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.

  • 1
    $\begingroup$ 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. $\endgroup$ Mar 24 at 1:54
  • $\begingroup$ 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 $a,b,c,...$? $\endgroup$
    – Aley20
    Mar 24 at 8:35
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    $\begingroup$ 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. $\endgroup$ Mar 27 at 6:54

1 Answer 1


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 $P=$


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.


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