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$$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right], \left[\begin{matrix}1&4\\0&3\end{matrix}\right]\bigg\}$$

How can I determine if a set of five $2\times2$ matrices are independent?

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    $\begingroup$ They can't be, View them as length-4 vectors and conclude. $\endgroup$ – Parcly Taxel Jul 25 '18 at 12:59
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Since the space of all $2\times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.

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    $\begingroup$ @Mi_11 consider accepting it as valid $\endgroup$ – Ander Biguri Jul 25 '18 at 15:43
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As has been pointed out, four matrices form a basis for the $2\times2$ matrices (the easiest would be $$ \left[\begin{matrix}1&0\\0&0\end{matrix}\right], \left[\begin{matrix}0&1\\0&0\end{matrix}\right], \left[\begin{matrix}0&0\\1&0\end{matrix}\right], \left[\begin{matrix}0&0\\0&1\end{matrix}\right] $$) so five matrices cannot be linearly dependent.

In your case the dependence is $$ \left[\begin{matrix}1&2\\2&1\end{matrix}\right] + \left[\begin{matrix}2&1\\-1&2\end{matrix}\right] + \left[\begin{matrix}0&1\\1&2\end{matrix}\right] - 2\left[\begin{matrix}1&0\\1&1\end{matrix}\right] - \left[\begin{matrix}1&4\\0&3\end{matrix}\right] = \left[\begin{matrix}0&0\\0&0\end{matrix}\right]. $$

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As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2\times 2$ matrices is $4$.

More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.

$$ a_1X_1 + a_2X_2 + \dotsb+ a_nX_n = 0 $$ If the only possible solution is $a_1 = a_2 = \dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.

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Stretch out the matrices to complete the rows of the following matrix $$\newcommand{\adj}{\operatorname{adj}} M(v)=\begin{bmatrix} v_1&1&2&2&1\\ v_2&2&1&-1&2\\ v_3&0&1&1&2\\ v_4&1&0&1&1\\ v_5&1&4&0&3 \end{bmatrix}\tag1 $$ Note that the top row of the adjugate of $M(v)$ $$ \begin{bmatrix} -14&-14&-14&28&14 \end{bmatrix}\tag2 $$ is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $\adj M(v)$. By Laplace's Formula, $$ \det M(v)=u\cdot v\tag3 $$ Setting $v$ to be any of the fixed columns of $M(v)$ gives $\det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.

We can rewrite the dot product of $-\frac1{14}u$ with the fixed columns of $M(v)$ as $$ 1\ \overbrace{\begin{bmatrix} 1&2\\2&1 \end{bmatrix}}^\text{row $1$} + 1\ \overbrace{\begin{bmatrix} 2&1\\-1&2 \end{bmatrix}}^\text{row $2$} + 1\ \overbrace{\begin{bmatrix} 0&1\\1&2 \end{bmatrix}}^\text{row $3$} -2\ \overbrace{\begin{bmatrix} 1&0\\1&1 \end{bmatrix}}^\text{row $4$} - 1\ \overbrace{\begin{bmatrix} 1&4\\0&3 \end{bmatrix}}^\text{row $5$} = \begin{bmatrix} 0&0\\0&0 \end{bmatrix}\tag4 $$

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