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If I have two matrices, for example: $\begin{pmatrix}1&0\\2&1 \end{pmatrix}$ and $\begin{pmatrix} 1&2\\4&3\end{pmatrix},$ how do I determine if they are linearly independent or not in $\mathbb{R}^4$?

I am familiar with checking for independence with vectors, such as by checking the determinant to be non-zero, or using the definition of linear independence such as $a(1,2)+b(2,3)=(0,0)$ and checking if $a=b=0$ is the only solution.

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How is the matrix case different? – Vedran Šego Jul 20 '13 at 21:36
The concept is the same. If you think about it a bit you will notice that they are lin. dependent iff $a_{i,j } =\alpha \cdot b_ {i,j } \ \ \ \ 1\leq i,j\leq 4$ $\ \ \alpha \neq 0 \ \ $ where $A$ and $B$ are the two matrices – sigmatau Jul 20 '13 at 21:38
You are viewing these matrices as members of $\mathbb R^4$, so you can just straighten them as $4$-dimensional vectors. – Tunococ Jul 20 '13 at 21:39
Amire meant $1\le i,j\le 2$ of course. – Cameron Buie Jul 20 '13 at 21:55
@CameronBuie yes.Sorry. – sigmatau Jul 20 '13 at 22:03
up vote 7 down vote accepted

To show if two matrices are independent, you do exactly what you always do: if your matrices are $A$ and $B$, you want to show that $\alpha A+\beta B=0$ for $\alpha,\beta\in\mathbb{R}$ (or $\mathbb{C}$, depending) if and only if $\alpha=\beta=0$.

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Thank you for your help. – Sujaan Kunalan Jul 21 '13 at 13:59
Equivalently, if there are $\alpha$ and $\beta$ such that at least one of them is nonzero, then if $A$ and $B$ are nonzero this forces the other coefficient to be nonzero. So solving, we get $\gamma A = B$ where $\gamma = -\alpha / \beta$. In other words, two non-zero matrices are linearly dependent if and only if one is a multiple of the other. – Squirtle Jul 22 '15 at 17:32

Remember that a matrix $X = (x_{ij})$ can be replaces by the vector given by reading the rows one after another. Your two matrices can be indentified with the vectors $(1,0,2,1)$ and $(1,2,4,3)$.

Let $M:=(m_{ij})$ and $N:=(n_{ij})$ be your two matrices. If you can find a unique $\lambda$ for which $M = \lambda N$ then $M$ and $N$ are not linearly independent. You can compare the matrices entry-by-entry. Look at the corresponding entries in both matrices and ask yourself:

Is there a unique value of $\lambda$ for which $m_{ij} = \lambda \, n_{ij}$ (for all $i$ and $j$) ?

Keep asking this question entry by entry. If you manage to get through all of the entries and there is a single value of $\lambda$ for which $m_{ij} = \lambda n_{ij}$ then $M$ and $N$ are linearly dependent. If, at any time, you need a new value for $\lambda$ then $M$ and $N$ are linearly independent.

Let's look at your example: $m_{11} =1$ and $n_{11} = 1$ so $m_{11} = \lambda n_{11} \iff \lambda = 1$. Next, look at $m_{12} = 0$ and $n_{12} = 2$. We have $m_{12} = \lambda n_{12} \iff \lambda = 0$. We a non-unique value of $\lambda$ meaning $M$ and $N$ are linearly independent.

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