Let V be a vector space and $f,g_1,g_2,\ldots,g_m\in V^*$ its dual, prove that f is a linear combination of $g_i$'s iff $\cap ker g_i \subseteq kerf$

Question

Let $V$ be a vector space and $f,g_1,g_2,\ldots,g_m\in V^*$ its dual, prove that $f$ is a linear combination of $g_i$'s iff $\cap \text{ker} g_i \subseteq \text{ker}f$

I already understand that if the intersection of $g_i$'s is empty, then it is a basis for the dual, and therefore $f$ is a linear combination of $g_i$'s. And I also managed to prove the foward direction (if $f$ is a linear combination of $g_i$, then $\cap \text{ker} g_i \subseteq \text{ker} f$, since that for any $v \in \cap ker g_i$ $f(v)=0=a_1g_1(v)=g_2(v)=\ldots=g_m(v)=0$). But I'm not finding a way to prove the reverse direction. Tried chatgpt but got only nonsense. Can someone give me a hint or post a duplicate (really didn't find it out here).

Thanks in advance.

The exercises doesnt poses any condition, but assume that V is finite dimensional. — Victor Hugo, Apr 10 '23 at 10:33
Dint got it. It wanst supposed that g and f maps vectors on V to scalars on the field? — Victor Hugo, Apr 10 '23 at 10:40
@Jorge The dual space is the space of linear maps $V \rightarrow F$ so $e_1$ and $e_1 + e_2$ are not valid outputs. — David Lui, Apr 10 '23 at 10:56
This already answered your (frequently asked) question: Intersection of kernels and linear dependence of functionals (found using Approach$0$). Here is another duplicate: How to prove this beautiful statement on linear forms using induction?. — Anne Bauval, Apr 11 '23 at 11:29

score 1 · Accepted Answer · answered Apr 10 '23 at 11:10

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In a comment it says we can assume $V$ is finite dimensional, so choose a basis $v_1, ... v_n$ of $V$, and $v_1^*, ... v_n^*$ is the dual basis. Now, both $V$ and $V^*$ are isomorphic to $F^n$, and the evaluation map $V^* \times V \rightarrow F$ is the dot product.

We now have a system of linear equations $f = \sum_{i=1}^m a_i g_i$ with $n$ equations and $m$ variables. Gaussian elimination either gives a solution to this equation (which shows that $f$ is a linear combination of the $g_i$), or outputs some vector whose dot product with $f$ is nonzero, but whose dot product with $g_i$ are all zero. This is an element in $\cap ker(g_i)$ but not in $ker(f)$.

answered Apr 10 '23 at 11:10

David Lui

6,295

Got everything less the last part. I dont understand how if the system doesnt have a solution it outputs some vector (whose coordinates are $a_i$ )whose dot product with f is non zero but zero with all $g_i$. I will think more a bit. – Victor Hugo Apr 10 '23 at 12:26
Got it thanks! Will comment the details later. – Victor Hugo Apr 10 '23 at 13:03
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Gaussian elimination does row operations to try to find a solution. If there is no solution, then Gaussian elimination produces a list of row operations that when applied to the matrix, results in zero, but when applied to the solution vector, is not zero. However, row operations corresponds to left multiplication by a row vector, and left multiplying by a row vector means taking the dot product of that vector and the columns of the matrix. – David Lui Apr 10 '23 at 15:11
Better look for duplicates (see above) than duplicate answers. And actually, finite dimensionality is dispensable. – Anne Bauval Apr 11 '23 at 11:35

Let V be a vector space and $f,g_1,g_2,\ldots,g_m\in V^*$ its dual, prove that f is a linear combination of $g_i$'s iff $\cap ker g_i \subseteq kerf$

1 Answers1