Linear Algebra : Eigenvalues and rank 1) A $4\times4$ square matrix has distinct eigenvalues $\{0, 1, 2, 3\}$. What is its rank?
2) Let $a,b\in\mathbb{R}^n$ be two non-zero linearly independent vectors, and let $\alpha,\beta\in\mathbb{R}$ be two non-zero scalars.
i) What is the rank of the matrix $M = \begin{bmatrix}a&\alpha a&b&\beta b\end{bmatrix}$?
ii) Can you name two linearly independent non-zero vectors $x_1, x_2\in\mathbb{R}^4$ in the null space of $M$? (i.e., $Mx_1 = Mx_2 = 0$)
For question 1, is the answer $3$? It seems that the rank will correspond to the number of non-zero eigenvalues.
For question 2 i), is the answer $n$? Besides, what is null space? I would be grateful if someone can help .
 A: *

*Your answer is correct (but it seems for the wrong reason; see below).  The equation $A\mathbf{x}=0\mathbf{x}$ (implying $\mathbf{x}$ is an eigenvector with eigenvalue $0$, or $\mathbf{x}=\mathbf{0}$) is the same as $A\mathbf{x}=\mathbf{0}$ (implying $\mathbf{x}$ is in the null space of $A$).  In other words:

The eigenspace corresponding to eigenvalue $0$ is the null space of the matrix.

The eigenvalue $0$ has algebraic multiplicity $1$ (since the characteristic polynomial will have $4$ roots, and three of them are non-zero) and hence has geometric multiplicity $1$ (since $1 \leq$ geometric multiplicity $\leq$ algebraic multiplicity).  Hence the null space is $1$-dimensional.  The Rank-Nullity Theorem implies the rank is therefore $3$.

*i. No, the column space of $M$ is $\mathrm{span}\{a,\alpha a,b,\beta b\}=\mathrm{span}\{a,b\}$ which is $2$-dimensional, since $a$ and $b$ are linearly independent.  Hence the rank is $2$.
ii. For example: $(-\alpha,1,0,0)^T$ and $(0,0,-\beta,1)^T$.  These can be found by inspecting the linear dependencies among the columns of $M$.


It seems that the rank will correspond to the number of non-zero eigenvalues.

This is untrue in general; a counter-example is $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ has rank $2$ but characteristic polynomial $x^3(x-1)$, so only one non-zero eigenvalue (even when multiplicities are accounted for).
Another way to phrase this is that the algebraic multiplicity of $0$ is $3$, whereas the geometric multiplicity of $0$ (i.e., the nullity) is $2$.
