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In Page 115 of Axler's book Linear Algebra Done Right (third edition) there is an exercise (numbered 30). See blow:

Suppose $V$ is finite-dimensional and $\phi_1,\dots,\phi_m$ is a linearly independent list in $V'.$ Prove that $$\dim\big((\text{null}\phi_1)\cap\cdots\cap(\text{null}\phi_m)\big)=(\dim V)-m.$$

I have tried for a long time, but failed. In the case of $m=1,$ the result is trivial, just a consequence of the Fundamental Theorem of Linear Maps. But in the case of $m=2,$ I have tried as follows. Since $\phi_1, \phi_2$ is linearly independent, $$\dim\text{null}(\phi_1)=\dim\text{null}(\phi_2)=\dim V-1.$$ Thus \begin{align*} &\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\\ =&\dim(\text{null}(\phi_1))+\dim(\text{null}(\phi_2))-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\ =&\dim V-1+\dim V-1-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\ \geq &\dim V-1+\dim V-1-\dim V\\ =&\dim -2. \end{align*} But how to show the reversed inequality, that is, $\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\leq \dim -2?$ Moreover, it seems that the method is hardly valid for $m\geq 3.$ Can anyone help me?

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    $\begingroup$ Hint: Take a look at the linear map $V \rightarrow K^m,~v \mapsto (\phi_1(v),...,\phi_m(v))$ (where $K$ is the ground field of your vector space). What do you know about the kernel, what do you know about its rank? $\endgroup$ Commented Mar 31, 2017 at 9:06
  • $\begingroup$ @SebastianSchoennenbeck. Thanks. I've got it. $\endgroup$
    – azc
    Commented Mar 31, 2017 at 14:19
  • $\begingroup$ In that case you should consider writing an answer to your question yourself and then accepting it. $\endgroup$ Commented Mar 31, 2017 at 14:22
  • $\begingroup$ I've post the proof according to your suggestion. Please check if it is right. $\endgroup$
    – azc
    Commented Mar 31, 2017 at 14:39

2 Answers 2

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Proof. $\quad$ Consider the mapping \begin{gather*} f\colon V\to\mathbb{F}^m, \end{gather*} such that \begin{gather*} f(v)=\big(\phi_1(v),\dots,\phi_m(v)\big) \end{gather*} for all $v\in V.$ It is clear that $ \text{null} f=\text{null}(\phi_1)\cap\cdots \text{null}(\phi_m).$ Indeed, let $v\in \text{null}f.$ Then $f(v)=0,$ which implies that \begin{gather*} \big(\phi_1(v),\dots,\phi_m(v)\big)=0, \end{gather*} and thus $\phi_j(v)=0$ for $j=1,\dots,m.$ So we have $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Hence we have proved $ \text{null} f\subset \cap_{j=1}^n \text{null} \phi_j.$ In the other direction of inclusion, suppose $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Then $\phi_j(v)=0$ for all $j=1,\dots, m.$ Thus $f(v)=0,$ which implies that $v\in \text{null} f.$ Hence $\cap_{j=1}^m \text{null}(\phi_j)\subset \text{null} f.$ Therefore $\cap_{j=1}^m \text{null}(\phi_j)= \text{null} f.$

We then show that $\dim \text{range} f=m.$ Let $v_1,\dots, v_n$ be a basis of $V$ and $e_1,\dots, e_m$ be the standard basis of $\mathbb{F}^m.$ Then \begin{gather*} f(v_k)= \big(\phi_1(v_k),\dots, \phi_m(v_k)\big) = \sum_{j=1}^{m}\phi_j(v_k)e_j. \end{gather*} Thus the matrix of $f$ with respect to the given bases is $A:=\big(\phi_j(v_k)\big)_{m\times n}.$ Therefore, by 3.117 and 3.118, \begin{gather*} \dim \text{range} f=\text{column}\text{ rank }{\mathcal{M}(f)}=\text{row rank } \mathcal{M}(f). \end{gather*} We claim that the list of rows of $A$ is linearly independent. Indeed, let $a_1,\dots, a_m\in\mathbb{F}.$ Suppose $a_1A_{1,\cdot}+\cdots+a_mA_{m,\cdot}=0.$ That is, \begin{gather*} a_1\big(\phi_1(v_1),\dots, \phi_1(v_n)\big)+\cdots+a_m\big(\phi_m(v_1),\dots,\phi_m(v_n)\big)=0. \end{gather*} We have \begin{gather*} \big(a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1),\dots, a_1\phi_1(v_n)+\cdots+a_m\phi_m(v_n)\big)=0. \end{gather*} Thus \begin{gather*} a_1\phi_1(v_j)+\cdots+a_m\phi_m(v_j)=0,\qquad \text{for all $j=1,\dots, n.$} \end{gather*} Particularly, \begin{gather*} a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1)=0.\tag{1} \end{gather*} Because $\phi_1,\dots, \phi_m$ is linearly independent, from (1) it follows that $a_1=\cdots=a_m=0.$ Thus the list of rows of $A$ is linearly independent. As a result, we have the row rank of $A$ is $m.$ Thus $\dim \text{range} f=m.$ Finally, by the Fundamental Theorem of Linear Maps, we have \begin{align*} &\dim\big( \text{null}(\phi_1)\cap\cdots\cap \text{null}(\phi_m)\big)=\dim \text{null} f\\ =&\dim V-\dim \text{range} f\\ =&\dim V-m. \end{align*}

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  • $\begingroup$ Looks good. Very thorough. $\endgroup$ Commented Mar 31, 2017 at 21:26
  • $\begingroup$ I am curious about (1), which follows after "Particularly, ... " I thought that we should instead say that because it holds for v1, .. vn, therefore if holds for all v in V, and then we only apply the linear independence of phi. Because when we say the phi's are linearly independent, we view them as maps and not just look at particular values. $\endgroup$ Commented May 10, 2023 at 11:03
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Let $W = \text{null}\:\varphi_{1}\cap\dots\cap \text{nulll}\:\varphi_{m}$. We need to prove that $\text{dim}\:W^0 = m$. It is clear that $\{\varphi_{1},\dots,\varphi_{m}\}$ is a linearly independent subset of $W^{0}$, so $\text{dim}\:W^{0}\geq m$. It can be shown that for any subspaces $U_1$ and $U_2$ of a finite-dimensional vector space $V$ one has $(U_1\cap U_2)^{0} = U_1^{0}+U_2^{0}$. Therefore, \begin{align} \text{dim}\:W^0 &= \text{dim}\:(\text{null}\:\varphi_{1}^0+\dots+\text{null}\:\varphi_{m}^{0})\\ &\leq \text{dim}\:\text{null}\:\varphi_{1}^{0}+\text{dim}\:\text{null}\:\varphi_{m}^{0}\\ &= 1+\dots+1\\ &= m. \end{align}

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