Let $$A = \begin{pmatrix} 1 & 1 & -3 \\ -3 & -2 & -2 \\ -7 & -5 & 1 \end{pmatrix}$$ be a $3$ by $3$ matrix and $$b = \begin{pmatrix} 4 \\ 6 \\ 8 \end{pmatrix}$$ be a column matrix. Are there any solutions to $Ax = b$ satisfying $\Vert x \rVert = 1$? Please explain, I am a bit confused on how to approach this, I know it's connected to the min norm solutions.
Any help will be appreciated.
This is a problem with norms. You need to know that the norm of a matrix $A$ is defined as the maximum norm you get for a vector $Ax=b$ when $x$ is a vector with norm $1$.
Let us use the $L_1$ vector norm which is the sum of the absolute values of the components, meaning in this case $4+6+8=18$ for your target product vector.
Now the corresponding $L_1$ norm for the matrix $A$ is the maximum of the norms of its column vectors. Here the first column has $L_1$ norm equal to $11$, the second column gives $8$ and the third column $6$. So the overall $L_1$ norm of your matrix is only $11$, thus the norm of the product $Ax$ with $x$ having unit norm can't reach the value of $18$ for your given $b$ vector. Hence our answer based on this norm is No.