Answer to Question #150821 in Linear Algebra for shanto

Question

Find a basis and the dimension of the subspace w of V spanned by the matrices

A=[ 1 2
-1 3]
B=[ 2 5
1 -1]
C=[3 4
-2 5]

Answer 1

Let us find a basis and the dimension of the subspace $W$ of $V$ spanned by the matrices:

$A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]$, $B=\left[\begin{array}{cc} 2 & 5 \\ 1 & -1 \end{array}\right]$, $C=\left[\begin{array}{cc} 3 & 4 \\ -2 & 5 \end{array}\right]$ .

Let us consider a linear combination $kA+mB+nC=0.$ It follows that

$k\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+m\left[\begin{array}{cc} 2 & 5 \\ 1 & -1 \end{array}\right]+n\left[\begin{array}{cc} 3 & 4 \\ -2 & 5 \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]$

$\left[\begin{array}{cc} k+2m+3n & 2k+5m+4n \\ -k+m-2n & 3k-m+5n \end{array}\right] =\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]$

Therefore, we have the following system:

$\begin{cases}k+2m+3n=0\\ 2k+5m+4n=0 \\ -k+m-2n=0 \\ 3k-m+5n=0 \end{cases}$ which is equivalent to $\begin{cases}k+2m+3n=0\\ m-2n=0 \\ 3m+n=0 \\ -7m-4n=0 \end{cases}$ and to

$\begin{cases}k+2m+3n=0\\ m-2n=0 \\ 7n=0 \\ -18n=0 \end{cases}$

It follows that $n=m=k=0$. Consequently, the matrices are linearly independent, and thus form a basis of the subspace $W$ of $V$ spanned by them. Also $\dim W=3.$