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I have the matrix $\left( \begin{array} {cc} 1 && 1 \\ -1 && -1 \end{array} \right)$. When applied to the bases $\left( \begin{array} {c} 1 \\ -1 \end{array} \right)$ and $\left( \begin{array} {c} 1 \\ 0 \end{array} \right)$ I get a resulting matrix with respect to this basis which is $\left( \begin{array} {cc} 0 && 1 \\ 0 && 0 \end{array} \right)$.

My question is: what does it mean to have a matrix with respect to another basis and why is it significant?

I know that this does not mean the matrices are equivalent, otherwise a simple Gaussian elimination would have done that.

Furthermore, applying each of the matrices to, say, the vector $\left( \begin{array} {c} 1 \\ 2 \end{array} \right)$ you get different answers.

So if my resulting matrix is not just a simplified version of the other, what is it exactly?

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  • $\begingroup$ Matrix can be interpreted as a numerical record of linear map (and "vector" of yours - as a numerical record of vector, element of linear space). When you change basis of either source or destination linear space, you change numbers, but object remains the same (this can be seen similar to character encoding with different codepages or expressing the same price in different currencies). With change of basis, numerical record of vector changes as well. There are bases for which numerical record of given linear map is particularly simple. $\endgroup$ – Abstraction Nov 12 '15 at 10:41
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Suppose that $A$ has the matrix $B$ with respect to the basis $\{v_1,v_2\}$. What this means is that whenever $$ A(a v_1 + b v_2) = cv_1 + dv_2 $$ we also have $$ B \pmatrix{a\\b} = \pmatrix{c\\d} $$ Put another way, this means that "$v_1$ and $v_2$ act under $A$ in the exact same way that the standard basis vectors $(1,0)$ and $(0,1)$ under $B$". Yet another way of saying this: "$B$ is the same linear transformation as $A$, with $v_1$ 'relabeled' as $(1,0)$ and $v_2$ 'relabeled' as $(0,1)$".

Note that all of this only really makes sense when $A$ is a square matrix. In particular, as a linear transformation, it takes the span of $v_1$ and $v_2$ (which is $\Bbb R^2$) to the span of $v_1$ and $v_2$, which means that both the inputs and outputs can be (and are) relabeled in the same way.

So, for example, if $A$ is equal to $B$, then $A$ and $B$ will have the same rank, determinant, trace, and as you will soon discover, eigenvalues. If $A$ is row-reduced to get $B$, then all $A$ and $B$ have in common is their null space.

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