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I have this assignment where I have three matrices:

$$ A_1 = \pmatrix{1 & 1 \\ -1 & 2},\quad A_2 = \pmatrix{-1&-4 \\ 1&1},\quad A_3 = \pmatrix{1&2\\-1&1}$$

The first I have to do is to determine whether they are linear independent or not:

I should therefore prove this: $c_1A_1+c_2A_2+c_3A_3=0$. If I put this together I get the matrix: $$\pmatrix{c_1-c_2+c_3&c_1-4c_2+2c_3\\-c_1+c_2-c_3&2c_1+c_2+c_3} = \pmatrix{0&0\\0&0}$$

I could put $A_1, A_2$ and $A_3$ as colonvectors like this:

$$\pmatrix{1&-1&1\\1&-4&2\\-1&1&-1\\2&1&1}\sim\pmatrix{1&0&\frac{2}{3}\\0&1&-\frac{1}{3}\\0&0&0\\0&0&0}$$

I know that this should not be linear independent, but I am not really sure how I know that... Is it because of the zero-lines? Or is it because there is two non-zero values in the same column?

The other part of the assignment is to determine the $\operatorname{span}(A_1, A_2, A_3)$ and the dimension of the span. How can I continue with that?

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    $\begingroup$ Remember that the dimension of the space spanned by a set $\cal A$ of vectors is the maximum number of linearly independent vectors within $\cal A$. Your three matrices are linearly dependent, so their span has dimension $<3$. Can you find two linearly independent matrices among those three? $\endgroup$ – Andrea Mori Feb 19 '14 at 10:00
  • $\begingroup$ Is the entry in the second row, first column of $A_2$ supposed to be $1$ instead of $-1$? If so, your row reduced matrix is correct and my answer applies. $\endgroup$ – Christoph Feb 19 '14 at 10:14
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    $\begingroup$ Yes, I have edited it now, sorry! $\endgroup$ – theva Feb 19 '14 at 10:25
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In the row reduced matrix you can see that $$\frac 2 3\cdot\text{first coloumn}-\frac 1 3\cdot\text{second column}=\text{third column}.$$ Since you only used row-operations, the same is true for the non-recuded matrix, so $$\frac 2 3 A_1 - \frac 1 3 A_2 = A_3.$$ The matrices are linearly dependent indeed.

This also helps us to find $\operatorname{span}(A_1, A_2, A_3)$, since the above means $A_3\in\operatorname{span}(A_1, A_2)$, thus $$ \operatorname{span}(A_1, A_2, A_3) = \operatorname{span}(A_1, A_2) = \left\{ \lambda_1\pmatrix{1 & 1 \\ -1 & 2} + \lambda_2\pmatrix{-1 & -4 \\ 1 & 1} \ \Bigg|\ \lambda_1,\lambda_2\in\mathbb R \right\}. $$ Since $A_1, A_2$ are linearly independent (check this), the dimension of the span is $2$.

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    $\begingroup$ the row reduced matrix has a mistake in the third line. It should be $-1 -1 -1$ instead of $-1 +1 -1$ as you also noticed $\endgroup$ – Jimmy R. Feb 19 '14 at 10:05
  • $\begingroup$ Sorry, there was a minus too much in $A_2$... I have edited it now. $\endgroup$ – theva Feb 19 '14 at 10:23
  • $\begingroup$ Is $span(A_1,A_2,A_3) = span(A_1,A_3)$ also correct? Does the vectors that span also form a basis for the set? $\endgroup$ – theva Feb 19 '14 at 10:38
  • $\begingroup$ Yes in your case any two of the three matrices are linearly independent and so any two form a basis of the span. $\endgroup$ – Christoph Feb 19 '14 at 10:38
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Hint: Why don't you solve the system that you found \begin{cases} c_1-c_2+c_3=0 \\ -c_1-c_2-c_3=0 \\ c_1-4c_2+2c_3=0 \\ 2c_1+c_2+c_3=0 \end{cases} to establish that it has no solutions other than $c_1=c_2=c_3=0$. Note: you have a mistake in the second equation (down left of the matrices). The matrices are independent! The second equation (down left) should read $-c_1-c_2-c_3$ instead of $-c_1+c_2-c_3$ as you have it.

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As your colonvectors manipulation shows, the matrices are not linearly independent. This is because you have only $2$ non-zero rows and $3$ columns. It's also a good idea to understand why this means linear dependece. Really, what you did was ask if there exist such nontrivial $c_1,c_2,c_3$ so that $$\pmatrix{1&-1&1\\1&-4&2\\-1&1&-1\\2&1&1}\pmatrix{c_1\\c_2\\c_3} = \pmatrix{0\\0\\0\\0}.$$

By manipulating the equations, you get an equivalent set of equations. You then know that you can find the $c_1,c_2,c_3$ if and only if you can find such $d_1,d_2,d_3$ that $$\pmatrix{1&0&\frac{2}{3}\\0&1&-\frac{1}{3}\\0&0&0\\0&0&0}\pmatrix{d_1\\d_2\\d_3} = \pmatrix{0\\0\\0\\0}.$$

These $d_i$ can obviously be found. Choose any $d_3$, then from $d_2 - \frac{1}{3} d_3 = 0$, you get $d_2$, and from $d_1 +\frac23 d_3 = 0$, you get $d_1$.

As for the span, it is easy to show that any two of the matrices are linearly independent, meaning any $2$ of them form the basis of the span (which, therefore, has dimension $2$).

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    $\begingroup$ the matrix had a mistake. The third line should be $-1, -1, -1$ instead of $-1, +1, -1$. $\endgroup$ – Jimmy R. Feb 19 '14 at 10:00
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You almost completed the task.

You should further your knowledge of "rank". The last matrix that you wrote has rank=2 (take the identity sub-matrix $(a_{ij})_{i,j=1,2}$).

It means that $<A_1,A_2,A_3>=<A_1,A_2>$.

You can say that because of null row (or columns). Every time you see a null row vector and\or a null column vector you must decrement rank by 1, and removing the corresponding item in the base.

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