Suppose \(A\) is a \(3 \times 3\) matrix and \(\vec{v}_1, \vec{v}_2, \vec{v}_3 \in \mathbb{R}^3\). Prove the following.
If \(\vec{v}_1, \vec{v}_2, \vec{v}_3\) are linearly dependent, then \(A\vec{v}_1, A\vec{v}_2, A\vec{v}_3\) are linearly dependent.
If \(\vec{v}_1, \vec{v}_2, \vec{v}_3\) are linearly independent and \(A\vec{v}_1, A\vec{v}_2, A\vec{v}_3\) are linearly dependent, then \(A\) is singular.
<1>2. The vectors \(Av_1,Av_2,Av_3\) are linearly dependent.
<1>3. Now assume \(v_1,v_2,v_3\) are linearly independent, while \(Av_1,Av_2,Av_3\) are linearly dependent. Then there exist scalars \(c_1,c_2,c_3\), not all zero, such that \[ c_1Av_1+c_2Av_2+c_3Av_3=0. \]
<1>4. Set \[ w=c_1v_1+c_2v_2+c_3v_3. \] Then \(w\ne0\) and \(Aw=0\).
<1>5. Therefore \(A\) is singular.
Solution
1. Assume \(v_1,v_2,v_3\) are linearly dependent. Then there exist scalars \(c_1,c_2,c_3\), not all zero, such that\[ c_1v_1+c_2v_2+c_3v_3=0. \]
Applying the linear map determined by \(A\) to the relation in 1 gives \[ c_1Av_1+c_2Av_2+c_3Av_3 =A(c_1v_1+c_2v_2+c_3v_3) =0. \] The coefficients are not all zero, so this is a nontrivial linear relation among the three images.
Because \(v_1,v_2,v_3\) are linearly independent and the coefficients are not all zero, we have \(w\ne0\). By linearity and 3, \[ Aw=c_1Av_1+c_2Av_2+c_3Av_3=0. \]
The linear map represented by \(A\) has the nonzero vector \(w\) in its kernel, so it is not injective. A square matrix is invertible if and only if its associated linear map is injective; hence \(A\) is not invertible, i.e. it is singular.