Does this vector lie in the vector subspace $U$?

Question

Given are the vectors $v_{1}=\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix}, v_{2}=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix}, v_{3}=\begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ from $\mathbb{R}^{3}$.

The vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3}\right\}$

Does the vector $\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$ lie in $U$?

I'm not sure how this is done correctly but I think it's done by checking if the vector $\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$ is a linearly combination of $v_{1},v_{2},v_{3}$.

So I have written it all like that:

Let $a,b,c \in \mathbb{R}$.

$$a\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix}+b\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix}+c\begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}=\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$$

And then calculated each variable by using Gauss. In the end I had a wrong statement aka no solution and from this I conclude that the vector $\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$ doesn't lie in $U$.

I don't want write down all the calculation steps with Gauss because it would be too long, but is the way I did the correct way? Or is it done completely different?

Answer 1

The complete matrix of the system, with Gaussian elimination: \begin{align} \begin{bmatrix} 1 & 4 & 6 & 2 \\ 3 & 5 & 4 & 4 \\ 5 & 6 & 2 & 5 \end{bmatrix} &\to \begin{bmatrix} 1 & 4 & 6 & 2 \\ 0 & -7 & -14 & -2 \\ 0 & -14 & -28 & -5 \end{bmatrix} && \begin{aligned}R_2&\gets R_2-3R_1 \\ R_3&\gets R_3-5R_1\end{aligned} \\&\to \begin{bmatrix} 1 & 4 & 6 & 2 \\ 0 & -7 & -14 & -2 \\ 0 & 0 & 0 & -1 \end{bmatrix} && R_3\gets -2R_2 \end{align} The system has no solution. I don't find this too long, do you?

Anyway, your method is correct, as far as I can see.

Answer 2

Equivalently, we want a+ 4b+ 6c= 2, 3a+ 5b+ 4c= 4, and 5a+ 6b+ 2c= 5. Subtract the second equation from twice the third: 7a+ 7b= 6. Subtract the first equation from three times the third: 14a+ 14b= 13. Multiplying 7a+ 7b= 6 by two, 14a+ 14b= 12. a and b cannot satisfy both 14a+ 14b= 13 and 14a+ 14b= 12. There are no numbers a and b that make the vector equation true. $\begin{pmatrix} 2 \\ 4 \\ 5\end{pmatrix}$ cannot be written as a linear combination of $v_1$, $v_2$, and $v_3$ so is not in subspace U.