I'm stuck trying to solve this:

Given $U = \{\vec{u_1}, \vec{u_2}, \vec{u_3}, \vec{u_4}\}$ is a linearly independant set of vectors in $\mathbb{R}^5$ and vectors: \begin{equation*} \begin{alignedat}{4} & v_1 & {}={} & 8au_1 {}+{} & 2u_2 {}+{} & u_3 \\ & v_2 & {}={} & & 16au_2 {}+{} & u_4 \\ & v_3 & {}={} & u_1 {}-{} & \frac{1}{2}u_3 {}+{} & au_4 \\ & a \in \mathbb{R} \end{alignedat} \end{equation*} Find all $a$ such that $V = \{v_1, v_2, v_3\}$ is linearly dependent.

I've got as far as spelling out what $v_i$ would look like:

\begin{equation} b(8au_1 + 2u_2 + u_3) + c(16au_2 + u_4) + d(u_1 - \frac{1}{2}u_3 + au_4) = 0 \end{equation}

$b$, $c$ and $d$ not all zero.

Which then gives this matrix:

\begin{bmatrix} 8ab & d & k_1 \\ 2b & 16ac & k_2 \\ b & -\frac{1}{2}d & k_3 \\ c & da & k_4 \end{bmatrix}

And this is where I'm stuck. I must be going in the wrong direction because it doesn't even seem that this matrix has the right dimensions for the thing I'm looking for.

I would appreciate a hint rather than the complete solution (this still being a homework). If you could say what exactly am I to search for in order to find $v_i$, that would be best.

  • $\begingroup$ Hint: write down the matrix whose columns are the coefficients of the three vectors $v_1,v_2,v_3$ as written in the basis $\{u_j\}$ (this is how they're given to you). Why will $\{v_1,v_2,v_3\}$ be linearly dependent if and only if that matrix has rank less than $3$? How do you determine when that matrix has rank $3$ or smaller rank? $\endgroup$ – Greg Martin Nov 1 '14 at 20:11
  • $\begingroup$ @GregMartin would I had the matrix, I'd know how to find the rank and how that affects that solutions are linearly dependant. (This is from definition of rank: it is the number of the independent equations / vectors, so matrices with rank 1 and 2 would be the solutions). What I'm still struggling with is building the matrix you probably mentioned first. Sorry, I just don't understand what are the coefficients of the vectors. I'll probably need to sleep on it :S $\endgroup$ – wvxvw Nov 1 '14 at 20:36
  • $\begingroup$ Ok: for example, the entries in the first column would be $8a,2,1,0$, since $v_1 = 8au_1+2u_2+1u_3+0u_4$. $\endgroup$ – Greg Martin Nov 1 '14 at 23:16
  • $\begingroup$ @GregMartin Oh, thank you. Still, I think I'm just going through the motions, but let me just make sure. Suppose I've written this matrix, triangularized it, and then I have a row in it $(0, 0, 32a^2-2)$. Then solving $32a^2-2$ will be a mandatory requirement for this to be a linear combination, is that right? I'm still not sure on why / what is the meaning of concatenating coefficients in this way. Does this procedure have a special name? I'd like to read more about this, if possible. $\endgroup$ – wvxvw Nov 2 '14 at 19:33

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