Eigenvalues and eigen vectors of $A^TA $ Let $A $ be some matrix in $M_{1×3} $:
a) find the eigenvalues of $A^TA$
b) for $A=(1, 2, 3) $ find the eigenvectors of $A^TA $
I am kind of confused here. How am I supposed to find the eigenvalues if I don't know the matrix elements? Any help is appreciated. 
 A: 
Let $A $ be some matrix in $M_{1×3} $:
a) find the eigenvalues of $A^TA$

If $A \in M_{1×3}$, then $A = \begin{pmatrix}a & b & c\end{pmatrix}$ for some $a,b,c \in \mathbb{R}$ and:
$$A^TA = \begin{pmatrix}a \\ b \\ c\end{pmatrix}\begin{pmatrix}a & b & c\end{pmatrix} = \begin{pmatrix}a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2\end{pmatrix}$$
Can you continue?
A: Hint:
$$
A^TA= \begin{pmatrix}1\\2\\3 \end{pmatrix}\begin{pmatrix}1&2&3 \end{pmatrix}=
\begin{pmatrix}1&2&3\\2&4&6\\3&6&9 \end{pmatrix}
$$
A: For general case without calculating eigenvalues from characteristic polynomial which is a big pain!
For a nonzero row-vector $A \in R^n $, you can easily show that the matrix $A^T A$ is symmetric with rank $1$. So it has the  eigenvalue $\lambda =0$ with degree $n-1$, and one not zero eigenvalue which has to be $\lambda'=trace(A^TA) = \sum_{i=1}^{n} a_i^2  $ 
Now you can find all Eigenvectors by solving $(A^TA) x=0$ and $(A^T A)x=(\sum_{i=1}^{n} a_i^2) x.$  
A: Compute $A^TA=(1, 2, 3)^T*(1, 2, 3)$. This gives a matrix $ \in M_{3×3}$.
A: First of all we note that due to how $A^T A$ is constructed (all it's columns and rows are just scaled version of each other) it's determinant will be zero. That is you can find eigenvectors from the equation $Ax = 0$ (all vectors that are perpendicular to $A$, ie a plane of eigenvectors).
Also note that $(A^TA) x = (A^T) (Ax) = (Ax) A$ so if $x=A^T$ we have $(A^TA)x = (A^Tx) x$ which makes $x$ an eigenvector with eigenvalue $A^Tx$.
We also can see that there are no more eigenvectors since such a vector could be written as $cA^T + u$ where $u\ne 0$ is perpendicular to $A$ and we have that $$(A^TA)(cA^T+u) = cA^T(AA^T) + A^T(Au) = cA^T(AA^T) \ne \lambda A^T + \lambda u$$
A: Better it would be to denote $A$ as $v^T$ then the matrix $A^TA=vv^T$ is known as a scaled projection matrix (into a line determined by the vector $v$ ).  
Non-scaled projection matrix would be $U=uu^T$ where $u=\dfrac {v}{\Vert v \Vert}$ - here vector $u$ has unit length,
so in this case $vv^T={\Vert v \Vert}^2 uu^T$.    
Projection into a 3-dimensional line  has one $1$ eigenvalue (because for  any vector $ku$ we have $ku=uu^T(ku)$ and two $0$ (2-dimensional space orthogonal to u, i.e. plane where for vectors $w$ from this plane it holds $uu^tw=0=0w$).
Scaling factor here is ${\Vert v \Vert}^2$ hence the first eigenvalue for $vv^T$ is just ${\Vert v \Vert}^2$, the remaining stay as zeros.
