Adding a constant to all elements of a matrix and solving

Question

Will solving $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1\\ -1 & 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 4 & 9 \\ -1 & 0 & 1 & 8 & 27 \\ 1 & 0 & 1 & 16 & 81 \\ \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \\ w_5 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}, $$ produce the same values for $w_i$ as solving for $$\begin{pmatrix} 1+a & 1+a & 1+a & 1+a & 1+a\\ -1+a & 0+a & 1+a & 2+a & 3+a \\ 1+a & 0+a & 1+a & 4+a & 9+a \\ -1+a & 0+a & 1+a & 8+a & 27+a \\ 1+a & 0+a & 1+a & 16+a & 81+a \\ \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \\ w_5 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}?$$ And if so, by which law of matrices ? Thank you.

Answer 1

This happens to be true for the particular matrix and vector that are given. Define the following matrix and vectors: $$ M = \pmatrix{1 & 1 & 1 & 1 & 1\\ -1 & 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 4 & 9 \\ -1 & 0 & 1 & 8 & 27 \\ 1 & 0 & 1 & 16 & 81}, \quad e = \pmatrix{1\\1\\1\\1\\1}, \quad e_2 = \pmatrix{0\\1\\0\\0\\0}, \quad w = \pmatrix{w_1\\w_2\\w_3\\w_4\\w_5}. $$ We are given that $Mw = e_2$. The first entry of $Mw$ is equal to the product $e^Tw$, so from $Mw = e_2$ we see that it must hold that $e^Tw = 0$.

The new equation that you present can be written as $$ (M + a ee^T)w = e_2. $$ We can see that this equation must hold because $$ (M + aee^T)w = Mw + aee^Tw = Mw + ae(e^Tw) = Mw + 0 = Mw = e_2. $$