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Today, the teacher in my class said that any similarity transform of a matrix is essentially a change of basis. So as a result, we end up with the same transformation, just with respect to a different basis. I did not follow that, could someone please explain the intuition behind how multiplying a matrix by $S$ from the left and $S^{-1}$ produces the same transformation but wrt to a different basis?

zvdsgf
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1 Answers1

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An $n\times n$ matrix is invertible if and only if the columns are a basis for $\mathbb{R}^n$. So you can view $S^{-1}$ as the change-of-basis matrix that translates from the basis whose vectors are the columns of $S^{-1}$ to the standard basis. Then $S$ is the matrix that translates from the standard basis to the basis whose vectors are the columns of $S^{-1}$.

So when you perform $SAS^{-1}$ you can view it as follows: you give it a coordinate vector in terms of the basis $\beta$ made up of the columns of $S^{-1}$. Then $S^{-1}$ translates this into the standard basis; then you apply $A$ as usual; then you apply $S$ and translate it back into the basis $\beta$. So you can view $SAS^{-1}$ as performing $A$, but in terms of the basis $\beta$.

Arturo Magidin
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  • excellent explanation! – user1816847 Oct 07 '14 at 23:14
  • In other words, if $SAS^{-1}$ is the matrix of the linear operator $T$ under the basis consisting of columns of $S^{-1}$, then $A$ is the matrix of $T$ under the standard basis and vice versa. – cxh007 Nov 20 '23 at 13:37
  • And if $SAS^{-1}$ is the matrix of $T$ under the standard basis, then $A$ is the matrix of $T$ under the basis consisting of columns of $S$. – cxh007 Nov 20 '23 at 13:55