I have a linear transformation, T, such that;


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

  • $\begingroup$ $\newcommand{\B}{\mathcal{B}}$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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.