Change of basis and diagonalization is a hassle for anyone new to the world of linear algebra; anyway for me. I was thinking I could post my full interpretation and then you guys could correct me if I'm wrong in fill in the gaps if I've missed anything etc. Maybe it also could work as future reference for others.

Current questions:

  1. First, are my interpretations below correct? Anything that's vague or indefinite?
  2. I'm especially unsure about the use of diagonalization for system of equations. Does it really have the same equal solution or is it dependent on the "new" basis or something like that? What are the further uses of diagonalization?
  3. How would one go about to find the change of basis matrix from $\mathcal A$ to $\mathcal B$? From e.g. base $\mathcal C$ to the standard basis it's simple, the transformation matrix consists of the basis vectors of $\mathcal C$ (although expressed in the standardbasis I guess, but it's simpler to grasp). Must one "go through" the standard basis if I want to go from $\mathcal A$ to $\mathcal B$?

Change of basis

Let $\mathcal A = \{\overrightarrow{a_1},\cdots,\overrightarrow{a_n}\}$ and $\mathcal B = \{\overrightarrow{b_1},\cdots,\overrightarrow{b_n}\}$ be two bases in $\mathbb R^n$. Then, the change of basis matrix from $\mathcal A$ to $\mathcal B$ is the matrix $T_{\mathcal A \rightarrow \mathcal B}$ which columns is the $\mathcal A$-basis vectors expressed in $\mathcal B$: $$T_{\mathcal A \rightarrow \mathcal B} = \begin{bmatrix}[\overrightarrow{a_1}]_\mathcal B&\cdots&[\overrightarrow{a_n}]_\mathcal B\end{bmatrix}$$ Then, for all vectors $\overrightarrow{x}$ in $\mathbb R^n$ the following applies: $$T_{\mathcal A \rightarrow \mathcal B}[\overrightarrow{x}]_\mathcal A=[\overrightarrow{x}]_\mathcal B$$

I.e. the change of basis matrix T that transforms a vector $\overrightarrow{x}$ from $\mathcal A$ to $\mathcal B$ is obtained by inserting $\mathcal A$'s base vectors expressed in $\mathcal B$ as columns. "The old base is expressed in the new".


  • $T_{\mathcal A \rightarrow \mathcal B} = (T_{\mathcal B \rightarrow \mathcal A})^{-1}$ (invers)

  • $T_{\mathcal A \rightarrow \mathcal C} = T_{\mathcal B \rightarrow \mathcal C}T_{\mathcal A \rightarrow \mathcal B}$ (multiple basis changes, note the order)

  • The above is in some sense also true for a linear transformation, since the transformation matrix have the images of its basis as columns. Though, a linear transformation can be from $\mathbb R^n$ to $\mathbb R^k$ and is not always invertible.


A square matrix $A$ can be decomposed to special form $A=SDS^{-1}$ (or the more commonly written and equal $D=S^{-1}AS$), where $S$ is the matrix consisting of eigenvectors to $A$ and $D$ is the diagonal matrix with its corresponding eigenvalues.

To do this one must find the eigenvalues and eigenvectors and if there are as many linear independent eigenvectors as the basis of the space which $A$ exists in, diagonalization is possible.

In other words, to diagonlize a matrix $A$ one must find a new basis which consists of its eigenvectors.

So, diagonalization is a way of expressing the same linear transformation but in another basis, which is easier to understand and visualize since they are mapped on themselves by a scalar of their corresponding eigenvalues.

Use in system of equations

  1. An initial equation $AX=Y$ can then be written as: $$SDS^{-1}X=Y$$
  2. By multiplying both sides with $S^{-1}$ we get: $$DS^{-1}X=S^{-1}Y$$
  3. Since the same linear transformation $S^{-1}$ is applied to both $X$ and $Y$ the initial equation is equivalent to the transformed one: $$DX'=Y'$$ Where $X'=S^{-1}X$ and $Y'=S^{-1}Y$
  4. So, to solve for $X$ one must first obtain $X'$ from $DX'=Y'$, which becomes easy since $D$ is diagonal. $X$ is then obtained from $X=SX'$

Therefore diagonalization is a way solving a system of equations.

  • $\begingroup$ Consider a punctured space with the metric in polar coordinates $ds^2=dr^2+r^2 d\Theta^2$. The old basis is the polar coordinates at the point of origin. The new basis is Cartesian coordinates shifted away from the point of origin. Can you express the old base in the new? $\endgroup$ – safesphere Dec 19 '17 at 15:39

Everything you wrote is correct, and clear.

As for solving the system, after you solve $DX'=Y'$, you actually want $X$. Since $X'=S^{-1}X$, you have $X=SX'$.

  • $\begingroup$ Okey, so to get X I must first obtain X' from DX'=Y'? Since it's diagonalized, that becomes easy I guess. Then I take S times X' to get the original X? Just to be clear, X' does not equal to X then? $\endgroup$ – Defindun Dec 19 '17 at 13:55
  • $\begingroup$ Exactly. $\ \ $ $\endgroup$ – Martin Argerami Dec 19 '17 at 14:19
  • $\begingroup$ In general, you won't be able to invert $D$. Solving the system is still trivial, though. $\endgroup$ – Martin Argerami Dec 19 '17 at 15:55
  • $\begingroup$ Okey, since multiplication with a diagonal matrix leaves us with a vector/solution with only one variable in each row corresponding to the value in Y? $\endgroup$ – Defindun Dec 19 '17 at 16:02
  • $\begingroup$ Yes. It is trivial. But $D$ might have zeroes in its diagonal and be not invertible. $\endgroup$ – Martin Argerami Dec 19 '17 at 16:03

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