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Explain why the columns of a $3 \times 4$ matrix are linearly dependent

I also am curious what people are talking about when they say "rank"? We haven't touched anything with the word rank in our linear algebra class.

Here is what I've came up with as a solution, will this suffice?

I know that the columns of a matrix $A$ are linearly independent iff the equation $Ax = 0$ has only the trivial solution. $\therefore$ If the equation $Ax= 0$ does not have only the trivial solution $\implies$ that the columns of the matrix $A$ are linearly dependent?

UPDATE I don't understand why a $3x4$ matrix is always linearly dependent.. what about $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$

where $x_1 = 0$ now $x_1= x_2 = x_3...$ then we can see that $x_1v_1 + x_2v_2 + x_3.. = 0 $ and we have the trivial solution?

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  • $\begingroup$ Re: rank, you're probably looking at a separate question altogether, but I like khan academy. TL;DR it's the dimension of the space spanned by your vectors. $\endgroup$ – janezdu Jun 28 '16 at 18:15
  • $\begingroup$ The last column is the 0 vector, which renders a matrix dependent automatically. $\endgroup$ – janezdu Jun 28 '16 at 18:19
  • $\begingroup$ Why does that render the matrix dependent automatically? $\endgroup$ – Yusha Jun 28 '16 at 18:21
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    $\begingroup$ See number 2: math.ku.edu/~lerner/LAnotes/Chapter11.pdf $\endgroup$ – janezdu Jun 28 '16 at 18:23
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    $\begingroup$ Great :) glad to help! $\endgroup$ – janezdu Jun 28 '16 at 18:25
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Why are the colums of

$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$ linearly dependent?

Because there exists non-zero $x$ such that

$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix} x = 0$

i.e.

$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}\begin{bmatrix} 0\\0\\0\\1\end{bmatrix} = \begin{bmatrix} 0\\0\\0\end{bmatrix}$

How do you prove that any $3\times4$ matrix has linearly dependent columns?

Suppose the columns of your matrix are $\mathbf v_1,\mathbf v_2,\mathbf v_3,\mathbf v_4.$ And suppose that $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are linearly independent. Then we want to show that there exists and $a,b,c$ such that $a\mathbf v_1 + b\mathbf v_2 + c\mathbf v_3 = \mathbf v_4$

How to do that? It might help to show that there exists $a_1,b_1,c_1$ such that:

$a_1\mathbf v_1 + b_1\mathbf v_2 + c_1\mathbf v_3 = \begin{bmatrix} 1\\0\\0\end{bmatrix}$

and similarly there is $a_2,b_2, c_2$ and $a_3, b_3, c_3$ such that

$a_2\mathbf v_1 + b_2\mathbf v_2 + c_2\mathbf v_3 = \begin{bmatrix} 0\\1\\0\end{bmatrix}$ and

$a_3\mathbf v_1 + b_3\mathbf v_2 + c_3\mathbf v_3 = \begin{bmatrix} 0\\0\\1\end{bmatrix}$

And certainly $\mathbf v_4$ can be composed as a combintation of $\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix},\begin{bmatrix} 0\\0\\1\end{bmatrix}$

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Your solution is almost there, you just need to incorporate the original question in. In other words, you need to show that $Ax=0$ has more than one solution when $A$ is $3 \times 4$.

I also like to look at this problem as working with 4 vectors in $\rm I\!R^3$. You can definitely construct one of them with a linear combination of the others, or one of them must be the $0$ vector.

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  • $\begingroup$ Like can I just show any example with numbers? $\endgroup$ – Yusha Jun 28 '16 at 18:11
  • $\begingroup$ What do you mean by that exactly? Your answer needs to be general, so you can only really use the numbers in the context of "3 rows", "4 columns", etc. $\endgroup$ – janezdu Jun 28 '16 at 18:11
  • $\begingroup$ Like I know how to show an example of it; but I don't know how to like "generalize it" $\endgroup$ – Yusha Jun 28 '16 at 18:12
  • $\begingroup$ Its like hes saying explain why instead of show an example, idk how to explain why really $\endgroup$ – Yusha Jun 28 '16 at 18:12
  • $\begingroup$ Ya i don't understand why a $3x4$ matrix is always linearly dependent.. what about $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$ $\endgroup$ – Yusha Jun 28 '16 at 18:16
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Without knowing how far you've gotten in your linear algebra class it's hard top produce a proof at the right level.

What's really going on here is that the four columns of a matrix with three rows are vectors in three dimensional space. Since the dimension of the space is three, any set with more than three vectors must be dependent. Of course you can't use that as a proof unless you've gotten that far in your studies.

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  • $\begingroup$ Yea, see thats the problem. That's why I'm confused on what he is wanting. $\endgroup$ – Yusha Jun 28 '16 at 18:10

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