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Let's say we have some transformation matrix:

$$\begin{pmatrix} 2 & 2\\ 2 & 5 \end{pmatrix}$$

The eigen vectors are: $$\lambda_1 = 6 \quad \lambda_2 = 1$$

With eigenvectors: $$\begin{pmatrix} 0.5 \\ 1 \end{pmatrix} \quad \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$

And we know that, if we multiply this matrix by a (1,1) vector, we get: $$\begin{pmatrix} 2 & 2\\ 2 & 5 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$$

I have the question of "what do the eigenvalues and eigenvectors tell us about the transformation matrix". I know that the eigenvalue means scale and eigenvector means direction, but how do I get the (4,7) if I JUST know the eigenvalues and eigenvectors?

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If you only know the eigenvalues and the eigenvectors, then you can write ${1\choose 1}$ as a linear combination of these eigenvectors, and then you can use the fact that the given matrix acts as a linear operator on these vectors, and so you can apply the appropriate scalar to these vectors.

You need to solve $a\begin{pmatrix} 0.5 \\ 1 \end{pmatrix} + b\begin{pmatrix} -2 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, which is equivalent to

$$\begin{pmatrix} 0.5 & -2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$

So you can apply the inverse of this matrix to both sides and figure out what $a$ and $b$ are.

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    $\begingroup$ Thank you so much! This answers it well! $\endgroup$ – TheAkashain Oct 16 '20 at 0:39
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The two given eigenvectors span $\Bbb R^2$. So we can write $\left(\begin{smallmatrix}1\\1\end{smallmatrix}\right)$ as a linear combination of the two: $$ \begin{pmatrix}1\\1\end{pmatrix} = 1.2\begin{pmatrix}0.5\\1\end{pmatrix} - 0.2\begin{pmatrix}-2\\1\end{pmatrix} $$ Now we can apply your matrix to both sides here. Using linearity on the right-hand side, the result must be $$ \lambda_1\cdot1.2\begin{pmatrix}0.5\\1\end{pmatrix} - \lambda_2\cdot0.2\begin{pmatrix}-2\\1\end{pmatrix} $$ Just insert and calculate, and you get the answer.

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  • $\begingroup$ Oh! Thank you so much! $\endgroup$ – TheAkashain Oct 16 '20 at 0:39

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