Let's say that we have two sets $B=(b_1, b_2, b_3)$ and $C=(c_1, c_2, c_3)$. These two sets represent two different bases of real vector space $V$. We know that these two sets (their elements to be precise) satisfy the following:
$$ c_1 = 2b_1 - b_2 - b_3 \\ c_2=-b_2\\ c_3 = 2b_2 + b_3 $$
Find all vectors that have the same coordinates with respect to both bases.
This is what i have so far:
Let $x \in V$ ( $x$ is a random vector from $V$), if $x=(x_1, x_2, x_3)$ then this vector meets the requirement given here if:
$$x_1 c_1 + x_2 c_2 + x_3 c_3=x_1 b_1 + x_2 b_2 + x_3 b_3 $$
if we express coefficients of second base in terms of the first one we have:
$$2x_1 b_1 + (-x_1-x_2+2x_3) b_2 + (-x_1+x_3) b_3=x_1 b_1 + x_2 b_2 + x_3 b_3$$
which leads me to conclusion that $x_1=0, x_2=x_3$
Which means that any vector from set $span{(0,1,1)}$ meets these requirements. This might be correct, but i have no intuition on this, how come this is possible, this means that it can happen that we have one same vector and we choose two different bases but we get same coordinates, this confuses me, i'd appreciate any intuitive explanation of this.
The base $C$ is obtained from the base $B$ using the matrix: $$ M=\begin{pmatrix} 2&0&0\\ -1&-1&2\\ -1&0&1 \end{pmatrix} $$ and you want the eigenvectors of this matrix for the eigenvalue $\lambda=1$ ( if they exists).
In this case $\lambda=1$ is really an eigenvalue of the matrix and you have found the corresponding eigenspace.