Dual Vector Space and Dual Basis, Some Equality

Linear Algebra Problems and Solutions

Problem 282

Let $V$ be a finite dimensional vector space over a field $k$ and let $V^*=\Hom(V, k)$ be the dual vector space of $V$.
Let $\{v_i\}_{i=1}^n$ be a basis of $V$ and let $\{v^i\}_{i=1}^n$ be the dual basis of $V^*$. Then prove that
\[x=\sum_{i=1}^nv^i(x)v_i\] for any vector $x\in V$.

 
LoadingAdd to solve later

Sponsored Links

Proof.

Recall that the dual basis $\{v^i\}_{i=1}^n$ consists of vectors $v^i \in V^*$ satisfying
\[v^j(v_i)=\delta_{i,j}, \tag{*}\] where $\delta_{i,j}$ is the Kronecker delta function that is $1$ if $i=j$ and $0$ if $i\neq j$.

Let $x$ be an arbitrary vector in $V$.
Since $\{v_i\}_{i=1}^n$ is a basis of $V$, we express $x\in V$ as a linear combination of the basis. We have
\[x=\sum_{i=1}^nc_iv_i,\] where $c_i$ is a scalar (an element in $k$) for $i=1, \dots, n$.
For a fixed $j$, we have
\begin{align*}
v^j(x)&=v^j \left(\sum_{i=1}^nc_iv_i \right)\\
&=\sum_{i=1}^nc_iv^j(v_i) && \text{ by the linearity of $v_j$}\\
&=\sum _{i=1}^nc_i \delta_{i,j} && \text{ by (*)}\\
&=c_j.
\end{align*}

Thus we have obtained $c_j=v^j(x)$ for any $j$. Substituting this into the linear combination of $x$, we have
\[x=\sum_{i=1}^nv^i(x)v_i\] as required.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Please Login to Comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Introduction to Linear Algebra at the Ohio State University quiz problems and solutions
Quiz 3. Condition that Vectors are Linearly Dependent/ Orthogonal Vectors are Linearly Independent

(a) For what value(s) of $a$ is the following set $S$ linearly dependent? \[ S=\left \{\,\begin{bmatrix} 1 \\ 2 \\...

Close