# How to Tell If Matrices Are Linearly Independent

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.

• How is the matrix case different? Commented Jul 20, 2013 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 Commented Jul 20, 2013 at 21:38
• You are viewing these matrices as members of $\mathbb R^4$, so you can just straighten them as $4$-dimensional vectors. Commented Jul 20, 2013 at 21:39
• Amire meant $1\le i,j\le 2$ of course. Commented Jul 20, 2013 at 21:55
• @CameronBuie yes.Sorry. Commented Jul 20, 2013 at 22:03

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 $\alpha=\beta=0$.

• 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. Commented Jul 22, 2015 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.

Another alternative for testing is to check for the determinant for each matrices (this may look tedious for a complicated matrix system), If the determinant is non zero, It is said to be Linearly Independent, and if the determinant is zero, it is Linearly dependent

it is clear from the matrix system above that the determinant of the first matrix is [(1*1) - (0*2)] = 1

and the Second [(1*3) - (2*4)] = -5 hence , both are Linearly independent