Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.

Let's say I have the following matrices: $$A=\begin{bmatrix}-1&6\\-2&6\end{bmatrix}; B=\begin{bmatrix}1&2\\-1&4\end{bmatrix}$$ I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=\begin{bmatrix}2&0\\0&3\end{bmatrix}$.

If we let $x=\begin{bmatrix}2\\3 \end{bmatrix}$, we have: $$Ax=\begin{bmatrix}16\\14\end{bmatrix}; Bx=\begin{bmatrix}8\\10\end{bmatrix}$$

I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?

Thanks for help and sorry if an elementary question.


When you write

$$x = \begin{bmatrix} 2\\3\end{bmatrix}$$

what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $\mathbb{R}^2$. Usually, we assume that

$$v_1 = \begin{bmatrix} 1\\0\end{bmatrix}\qquad v_2 = \begin{bmatrix} 0\\1\end{bmatrix}, $$

but we don’t need to. Given a vector $x$ and a basis $\alpha = \{\alpha_1, \alpha_2\}$, if $x = a\alpha_1 + b\alpha_2$, I’ll be explicit and write that

$$[x]_{\alpha} = \begin{bmatrix} a\\b\end{bmatrix}. $$

Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $\alpha$ and $\beta$ such that given any $x\in\mathbb{R}^2$

$$A[x]_{\alpha} = B[x]_{\beta}.$$

  • $\begingroup$ Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other. $\endgroup$ – Andrew Jan 4 '19 at 9:20
  • $\begingroup$ You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{\alpha} = B[x]_{\beta}$. For two non-similar matrices, this won’t be possible. $\endgroup$ – Santana Afton Jan 4 '19 at 14:41

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