2
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ 1) Do you know that eigen vectors corresponds to different eigen values are linearly independent? $\endgroup$ – user9077 Oct 13 '13 at 6:20
4
$\begingroup$
  1. 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 $3$.

  2. 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 $2$.

$\endgroup$
  • 2
    $\begingroup$ Perhaps a simpler counter-example is $$\pmatrix{0&1\cr0&0\cr}$$ which has rank 1 and no non-zero eigenvalue. $\endgroup$ – Gerry Myerson Feb 7 '14 at 11:33

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.