# Find the linear transformation in terms of the canonical base

Q. Consider the following direct sum decomposition of subspaces of $$\mathbb{C}^4 = V_1 \oplus V_2\oplus V_3$$. The subspaces are defined by the canonical basis of $$\mathbb{C}^4$$ by the expressions $$V_1 =\text{span} \{ \vec{e_2}, \vec{e_4} \}$$, $$V_2 = \text{span} \{\vec{e_1} + \vec{e_3}\}$$, and $$V_3 =\text{span} \{\vec{e_1} − \vec{e_3}\}$$. If $$\vec{x}= \vec{x_1} + \vec{x_2}+ \vec{x_3}$$ is the decomposition of a vector $$\vec{x} \in \mathbb{C}^4$$ With respect to the previous sum, the linear operator is defined $$f(\vec{x}) = −\vec{x_1} + i\vec{x_2} − i\vec{x_3}$$. Find the linear operator $$f$$ in the canonical base.

This problem is quite different from what I'm used to be getting it confuses me and I'm having troubles with finding where to even start, I know that the direct sum means $$V_1,V_2$$ and $$V_3$$ have to be linearly independent. Which then I can only guess that we can substitute: $$f(\vec{x})=f(\vec{x_1}+\vec{x_2}+\vec{x_3})=-\vec{x_1}+i\vec{x_2}-i\vec{x_3}$$ and maybe go from there? But I'm pretty lost and it's supposedly the beginning of the problem that goes on.

• Start by expressing each $x_i$ in terms of the canonical basis. Then calculate $f(x)$ using that information. Finally reconstruct the matrix in terms of the canonical basis. Mar 24 at 1:54
• How do I do the first one? Like $\vec{x_1}=a\vec{e_2}+b\vec{e_4}$, $\vec{x_2}=c(\vec{e_1}+\vec{e_3})$ and so on in terms of some constants $a,b,c,...$? Mar 24 at 8:35
• Well $e_1=(e_1+e_3)+(e_1-e_3)$ so that $f(e_1)=f((e_1+e_3)+f(e_1-e_3)=i(e_1+e_3)-i(e_1-e_3)=2i e_3$, so there's the first column of you matrix; now $e_2\in V_1$ so $f(e_2)=-e_2$ and there's the second. I really think you should now do this yourself. Mar 27 at 6:54

There are two ways of doing this:

1. As others suggested in the comments --- which given the amount of structure this problem admits is probably the fastest --- one expresses each element of the new basis in terms of the old basis and evaluates $$f$$ on it using linearity. For example $$e_1=\frac12(e_1+e_3)+\frac12(e_1-e_3)$$ (and this decomposition is unique because $$\{e_2,e_4,e_1+e_3,e_1-e_3\}$$ is a basis of $$\mathbb C^4$$), so because $$f$$ is linear we compute \begin{align*} f(e_1)&=f\big(\tfrac12(e_1+e_3)+\tfrac12(e_1-e_3)\big)\\ &=\tfrac12f(e_1+e_3)+\tfrac12f(e_1-e_3)\\ &=\tfrac12 i(e_1+e_3)+\tfrac12 (-i(e_1-e_3))\\ &= (\tfrac12i-\tfrac12i)e_1+(\tfrac12i-(-\tfrac12i))e_3=0\cdot e_1+i\cdot e_3=ie_3\,. \end{align*} Doing the same with the other elements of the new basis yields $$f(e_2)=-e_2$$, $$f(e_4)=-e_4$$, and $$f(e_3)=$$

\begin{align*}=f\big(\tfrac12(e_1+e_3)-\tfrac12(e_1-e_3)\big)&=\tfrac12f(e_1+e_3)-\tfrac12f(e_1-e_3)\\&=\tfrac12 i(e_1+e_3)-\tfrac12 (-i(e_1-e_3))\\&= (\tfrac12i+\tfrac12i)e_1+(\tfrac12i-\tfrac12i)e_3=ie_1\,.\end{align*}

Finally the corresponding matrix has to achieve precisely this transformation, that is, $$e_1\mapsto f(e_1)=ie_3$$, $$e_2\mapsto f(e_2)=-e_2$$, and so on. This is what gives you the columns of this matrix, resulting in

$$\begin{pmatrix} 0&0&i&0\\ 0&-1&0&0\\ i&0&0&0\\ 0&0&0&-1\end{pmatrix}\,.$$

1. Another way is to work exclusively with matrices: The idea is to find a matrix which describes how one basis transforms into the other basis and then use it to "modify" the original representation matrix $$M=\begin{pmatrix} -1&0&0&0\\ 0&-1&0&0\\ 0&0&i&0\\ 0&0&0&-i \end{pmatrix}$$ to described how $$f$$ acts on the new basis. [Check for yourself that (a) this matrix indeed describes how $$f$$ transforms "the" old basis $$\{e_2,e_4,e_1+e_3,e_1-e_3\}$$, and (b) that --- although we arranged the old basis elements in a certain order here --- this does not matter for the new representation matrix because the order will be "corrected" by the basis transformation matrix]. Now to get more precise, if the matrix $$P$$ "maps the new basis to the old one" then the representation matrix we are looking for is given by $$P^{-1}MP$$. The reason this works can be sketched as follows: $$\substack{\text{vector in}\\\text{new coordinates}}\quad\overset{P}\longrightarrow\quad\substack{\text{vector in}\\\text{old coordinates}}\quad\overset{M}\longrightarrow \quad\substack{\text{image under }f\text{ in}\\\text{old coordinates}}\quad\overset{P^{-1}}\longrightarrow\quad\substack{\text{image under }f\text{ in}\\\text{new coordinates}}$$ As for the actual computation: Recall that the initial basis of $$V$$ is $$\{e_2,e_4,e_1+e_3,e_1-e_3\}$$ and the target basis is $$\{e_1,e_2,e_3,e_4\}$$. Then, because \begin{align*} \text{new basis vector}_1=e_1&=\frac12(e_1+e_3)+\frac12(e_1-e_3)\\ &=\frac12\text{old basis vector}_3+\frac12\text{old basis vector}_4\end{align*} the first column of $$P$$ ($$=Pe_{\bf 1}$$) is: $$\begin{matrix}0\\0\\\frac12\\\frac12\end{matrix}$$ Next $$e_2=\text{old basis vector}_1$$ so the second column of $$P$$ equals $$(1,0,0,0)^\top$$. Continuing this argument we eventually arrive at $$P=$$

$$\begin{pmatrix}0&1&0&0\\0&0&0&1\\\frac12&0&\frac12&0\\\frac12&0&-\frac12&0\end{pmatrix}$$

The last thing we need is the inverse of $$P$$. We can either compute $$P^{-1}$$ by inverting $$P$$ or --- because $$P^{-1}$$ describes how the old basis translates into the new one --- we can repeat the above process, but now the two bases are switched. For the latter, e.g., $$\text{old basis vector}_4=e_1-e_3=\text{new basis vector}_1-\text{new basis vector}_3$$ so the fourth column of $$P^{-1}$$ is given by $$(1,0,-1,0)^\top$$. Either way one finds $$P^{-1}=$$

$$\begin{pmatrix}0 &0&1&1 \\1 &0&0&0 \\0 &0&1&-1 \\0&1&0& 0 \end{pmatrix}$$

With this all that is left is to multiply these matrices in the correct order: computing $$P^{-1}MP$$ indeed yields the same matrix we found via the previous method.

As a final remark let me emphasize that in this example the first method is easier because $$f$$ in the old basis was already of a rather convenient form. For generic linear transformations the second method is better suited because expanding the image of $$f$$ in different bases can become quite cumbersome, and this method outsources this expansion step to a simple matrix multiplication.