Determine the values for a for which the set of matrices are linear independent

Question

I wm given a set of matricies:

$S= \left( \begin{array}{cc} 1 & 0 \\ 1 & a \\ \end{array} \right), \left( \begin{array}{cc} -1 & 0 \\ a & 1 \\ \end{array} \right), \left( \begin{array}{cc} 2 & 0 \\ 1 & 3 \\ \end{array} \right) $

and $a\in \mathbb R$. I am told to find all the values of $a$ such that $S$ is a linearly independent set.

That would imply the linear combination of those $3$ matrices should equal $0$. Namely,

$ \alpha_1\left( \begin{array}{cc} 1 & 0 \\ 1 & a \\ \end{array} \right) + \alpha_2\left( \begin{array}{cc} -1 & 0 \\ a & 1 \\ \end{array} \right) + \alpha_3\left( \begin{array}{cc} 2 & 0 \\ 1 & 3 \\ \end{array} \right)=0$

should be true. Now, I can write the set of matrices as a single matrix, or basically,

$\left( \begin{array}{ccc} 1 & -1 & 2\\ 0 & 0 & 0\\ 1 & a & 1\\ a & 1 & 3\\ \end{array} \right)$

I am stuck at this point. How would I find the values of a? I believe I can row reduce this matrix and do something but I am not really sure how to go about it from here. I got the suggestion that I could simply take out the 0 row from this matrix and compute the determinant but I am not sure how I would justify that. Is that even valid?

Answer 1

You have done a good job, putting them in a matrix. If you write it in a different order, you get

$\left( \begin{array}{ccc} 1 & -1 & 2\\ a & 1 & 3\\ 1 & a & 1\\ 0 & 0 & 0\\ \end{array} \right) \left( \begin{array}{ccc} \alpha_1\\ \alpha_2\\ \alpha_3\\ \end{array} \right)=0$

Now, you should search for the null-space of the matrix. If the null-space is only the vector $\{0\}$, then columns of the matrix are independent. Also notice that you can totally ignore the last row, as any combination of it gives zero anyway.

A nonzero vector $(\alpha_1,\alpha_2,\alpha_3)^T$ is in the null-space of the matrix, if and only if it is in the null-space of

$\left( \begin{array}{ccc} 1 & -1 & 2\\ a & 1 & 3\\ 1 & a & 1\\ \end{array}\right)$

But this is a square matrix and we can use the determinant for it, in order to find values $a$, for which the columns are independent.

Answer 2

They are exactly the values that make that matrix of rank $3$. Id est, exactly the values of $a$ such that $$\det\begin{pmatrix}1&-1&2\\0&0&0\\1&a&1\end{pmatrix}\ne0\vee \det\begin{pmatrix}1&-1&2\\0&0&0\\a&1&3\end{pmatrix}\ne0\vee\det\begin{pmatrix}1&-1&2\\1&a&1\\a&1&3\end{pmatrix}\ne0\vee \det\begin{pmatrix}0&0&0\\1&a&1\\a&1&3\end{pmatrix}\ne0$$

Since all these determinants except the third one are $0$, this is equivalent to the inequation $$\det\begin{pmatrix}1&-1&2\\1&a&1\\a&1&3\end{pmatrix}\ne0$$

Answer 3

Think of it this way: You can arrange the matrices as columns for the $4\times 3$ matrix as you did, or you could also arrange them like this: $\left( \begin{array}{ccc} 0 & 0 & 0\\ 1 & -1 & 2\\ 1 & a & 1\\ a & 1 & 3\\ \end{array} \right)$. The only thing that matters, is that the $(i,j)_{th}$ entry of every matrix is in the same horizontal line. From here I believe it is easier to see why you can justify deleting the zero row?