Consider the following problem.

Let $V$ be an $n$-dimnensional $\mathbb{K}$-vector space and let $\textbf{A} \in \mathbb{K}^{n\times n}$ be an invertible matrix. Show that there are basis $B_1, B_2$ such that $\textbf{A}$ is the change of basis matrix from $B_1$ to $B_2$.

I found this problem, elementary as it may be, quite challenging, since in attempting to solve it I found I had several misconceptions on the theory of linear algebra. For now, I would like to know if my following approach is correct. I will also leave the theoretical comments I made when reasoning on the problem, since that may reveal any flaws either in my understanding or my procedure.


I : Let $B_2 = \{\beta_1, \ldots, \beta_n\}, B_1 = \{\alpha_1, \ldots, \alpha_n\}$ be two basis of $V$. We know there are scalars such that

\begin{align*} &\alpha_1 = x_1^{(1)}\beta_1 + \ldots + x_n^{(1)}\beta_n \\ &\ldots \\ &\alpha_n = x_1^{(n)}\beta_1 + \ldots + x_n^{(n)}\beta_n .\end{align*}

and that they define a unique transformation matrix

\begin{align*} \textbf{T}_{B_1 \to B_2} = \begin{pmatrix} x_1^{(1)} & \ldots & x_1^{(n)} \\ \ldots \\ x_n^{(1)} & \ldots & x_n^{(n)} \end{pmatrix} .\end{align*}


II : Let $(\lambda_1, \ldots, \lambda_n)_{B_1}$ be the representation of a vector $\textbf{v} \in V$ under basis $B_1$. We know

\begin{align*} &(\lambda_1, \ldots, \lambda_n)_{B_1} \equiv (\delta_1, \ldots, \delta_n)_{B_2} \\ \iff & \lambda_1 \alpha_1 + \ldots + \lambda_n \alpha_n = \delta_1 \beta_1 + \ldots + \delta_n \beta_n .\end{align*}

III : Consider the equation

\begin{align*} \textbf{A}\begin{pmatrix} \lambda_1 \\ \vdots\\ \lambda_n \end{pmatrix} &= \begin{pmatrix} \delta_1\\\vdots\\ \delta_n \end{pmatrix} .\end{align*}

Since $\textbf{A}$ is invertible, this equation has a solution. Particularly, such solution is given by

\begin{align*} \begin{pmatrix} \lambda_1\\ \vdots\\ \lambda_n \end{pmatrix} &= \textbf{A}^{(-1)}\begin{pmatrix} \delta_1\\ \vdots\\ \delta_n \end{pmatrix} \\ &= \begin{pmatrix} a_{11}\delta_1 + \ldots + a_{1n}\delta_n \\ \ldots & \\ a_{n1}\delta_1 + \ldots + a_{nn}\delta_n\end{pmatrix} .\end{align*}

In other words, for any $\textbf{v} = (\delta_1, \ldots, \delta_n)_{B_2}$, we can find a vector with coordinates in $B_1$ whose linear transformation via matrix $\textbf{A}$ maps to the coordinates $(\delta_1, \ldots, \delta_n)$. We can also atest the coordinates of such vector are of the form

\begin{align*} \Big(\sum_{n=1}^{\infty} a_{1i}\delta_i, \ldots, \sum_{n=1}^{\infty} a_{ni} \delta_i \Big) .\end{align*}

where $a_{ij}$ is an entry of $\textbf{A}^{(-1)}$.

  • $\begingroup$ To clear one thing, your point $2$ implies that if the coefficients of $v\in V$ in $B_1$ are equal to the coefficients of $w\in V$ in $B_2$, then, $v=w$? $\endgroup$
    – Lilla
    Jan 7, 2023 at 0:27
  • $\begingroup$ What if you just take $B_1$ to be the standard basis? $\endgroup$ Jan 7, 2023 at 3:54

1 Answer 1


To make the theoretical view point clearer let us illustrate with a simple example in $\mathbb R^2$.

If you have two basis related say $$c_1=4b_1+6b_2,$$ $$c_2=3b_1+5b_2.$$ then the associated matrix is $A=\left[\begin{array}{cc}4&3\\6&5\end{array}\right]$.

But if is added information of how the 1st basis is expressed in the canonical basis, say $$b_1=e_1+3e_2,$$ $$b_2=3e_1+e_2,$$ then the associate matrix is $B=\left[\begin{array}{cc}1&3\\3&1\end{array}\right]$.

So, subbing this there above, you get $$c_1=4(e_1+3e_2)+6(3e_1+e_2),$$ $$c_2=3(e_1+3e_2)+5(3e_1+e_2),$$ which upon simplification, you'll have $$c_1=22e_1+18e_2,$$ $$c_2=18e_1+14e_2,$$ with matrix $C=\left[\begin{array}{cc}22&18\\18&14\end{array}\right]$.

Observe that $BA=C$ then $A=B^{-1}C$.

Once an example is assimilated it is no hard to get the corresponding abstraction.


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