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I'm trying to calculate the matrix corresponding to the linear transformation $T:V -> V$, where $V$ is a subspace of $R^3$ (three dimensional Euclidean space) of dimension 2. Let the basis of $V$ be given by $\{b_1,b_2\}$.

So my attempt was:

Let $v$ be a vector in $V$ so that: $T(v)=A*v=v'$, where A has to be of size $3\times3$, since $v$ is $3\times1$.

Let $v=c_1b_1+c_2b_2$, $v'=d_1b_1+d_2b_2$, and so we have:

$A(c_1b_1+c_2b_2)-d_1b_1-d_2b_2=0$

$(c_1A-d_1I)b_1+(c_2A-d_2I)b_2=0$

Since $b_1$ and $b_2$ are linearly independent (not sure about this step and the following) then:

$(c_1A-d_1I)b_1=0$

$(c_2A-d_2I)b_2=0$

So $A$ is the matrix that has eigenvectors $b_1$ and $b_2$ with eigenvalues $\lambda_1=d_1/c_1$ and $\lambda_2=d_2/c_2$ respectively. In order to obtain an explicit matrix representation of A I tried using diagonalization, $A=SMS^{-1}$, where $$M= \left( \begin{matrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3=?\\ \end{matrix}\right) $$ $$S= \left( \begin{matrix} b_1 & b_2 & b_3\\ \end{matrix}\right) $$ What should the value of $\lambda_3$ be? zero doesn't seem to be correct. Should I find the value of the vector $b_3$ so that $\{b_1,b_2,b_3\}$ forms a basis for $R^3$ and then obtain $S$? Is there any easier way to solve for $A$?

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  • $\begingroup$ It can be anything, since your problem does not specify what happenshe outside the subspace $\endgroup$ – Paul Jun 2 '17 at 21:46
  • $\begingroup$ Every vector in $V$ will map to $(x,y,0)$ after the change in basis is applied. So $\lambda_3$ becomes irrelevant. However, if you set $\lambda_3=0$ then if you apply $A$ to a vector not in $V$ the transformed vector will be projected onto $V.$ $\endgroup$ – Doug M Jun 2 '17 at 22:27
  • $\begingroup$ @DougM According to OP, $\operatorname{dom} T = V$. So it doesn't even make sense to talk about applying $T$ (or its matrix representation) to a vector outside of $V$. $\endgroup$ – user137731 Jun 2 '17 at 22:45
  • $\begingroup$ @Bye_World I realize that, but I was trying to point out what would be the implications of that choice of $\lambda_3$. It wouldn't be the first time someone changed the domain of a mapping after defining the map. $\endgroup$ – Doug M Jun 2 '17 at 22:47
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The matrix representation of $T$ will be a $2\times 2$ matrix (because $V$ is 2-dimensional). The most you'll be able to say about it though is that it will have the form $$[T]_{\{b_1,b_2\}} = \begin{bmatrix} | & | \\ [T(b_1)]_{\{b_1,b_2\}} & [T(b_2)]_{\{b_1,b_2\}} \\ | & |\end{bmatrix}$$

Without choosing a specific $T$ there's really not anything else you can say about its matrix representation.

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  • $\begingroup$ Thanks! I was looking for something like this. By the way, why would setting both parts to zero (when I argued that b1 and b2 where linearly independent) be wrong? $\endgroup$ – Cami77 Jun 3 '17 at 10:46
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    $\begingroup$ If $b_1$ and $b_2$ are linearly independent then $$a_1b_1 + a_2b_2 = 0 \implies a_1=0=a_2$$ where $a_1, a_2$ are scalars. If $a_1,a_2$ are matrices, then they change the directions of the $b_i$'s (generally speaking) and hence $a_1b_1$ and $a_2b_2$ are not necessarily linearly independent. All you can conclude from $a_1b_1+a_2b_2=0$ in this case is that $$a_1b_1 = -a_2b_2$$ $\endgroup$ – user137731 Jun 3 '17 at 13:05

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