Can the given vector be expressed as a linear combination of the other three

Question

Four vectors from $\mathbb{R}^4$ are given: $$v_1 = (1, -2, 0, 1), v_2 = (0, 1, -1, -3), v_3 = (1, 0, -2, -5), v_4 = (-3, 5, 1, 0)$$ Can the vector $v_4$ be expressed as a linear combination of the other three? The solution given is yes: $$v_4 = av_1 + (2a + 5)v_2 + (-a - 3)v_3 \qquad a ∈ \mathbb{R}$$ I tried to solve the system of linear equations $\alpha v_1 + \beta v_2 + \gamma v_3 = a v_4$, but this seems to get me nowhere. How can I express $v_4$ as a linear combination of the other three?

Answer 1

The general solution is:

$$ av_1+(5+2a)v_2+(-3-a) v_3 = v_4$$

for any $a \in \mathbb{R}$, as stated. Here's why...

Firstly write $\alpha v_1+\beta v_2+\gamma v_3 = v_4$ out as

$$ \left(\begin{matrix}\alpha+\gamma \\-2\alpha+\beta \\-\beta-2\gamma \\ \alpha-3\beta-5\gamma\end{matrix}\right) = \left(\begin{matrix}-3 \\5 \\1 \\ 0\end{matrix}\right) $$

(I've obviously swapped your row vectors for column vectors.)

If a solution exists for the whole vector, then a solution must exist for its first three elements, so we might as well just work with the following to start with:

$$ \left(\begin{matrix}\alpha+\gamma \\-2\alpha+\beta \\-\beta-2\gamma \\\end{matrix}\right) = \left(\begin{matrix}-3 \\5 \\1 \end{matrix}\right) $$

Re-writing this as a matrix equation, we have:

$$ \left(\begin{matrix}1 & 0 & 1 \\-2 & 1 & 0 \\0 & -1 & -2 \\\end{matrix}\right) \left(\begin{matrix}\alpha \\ \beta \\ \gamma \\\end{matrix}\right) = \left(\begin{matrix}-3 \\5 \\1 \end{matrix}\right) $$

The determinant of the matrix here is $0$, so we may put $\alpha=a$ where $a \in \mathbb{R}$ is an arbitrary constant. Continuing to solve, we find $\beta=5+2a$ and $\gamma=-3-a$. These values also work to make $\alpha-3\beta-5\gamma=0$ (the fourth element of the original vector equation), so we have the solution!

Answer 2

In your set: $\;v_4\;$ is a linear combination of the other three vectors iff when we form the corresponding matrix with $\;v_4\;$ the last row and we reduce it by rows, this last row becomes all zeros (why?! This must be clear in order to be sure what is going on here...):

$$\begin{pmatrix} 1&-2&0&1\\ 0&1&-1&-3\\ 1&0&-2&-5\\ -3&5&1&0 \end{pmatrix}\longrightarrow\begin{pmatrix} 1&-2&0&1\\ 0&1&-1&-3\\ 0&2&-2&-6\\ 0&-1&1&3 \end{pmatrix}$$

...and we're done as we can see the last row is the third one multiplied by $\;-\frac12\;$ .

Answer 3

Yes we can!

Because we have: $$v_4=-2v_1+v_2-v_3$$

Answer 4

The answer is yes.

Note that $$\det \begin{bmatrix} 1&-2&0&1\\ 0&1&-1&-3\\ 1&0&-2&-5\\ -3&5&1&0 \end{bmatrix} =0$$

Thus these vectors are linearly dependent, which means we can express one in terms of the others. We can see by inspection that $v_4 = -3v_1-v_2$