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I have a linear transformation, T, such that;

T:${M_{22}}$${M_{22}}$:

T$\left(\begin{bmatrix}{x_{11}} & {x_{12}}\\{x_{21}} & {x_{22}}\end{bmatrix} \right)= \begin{bmatrix}{{x_{12}}-5{x_{21}}-{x_{22}}} & {-{x_{11}}-2{x_{12}}+3{x_{21}}+4{x_{22}}}\\{-3{x_{21}}} & {-{x_{11}}-{x_{12}}+{x_{21}}+3{x_{22}}}\end{bmatrix}$

What is the matrix that represents this ${M_{22}}$${M_{22}}$ transformation?

Is it:

$ \begin{bmatrix}{0} & {1} & {-5} & {-1}\\{-1} & {-2} & {3} & {4}\\{0} & {0} & {-3} & {0}\\{-1} & {-1} & {1} & {3}\end{bmatrix}$

If so, how could this be multiplied by a 2x2 matrix to give another 2x2 matrix. (2x2 matrices cannot multiply with 4x4 matrices).

There is likely something small I am missing so some help would be great! Thanks

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  • $\begingroup$ $\newcommand{\B}{\mathcal{B}}$Remember, the matrix $A$ that represents $T$ (with respect to standard basis $\mathcal{B}$ in domain and codomain) is by definition such that $$[T(X)]_{\B} = A[X]_{\B}$$ for all $X$ in the domain of $T$, where $[X]_{\B}$ is the coordinate vector of $X$ with respect to the basis $\B$. So $A$ won't multiply a matrix, but instead will essentially multiply the vector representation of that matrix, and will output the vector representation of $T(X)$, rather than outputting $T(X)$ itself. ("Vector representation" meaning coordinate vector with respect to $\B$.) $\endgroup$ – Minus One-Twelfth May 22 at 10:11
  • $\begingroup$ $\newcommand{\B}{\mathcal{B}}\newcommand{\C}{\mathcal{C}}\newcommand{\b}{\mathbf{b}}$Also in general for a linear map $T:V\to W$, where $V$ is a finite dimensional vector space with basis $\B$ and $W$ is a finite dimensional vector space with basis $\C$, the definition of the matrix of $T$ (with respect to basis $\B$ in domain and $\C$ in codomain) is that it is such that $[T(\mathbf{x})]_{\C} = A[\mathbf{x}]_{\B}$ for all $\mathbf{x}\in V$. (This uniquely characterises $A$. As you may have learnt, if $\B = \{ \b_1,\ldots, \b_n\}$, then the $j$-th column of $A$ is given by $[T(\b_j)]_{\C}$.) $\endgroup$ – Minus One-Twelfth May 22 at 10:15
  • $\begingroup$ Ah yes, thankyou! Don't know why I didn't think of this $\endgroup$ – Oliver Murfett May 22 at 10:16
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You'd have to write the matrix in column form $\begin{pmatrix} x_{11} \\ x_{12} \\ x_{21} \\ x_{22} \end{pmatrix}$. After that, the matrix you gave will work. No 2x2 matrix will multiply to produce that transformation since the top-left entry contains $x_{22}$, which is impossible under matrix multiplication.

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