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Example question about linear transformations

Question: you are given a linear transformation \(T: \mathbb{R}^3\to\mathbb{R}^3\), with \(\vec{\mathbf v}_1=\begin{bmatrix}1\\2\\3\end{bmatrix}\), \(\vec{\mathbf v}_2=\begin{bmatrix}0\\-4\\2\end{bmatrix}\), \(\vec{\mathbf v}_3=\begin{bmatrix}-1\\0\\0\end{bmatrix}\), \(\vec{\mathbf w}_1=\begin{bmatrix}2\kappa+3\\14\\6\xi+10\end{bmatrix}\), \(\vec{\mathbf w}_2=\begin{bmatrix}-4\kappa\\14\\4\xi-16\end{bmatrix}\) and \(\vec{\mathbf w}_3=\begin{bmatrix}-3\\-1\\-2\end{bmatrix}\), where it is given that \(T({\vec{\mathbf{v}}_1})=\vec{\mathbf{w}}_1\), \(T({\vec{\mathbf{v}}_2})=\vec{\mathbf{w}}_2\) and \(T({\vec{\mathbf{v}}_3})=\vec{\mathbf{w}}_3\).

(a): Find the matrix \(A\) of this transformation (it should contain \(\kappa\) and \(\xi\)).

(b): From now on, take \(\kappa=\xi=0\) and find the eigenvalues of the matrix \(A\).

(c): Find the eigenvectors of the matrix \(A\).
Solution (a): In order to find the matrix corresponding to the transformation, a useful thing to do it to determine to which vectors the unit vectors map. In other words, we want to find out what are \(T\left(\begin{bmatrix}1\\0\\0\end{bmatrix}\right)\), \(T\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right)\) and \(T\left(\begin{bmatrix}0\\0\\1\end{bmatrix}\right)\).
(The matrix \(A\) will then consist of these three column vectors glued together in order.)

Let's start by observing that \(\vec{\mathbf v}_3=\begin{bmatrix}-1\\0\\0\end{bmatrix}\), is almost a unit vector.
Since this is a linear transformation, we can say the following: $$T\left(\begin{bmatrix}1\\0\\0\end{bmatrix}\right) = -T\left(\begin{bmatrix}-1\\0\\0\end{bmatrix}\right) = -T\left(\vec{\mathbf v}_3\right)=-\begin{bmatrix}-3\\-1\\-2\end{bmatrix}=\red{\begin{bmatrix}3\\1\\2\end{bmatrix}}$$ That's already our first transformed unit vector (of the first dimension)! We now know that the first column of \(A\) must be \(\begin{bmatrix}3\\1\\2\end{bmatrix}\).

Now we should do something with \(\vec{\mathbf v}_1\) and \(\vec{\mathbf v}_2\). We cannot immediately create another unit vector, so as an intermediate step, let's add \(\vec{\mathbf v}_1\) and \(\vec{\mathbf v}_3\) together such that we get more zeroes ("working towards a unit vector"): \begin{equation}T\left(\vec{\mathbf v}_1+\vec{\mathbf v}_3\right)=\boxed{T\left(\begin{bmatrix}\blue 0\\2\\3\end{bmatrix}\right)=\begin{bmatrix}2\kappa\\13\\6\xi+8\end{bmatrix}}=T\left(\vec{\mathbf v}_1\right)+T\left(\vec{\mathbf v}_3\right)\end{equation} Now, let's add two times the vector \(\begin{bmatrix} 0\\2\\3\end{bmatrix}\) to \(\vec{\mathbf v}_2=\begin{bmatrix}0\\-4\\2\end{bmatrix}\), such that the middle number becomes a zero: $$T\left(2\cdot\begin{bmatrix} 0\\2\\3\end{bmatrix}+\begin{bmatrix}0\\-4\\2\end{bmatrix}\right)=\boxed{T\left(\begin{bmatrix}0\\\blue0\\8\end{bmatrix}\right)= \begin{bmatrix}0\\40\\16\xi\end{bmatrix}} = 2\cdot\begin{bmatrix}2\kappa\\13\\6\xi+8\end{bmatrix}+\begin{bmatrix}-4\kappa\\14\\4\xi-16\end{bmatrix}$$ From the boxed equation we now infer (dividing by \(8\)): \begin{equation}T\left(\begin{bmatrix}0\\0\\1\end{bmatrix}\right)= \red{\begin{bmatrix}0\\5\\2\xi\end{bmatrix}}\end{equation} And that is the second unit vector whose image (under the transformation) we found! So, the red thing will be the third row of the matrix, because it is a unit vector of the third dimension.

Now all we need to do is to find the image of the unit vector of the second dimension.
Using the result of equations ?? and ??, we can write $$T\left(\begin{bmatrix} 0\\2\\3\end{bmatrix}-3\cdot\begin{bmatrix}0\\0\\1\end{bmatrix}\right)=\boxed{T\left(\begin{bmatrix}0\\2\\0\end{bmatrix}\right)= \begin{bmatrix}2\kappa\\-2\\8\end{bmatrix}} = \begin{bmatrix}2\kappa\\13\\6\xi+8\end{bmatrix}-3\cdot\begin{bmatrix}0\\5\\2\xi\end{bmatrix}$$ Divide by \(2\) in the boxed equation to obtain the image of the unit vector of the second dimension: $$T\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right)= \red{\begin{bmatrix}\kappa\\-1\\4\end{bmatrix}}$$ Now that we found the images under the transformation of all three unit vectors, we can glue them together (the red vectors) in the correct order to obtain our matrix: $$\boxed{\boxed{A=\begin{bmatrix} 3&\kappa&0\\ 1&-1&5\\ 2&4&2\xi \end{bmatrix}}}$$

Solution (b): Let's take \(\kappa=\xi=0\). The matrix \(A\) now becomes $$A=\begin{bmatrix} 3&0&0\\ 1&-1&5\\ 2&4&0 \end{bmatrix}$$ In order to find the eigenvalues, we must solve the equation \(\det(A-\lambda I)=0\). The solutions \(\lambda\) will be the eigenvalues. $$\det(A-\lambda I)=\det\begin{bmatrix} 3-\lambda&0&0\\ 1&-1-\lambda&5\\ 2&4&-\lambda \end{bmatrix}=0$$ $$(3-\lambda)((-1-\lambda)(-\lambda) - (4)(5)) = 0$$ $$(3-\lambda)(\lambda^2+\lambda-20)=0$$ $$(3-\lambda)(\lambda-4)(\lambda+5)=0$$ So we get the eigenvalues $$\boxed{\lambda_1=-5\quad\lambda_2=3\quad\lambda_3=4}$$

Solution (c): In order to find the eigenvectors, we have to substitute the eigenvalues in the equation \((A-\lambda I)\vec{\mathbf x}=\vec{\mathbf 0}\) and solve for \(\vec{\mathbf x}\).

Let's first do the calculation for the first eigenvalue, \(\lambda_1=-5\). We then get the matrix \(A-(-5)I=\begin{bmatrix} 8&0&0\\ 1&4&5\\ 2&4&5 \end{bmatrix}\) Since we are trying to solve \((A-\lambda I)\vec{\mathbf x}=\vec{\mathbf 0}\), we can set up the augmented matrix: $$\begin{bmatrix} 8&0&0 &\mid& 0\\ 1&4&5 &\mid& 0\\ 2&4&5 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} 8&0&0 &\mid& 0\\ 0&4&5 &\mid& 0\\ 0&4&5 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} 8&0&0 &\mid& 0\\ 0&4&5 &\mid& 0\\ 0&0&0 &\mid& 0 \end{bmatrix}$$ This results in the eigenvectors \({t\begin{bmatrix}0\\5\\-4\end{bmatrix}}\) for all \(t\in\mathbb{R}\).

Let's now do the calculation for the second eigenvalue, \(\lambda_2=3\). We then get the matrix \(A-3I=\begin{bmatrix} 0&0&0 \\ 1&-4&5 \\ 2&4&-3 \end{bmatrix}\) The augmented matrix becomes $$\begin{bmatrix} 0&0&0 &\mid& 0\\ 1&-4&5 &\mid& 0\\ 2&4&-3 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} 0&0&0 &\mid& 0\\ 1&-4&5 &\mid& 0\\ 3&0&2 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} 0&0&0 &\mid& 0\\ 3&-12&15 &\mid& 0\\ 3&0&2 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} 0&0&0 &\mid& 0\\ 0&-12&13 &\mid& 0\\ 3&0&2 &\mid& 0 \end{bmatrix}$$ By setting for example \(x_3\) free and solving for the others, we get the eigenvectors \(x_3\begin{bmatrix}-2/3\\13/12\\1\end{bmatrix}\) for all \(x_3\in\mathbb{R}\). Those eigenvectors can be written more nicely as \({t\begin{bmatrix}-8\\13\\12\end{bmatrix}}\) for all \(t\in\mathbb R\).

Finally, let us consider the eigenvectors corresponding to the third eigenvalue \(\lambda_3=4\). This time we get $$\begin{bmatrix} -1&0&0 &\mid& 0\\ 1&-5&5 &\mid& 0\\ 2&4&-4 &\mid& 0 \end{bmatrix}\sim\begin{bmatrix} -1&0&0 &\mid& 0\\ 0&-5&5 &\mid& 0\\ 0&4&-4 &\mid& 0 \end{bmatrix}$$ From which we get the eigenvectors \({t\begin{bmatrix}0\\1\\1\end{bmatrix}}\) for all \(t\in\mathbb R\).

We now did the calculations for all three eigenvalues. We can now conclude that the eigenvectors are the vectors $$\boxed{\boxed{\vec{\mathbf u}_1=\begin{bmatrix}0\\5\\-4\end{bmatrix}, \vec{\mathbf u}_2=\begin{bmatrix}-8\\13\\12\end{bmatrix}, \vec{\mathbf u}_3=\begin{bmatrix}0\\1\\1\end{bmatrix}}}$$ Any constant multiple of one of these vectors is also an eigenvector!