MHB Is there a more efficient way to determine the basis of a subspace?

[mathmari]

Hey!

We are given the vectors $\vec{a}=\begin{pmatrix}4\\ 1 \\ 0\end{pmatrix}, \vec{b}=\begin{pmatrix}2\\ 0 \\ 1\end{pmatrix}, \vec{c}=\begin{pmatrix}0\\ -2 \\ 4\end{pmatrix}$.

I have shown by calculating the deteminant $|D|=0$ that these three vectors are linearly dependent.

I want to give a basis of the subspace of $\mathbb{R}^3$ that is spanned by these three vectors.

To find the basis of the subspace $\text{ span }\{\vec{a}, \vec{b}, \vec{c}\}$ we have to find a minimal subset $V$ of $\{\vec{a}, \vec{b}, \vec{c}\}$ with $\text{ span }V=\text{ span } \{\vec{a}, \vec{b}, \vec{c}\}$.

We have to use the Gauss algorithm to find the linearly independent subset, right?

$\begin{bmatrix}
\begin{matrix}
4 & 2 & 0\\
1 & 0 & -2\\
0 & 1 & 4
\end{matrix}\left|\begin{matrix}
0\\
0\\
0
\end{matrix}\right.\end{bmatrix}\begin{matrix}
\\
2.\text{row} \leftrightarrow 3.\text{row}\\

\end{matrix} \longrightarrow \begin{bmatrix}
\begin{matrix}
4 & 2 & 0\\
0 & 1 & 4\\
1 & 0 & -2
\end{matrix}\left|\begin{matrix}
0\\
0\\
0
\end{matrix}\right.\end{bmatrix}\begin{matrix}
\\
\\
4\cdot 3.\text{row}-1.\text{row}
\end{matrix} \longrightarrow \begin{bmatrix}
\begin{matrix}
4 & 2 & 0\\
0 & 1 & 4\\
0 & -2 & -8
\end{matrix}\left|\begin{matrix}
0\\
0\\
0
\end{matrix}\right.\end{bmatrix}
\begin{matrix}
\\
\\
-\frac{1}{2}\cdot 3.\text{row}- 2.\text{row}
\end{matrix} \longrightarrow \begin{bmatrix}
\begin{matrix}
4 & 2 & 0\\
0 & 1 & 4\\
0 & 0 & 0
\end{matrix}\left|\begin{matrix}
0\\
0\\
0
\end{matrix}\right.\end{bmatrix}$

So, we get $4\lambda_1+\lambda_2=0, \lambda_2+4\lambda_3=0$, and so $\lambda_1=2\lambda_3, \lambda_2=-4\lambda_3$.

Therefore, $(\lambda_1, \lambda_2, \lambda_3)=\lambda_3 (2,-4,1), \lambda_3\in \mathbb{R}$.

For $\lambda_3=1$ we get the coefficients $(2,-4,1)$. That means that $2\vec{a}-4\vec{b}+\vec{c}=0 \Rightarrow \vec{c}=-2\vec{a}+4\vec{b}$.

Do we have to check if the vectors $\vec{a}, \vec{b}$ are linearly independent? Or can we just conclude from here that the basis that we are looking for is $\{\vec{a}, \vec{b}\}$ ? (Wondering)

 

[HallsofIvy]

Personally, I don't like to immediately go to matrices and determinants for a problem like this. To determine whether or not these three vectors are independent, I would use the definition of independent- suppose there exist 3 numbers, a, b, and c, such that [math]a\begin{bmatrix}4 \\ 1 \\ 0 \end{bmatrix}+ b\begin{bmatrix} 2 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}0 \\ -2\\ 4\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/math]- must a= b= c= 0?

That is the same as the three equations 4a+ 2b=0, a-2c= 0, and b+ 4c= 0. From the first equation b= -2a. From the second equation a= 2c so b= -2(2c)= -4c. But the third equation also give b= -4c. These vectors are dependent.

So the sub space they span is not three dimensional, it is either one or two dimensional. If it is one dimensional then any one vector of the given three will span it. If it is two dimensional, then any two will span it.

Try the first two vectors, [math]\begin{bmatrix}4 \\ 1 \\ 0 \end{bmatrix}[/math] and [math]\begin{bmatrix} 2 \\ 0 \\ 1\end{bmatrix}[/math]. Clearly they are independent because one is not a multiple of the other. Therefore this subspace is two dimensional and these two vectors span it.

 

[mathmari]

Personally, I don't like to immediately go to matrices and determinants for a problem like this. To determine whether or not these three vectors are independent, I would use the definition of independent- suppose there exist 3 numbers, a, b, and c, such that [math]a\begin{bmatrix}4 \\ 1 \\ 0 \end{bmatrix}+ b\begin{bmatrix} 2 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}0 \\ -2\\ 4\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/math]- must a= b= c= 0?

That is the same as the three equations 4a+ 2b=0, a-2c= 0, and b+ 4c= 0. From the first equation b= -2a. From the second equation a= 2c so b= -2(2c)= -4c. But the third equation also give b= -4c. These equations are dependent.

So the sub space they span is not three dimensional, it is either one or two dimensional. If it is one dimensional then any one vector of the given three will span it. If it is two dimensional, then any two will span it.

Try the first two vectors, [math]\begin{bmatrix}4 \\ 1 \\ 0 \end{bmatrix}[/math] and [math]\begin{bmatrix} 2 \\ 0 \\ 1\end{bmatrix}[/math]. Clearly they are independent because one is not a multiple of the other. Therefore this subspace is two dimensional and these two vectors span it.

I understand!

But my way would also be correct, or not? (Wondering)