Coordinate Vectors and Dimension of Subspaces (Span)

Problems and solutions in Linear Algebra

Problem 350

Let $V$ be a vector space over $\R$ and let $B$ be a basis of $V$.
Let $S=\{v_1, v_2, v_3\}$ be a set of vectors in $V$. If the coordinate vectors of these vectors with respect to the basis $B$ is given as follows, then find the dimension of $V$ and the dimension of the span of $S$.
\[[v_1]_B=\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, [v_2]_B=\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}, [v_3]_B=\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}.\]

 
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Solution.

Coordinate vectors

Let us first recall the definition of coordinate vectors.
Suppose that $V$ is a vector space over $\R$ and let $B=\{v_1, v_2, \dots, v_n\}$ be a basis of $V$ (hence the dimension of $V$ is $n$).

Then any element of $v$ of $V$ can be written as
\[v=c_1v_1+c_2v_2+\cdots +c_n v_n,\] where $c_1, c_2, \dots, c_n$ are scalars in $\R$.

This expression is unique and the coordinate vector of $v$ with respect to the basis $B$ is defined as
\[[v]_B=\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}.\] Thus the coordinate vector $[v]_B$ is just a usual $n$-dimensional vector.

Main part of the solution

Note that in the current problem, the coordinate vectors are $4$-dimensional vectors.
This implies that the basis $B$ consists of four vectors. Hence the dimension of $V$ is $4$.

By the correspondence of the coordinate vectors, the dimension of $\Span(S)$ is the same as the dimension of $\Span(T)$, where
\begin{align*}
T&=\{[v_1]_B, [v_2]_B, [v_2]_B\}\\
&=\left\{\, \begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix} \,\right\}.
\end{align*}

To find the dimension of $\Span(T)$, we need to find a basis of $\Span(T)$.
One way to do this is to note that the third vector is the sum of the first two vectors. Also, it’s clear that the first two vectors are linearly independent.

Thus, the set
\begin{align*}
\left\{\, \begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix} \,\right\}
\end{align*}
is a basis of $\Span(T)$, hence the dimension of $\Span(T)$ is $2$.
We conclude that the dimension of $\Span(S)$ is $2$ as well.
(We can also conclude that the set $\{v_1, v_2\}$ is a basis of $\Span(S)$.)

Another way to find a basis of $\Span(T)$

Here is another way to find a basis of $\Span(T)$. We can use the leading 1 method.
We form the matrix whose column vectors are the vectors in $T$:
\begin{align*}
\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Note that this matrix is already in reduced row echelon form.
The first two columns contain the leading 1’s. Thus the first two columns form a basis of $\Span(T)$ (this is the leading 1 method).

Remark: In general we apply elementary row operations to reduce the matrix into a matrix in reduced row echelon form.


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