectors $A^T$ and $X$ must be linearly dependent.

Let us analyze this situation in detail.

First of all, notice that every vector in $A^{\perp}$ is an eigenvector for $A^TA$ associated to the eigenvalue $0$. Therefore, we only need to find those eigenvectors outside this space.

Suppose that $\lambda = 0$, then $A^T(AX) = 0$ which in turns implies that either $A^T$ is the zero vector (and in such a case there is not so much to say) or $AX = 0$, but in this case $X$ would be orthogonal to $A$, which is a case we're ruling out. Therefore, the ony remaining case is in which $\lambda\neq 0$, but as I already mentioned this in turn implies that $A^T$ and $X$ are L.D. and therefore, the eigenspace in this case is that generated by $A^T$.

In summary, the eigenspaces are the one generated by $A^T$ and the orthogonal space to it.

$$\DeclareMathOperator{\tr}{tr}$$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.