# Problem with linear transformation from $\mathbb{R}^2$ to $M_{2\times 2}$

I'm trying to solve this problem, but at the end I find something's wrong with my work. Here is the problem:

We're given the bases: $$\beta = \bigg\{\begin{pmatrix}1\\1\end{pmatrix} \begin{pmatrix}1\\-1\end{pmatrix}\bigg\} \\ \gamma = \text{Canonical for }\mathbb{R}^4$$

We're given the transformation from the vectores of the basis to the image: $$T((1,1)) = \begin{pmatrix}1\\0\\0\\1\end{pmatrix} \\ T((1,-1)) = \begin{pmatrix}0\\1\\-1\\0\end{pmatrix}$$

And were asked to:

• Find $[T]_\beta^\gamma$
• If $x = (5,1)$, find $[x]_\beta$
• Determine $[T(x)]_\gamma$
• Find $T(x)$

The first part I think is simple, since all we need to do is find the transformation of the base for $\beta$, but it is already given, and since $\gamma$ is the canonical, the transformations should be $[T]_\beta^\gamma$.

The second part is the application of the definition, so we find that $(5,1)$ may be represented by the coordinate vector $(3,2)$.

Then, for number three, we use the theorem that states that $[T(x)]_\gamma = [T]_\beta^\gamma[x]_\beta$. That is $(3,2,-2,3)$.

Finally, we use the coordinate vector and use the entries as coefficients for $\gamma$ to find $T(x)$. This is $(3,2,-2,3)$, since $\gamma$ is the canonical basis.

The problem is that if I try to multiply $[T]_\beta^\gamma$ by the vectors of $\beta$, I don't get the transformation I'm expecting, according to what is given. What am I doing something wrong?

• If something is unclear, please do tell me. – Cehhiro May 17 '14 at 16:57
• Are you using $$\left[ \begin{array}{c} 1 \\ 1 \end{array}\right]_\beta=\left[ \begin{array}{} 1 \\ 0 \end{array}\right]\;\;?$$ – Zircht May 17 '14 at 16:59
• @ᛥᛥᛥ Yes. (You mean that $(1,1)$ as a combination of $\beta$ is of $(1,0)$, right?) – Cehhiro May 17 '14 at 17:02
• @ᛥᛥᛥ Is $[T]_\beta^\gamma$ the matrix that, when multiplied by a vector, gives you the result of the transformation, or that's completely wrong? – Cehhiro May 17 '14 at 17:35
• Let's say $\beta=\{v_1,v_2\}$. When you compute $[T]_\beta^\gamma v$ and $v=(a,b)^T$, you get the image of the vector $av_1+bv_2$. If you want the transformation $T$, you need to compute $[T]_\beta^\gamma [I]_{\textbf{C}}^\beta v$ where the matrix on the right is the chage of basis matrix from the canonical basis in $\Bbb R^2$ to $\beta$. – Zircht May 17 '14 at 17:50

## 1 Answer

With the help of ᛥᛥᛥ:

I was thinking of something else. The process here is correct.

My problem was that I was expecting $[T]_\beta^\gamma$ to do what $[I]_C^\beta$ does, which is to transform a vector $x$ to the image, or do what $T(x)$ does. $[I]_C^\beta$ is the matrix formed by the transform of the canonical basis of the domain with $T(x)$.