A set of linear algebra questions? Could you help me with these questions, I figured most of them out on my own, but I'm not completely sure if I'm correct.
a) $A=\begin{bmatrix}a^2&ab&ac\\ ab&b^2&bc\\ ac&bc&c^2\end{bmatrix}$ where $a,b,c\in \mathbb{R}\backslash \left\{0\right\}$.
Does $A$ have i) three real eigenvalues ii) complex eigenvalues iii) two non-zero eigenvalues iv) one non-zero eigenvalue. NOTE: you're not to calculate the eigenvalues.
My approach:
I can see that this matrice is symmetric, thus it's eigenvalues must be real, so option ii) is false, next it's singular, thus one of the eigenvalues is $0$, the first pivot $a^2$ is positive thus one of the eigenvalues must be positive, also the matrix is only positively semi-defined, it's second minor $d_2=0$ so my guess is that the answer is iv), and that the eigenvalue $\lambda=0$ has algebraic multiplicity 2.
QUESTION 1 Does the number of minors $d_i$ that turn out to be zero equal the algebraic multiplicity of $\lambda = 0$?
b) If $M=\begin{bmatrix}1&2&2\\ 0&2&2\\ 0&1&1\end{bmatrix}$ and $V=\left\{Mx:x\in \mathbb{R}^3\right\}$ whatis the dimension of V?
My approach $x=\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}$, so by multiplication we get: $$\begin{bmatrix}x_1+2x_3+2x_3\\ \:2x_2+2x_3\\ x_2+\:x_3\end{bmatrix}=x_1\begin{bmatrix}\:1\\ \:0\\ \:0\end{bmatrix}+\left(x_2+x_3\right)\begin{bmatrix}\:2\\ \:2\\ \:1\end{bmatrix}=span\left(\begin{bmatrix}\:1\\ \:\:0\\ \:\:0\end{bmatrix},\begin{bmatrix}\:2\\ \:\:2\\ \:\:1\end{bmatrix}\right)$$
Thus $dimV=2$
c)if $A=\begin{bmatrix}2&1&0\\ 0&2&0\\ 0&0&3\end{bmatrix}$ which of these is a subspace?

QUESTION 2 I think i) and ii) are subspaces, but what about the other two?
 A: obviously,
$A = C\times D=\begin{bmatrix}a &0 &0\\ 0&b&0 \\0&0&c \end{bmatrix} \times \begin{bmatrix}a &a&a \\b &b &b \\ c&c&c\end{bmatrix}$, where $rank(C)=3$ for $a\neq b\neq c$, and $rank(D)=1$. From the rank property, we can see $rank(A)=rank(CD)\leq \min(rank(C),rank(D))$. Then you can get your results. Hope to help you.
A: a) I agree.
I am not sure about your conjecture how the minors relate to the multiplicity of $\lambda=0$  However, I can eyeball two eigenvectors for $\lambda=0$
$v_1 = \begin{bmatrix} b\\-a\\0\end{bmatrix}, v_1 = \begin{bmatrix} c\\0\\-a\end{bmatrix}$
b) looks fine
c) For each $X$, Suppose $P,Q \in X$
then for example iii) if $trace (AP)=0$ and $trace (AQ)=0$
does $trace (kAP) = 0$ for scalar $k$
and does $trace (A(P+Q)) = trace (AP+AQ) = 0$? 
If so, then X is closed under addition and scalar multiplication and hence a subspace.
A: Hints:


*

*For a), it is useful to note that $A = xx^T$, where $x$ is the column-vector $x = (a,b,c)$.  No, the number of minors that are zero is not necessarily the multiplicity of the zero eigenvalue; if you wanted to use minors, you'd have to show that every $2 \times 2$ submatrix has determinant zero.  You should instead find the rank of the matrix through the usual methods (e.g. row-reduction).

*For b), your approach is fine

*For c), i) is indeed a subspace, and ii) is certainly a subspace since it is the entire space.  iii) is also a subspace (since the map $X \mapsto trace(AX)$ is linear) and iv) is not a subspace.

