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I've been scouring the web and my textbook to no avail. I understand how to transform a matrix with respect to the standard basis, but not to a basis comprised of matrices.

I need to find the matrix of T with respect to a basis. My transformation matrix is:

$$ A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \\ \end{bmatrix} $$

and I need to find it with respect to the basis:

$$ B = \left\{ \begin{bmatrix} 1& 0 \\ 0 &0 \\ \end{bmatrix} , \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 1 \\ \end{bmatrix}\right\} $$

I'm sorry my formatting is rough, the three B matrices are in the book as B = (Matrix 1, Matrix 2, Matrix 3).

I really don't understand this. My book has a column by column method that has completely stumped me.

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    $\begingroup$ What is $T$? What is its domain and codomain? What space is the basis you provided a basis of? $\endgroup$ – Santiago Canez Mar 27 '15 at 20:26
  • $\begingroup$ The question is very incomplete as it is. $\endgroup$ – Timbuc Mar 27 '15 at 20:32
  • $\begingroup$ I am very sorry Santiago and Timbuc. T(M) = AM. A is given above. I need to find A with respect to the basis B (first matrix given above, second matrix given above, third matrix given above). From U^2x2 to U^2x2 $\endgroup$ – user226942 Mar 27 '15 at 20:41
  • $\begingroup$ The question is also given here math.stackexchange.com/questions/325768/… $\endgroup$ – user226942 Mar 27 '15 at 20:52
  • $\begingroup$ @user226942 Then is this a duplicate? Or why are you asking this again? $\endgroup$ – Ana Galois Mar 27 '15 at 21:17
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I think your difficulty is purely "conceptual". A vector space is ANY set that obeys the axioms of a vector space (such a definition assumes an associated "field of scalars"). In particular, over a given field $F$, the set:

$\text{Hom}_F(U,V)$ = all linear transformations $U \to V$ is a vector space.

So "matrices" are vectors, too! Look, we can add them: if $A = (a_{ij}), B = (b_{ij})$ then $A + B = (c_{ij})$, where for each $i,j: c_{ij} = a_{ij} + b_{ij}$, and we can "multiply by a scalar":

$rA = (ra_{ij})$ (sometimes this is written as $(rI)A$).

One way to "ease the conceptual transition" is called the "vectorization" of matrices: we just string the columns "head-to-toe" into one long column, so that:

$A = \begin{bmatrix}1&2\\0&3\end{bmatrix}$ becomes:

$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix}$, transforming an element of $\text{Mat}_2(F)$ into an element of $F^4$ (you can take $F = \Bbb R$, for concreteness, if you primarily deal with real vector spaces).

Seen this way, it becomes clear that the second coordinate of your basis vectors is "unnecessary baggage", as it is always $0$. This is no different than identifying the subspace:

$U = \{(x,0,y,z) \in \Bbb R^4\}$ with $\Bbb R^3 = \{(x,y,z):x,y,z \in \Bbb R\}$

It should be clear that $\phi:U \to \Bbb R^3$ given by $\phi(x,0,y,z) = (x,y,z)$ is a bijective linear transformation.

So, in your situation, you want to find $a,b,c$ such that:

$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix} = a\begin{bmatrix}1\\0\\0\\0\end{bmatrix} + b\begin{bmatrix}0\\0\\1\\0\end{bmatrix} + c\begin{bmatrix}0\\0\\1\\1\end{bmatrix}$

Which is just a system of 3 linear equations in 3 unknowns:

$a = 1\\b+c = 2\\c = 3$

that you should be proficient in solving by now.

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    $\begingroup$ If I were a dictator, I would demand that a statue of your likeness be built upon all public grounds in my fine nation. Bless your soul. $\endgroup$ – user226942 Mar 28 '15 at 19:28
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From the nature of your question, I'm not sure you've transcribed it correctly for us. I suspect that the answer is this:

$$ \begin{bmatrix} {1} & {2} \\ {0} & {3} \\ \end{bmatrix} = 1 \cdot \begin{bmatrix} {1} & {0} \\ {0} & {0} \\ \end{bmatrix} + (-1) \cdot \begin{bmatrix} {0} & {1} \\ {0} & {0} \\ \end{bmatrix} + 3 \cdot \begin{bmatrix} 0 &1 \\ 0 & 1 \\ \end{bmatrix} $$

so that the coordinates of your matrix, $A$, with respect to the basis given, are $1, -1, 3$. Perhaps you're supposed to write that as a vector, in which case it'd be $$ \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}. $$

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  • $\begingroup$ Interesting. My guess at an interpretation was that the OP is looking at the linear transformation from the space of $2 \times 2$ upper-triangular matrices to itself given by multiplication by $A$, but perhaps your interpretation is correct. $\endgroup$ – Santiago Canez Mar 27 '15 at 20:42
  • $\begingroup$ Santiago, you are right. My sincere apologies for my sloppy question-asking. I very much appreciate both of your help. I know the answer, it is a 3x3 matrix, but I don't understand how to get there. $\endgroup$ – user226942 Mar 27 '15 at 20:43
  • $\begingroup$ Of course, Santiago...that makes perfect sense. I didn't see that multiplication by $A$ sent upper triangulars to themselves (D'oh!) or I would have interpreted it that way as well. Sigh. $\endgroup$ – John Hughes Mar 28 '15 at 1:40

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