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 $A^T$.

  • 1
    $\begingroup$ Great explanation. Thank you. $\endgroup$ – Dragan Zrilić Aug 8 '17 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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.