An example asks me to define $T: M_{2\times2}(F) \to M_{2\times2}$ by $T(A) = A^t$. Compute $[T]_\alpha$

$\alpha$ is the standard ordered basis of $2\times 2$ matrices.

To find the transformation I performed the transformation on all four elements of $\alpha$, for example $T(\begin{pmatrix} 1&0\\0&0\\\end{pmatrix}) = (\begin{pmatrix} 1&0\\0&0\\\end{pmatrix}) $ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\times4$ matrix:

$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$

I don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\alpha$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in $\alpha$ since one is a $4\times4$ and one is a $2\times 2$.

Also, why doesn't this mean that any arbitrary $2\times 2$ matrix can be raised to the power of $t$ by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).

  • 1
    $\begingroup$ I changed $T: M_{2\times2}(F) -> M_{2\times2}$ to $T: M_{2\times2}(F) \to M_{2\times2}$. $\TeX$ is not so crude that you have to resort to things like that. $\endgroup$ – Michael Hardy Oct 11 '12 at 23:43
  • $\begingroup$ Thanks - I couldn't find the command for it. I'll know for next time. $\endgroup$ – CodyBugstein Oct 11 '12 at 23:57

You need to distinguish here between the space $\mathbb F^2$ that the $2\times2$ matrices operate on and the space of the $2\times2$ matrices themselves. The former is two-dimensional, the latter is four-dimensional.

A matrix is a representation of a linear transformation with respect to bases of the domain and target of the transformation. If the linear transformation is an endomorphism, a linear map from a space to itself, the corresponding matrix is square, and one typically uses the same basis for the space in its two roles as domain and target.

The standard basis of the space of $2\times2$ matrices that you're using has four basis elements and spans the four-dimensional space of $2\times2$ matrices. A linear transformation has been defined on that space – the four-dimensional space of $2\times2$ matrices, not the two-dimensional space that they operate on.

Representing that linear transformation as a matrix with respect to the standard basis (both for the domain and the target) yields a $4\times4$ matrix. That's not a mismatch because the vectors and matrices that will be multiplied by this matrix aren't elements of $\mathbb F^2$; they're the vectors of coefficients in an expansion of the matrix they represent in the standard basis.

As an example, the matrix

$$ \pmatrix{2&3\\4&5} $$

has coefficients $2$, $3$, $4$ and $5$, respectively, in the standard basis, so with respect to that basis it's represented by the column vector

$$ \pmatrix{2\\3\\4\\5}\;. $$

Its transpose

$$ \pmatrix{2&4\\3&5} $$

is represented by the column vector

$$ \pmatrix{2\\4\\3\\5}\;. $$

And sure enough we have

$$ \begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\pmatrix{2\\3\\4\\5}=\pmatrix{2\\4\\3\\5}\;. $$

Thus this matrix really does represent the linear transformation $T$, transposition, in this basis.

  • $\begingroup$ So matrices can unravel to form vectors? I've played around with some and it makes sense. Thanks for the amazing answer. I still have a question though; shouldn't the transforming vector actually put your sample vector to the power of something? Isn't that the whole idea? $\endgroup$ – CodyBugstein Oct 11 '12 at 23:32
  • $\begingroup$ @Imray: I think you've misunderstood the notation. $A^t$ is merely a notational device to denote the transpose of $A$; nothing is being raised to a power. I prefer the slightly different notation $A^\top$ (A^\top) to avoid the impression that the $t$ is a variable used as an exponent. $\endgroup$ – joriki Oct 11 '12 at 23:34
  • $\begingroup$ @Imray: As an exercise, I suggest to write down the matrix of $T$ with respect to the basis $$ \left\{\pmatrix{1&0\\0&1},\pmatrix{1&0\\0&-1},\pmatrix{0&1\\1&0},\pmatrix{0&1\\-1&0}\right\}\;, $$ and then perhaps use it to apply $T$ to a couple of matrices and see whether the transpose comes out right. $\endgroup$ – joriki Oct 11 '12 at 23:38
  • $\begingroup$ yes right! I didn't realize that! Ok I've got it now, thank you very much for explaining it to me. $\endgroup$ – CodyBugstein Oct 11 '12 at 23:38
  • $\begingroup$ @Imray: You're welcome! $\endgroup$ – joriki Oct 11 '12 at 23:40

Just to make sure our definition of matrix representation is the same: Let $\alpha = (a_1, \ldots, a_k)$ be a basis for $V$, and let $T: V \rightarrow V$ be a linear map. Then matrix representation $[T]_{\alpha}$ is defined as follows: $$ [T]_{\alpha} = \begin{bmatrix} [T(a_1)]_{\alpha} & \ldots & [T(a_k)]_{\alpha} \end{bmatrix} $$

The matrix representation obeys the following formula: for every $v \in V$, $[T(v)]_{\alpha} = [T]_{\alpha} [v]_{\alpha}$, where $[v]_{\alpha}$ is the coordinate vector of $v$ with respect to the basis $\alpha$.

So now, to answer your question, it's true that $[T]_{\alpha}$ is a $4 \times 4$ matrix in your example, but you don't multiply it onto the $2 \times 2$ matrices; you multiply it onto the $2 \times 2$ matrices after they have been converted into column vectors via the coordinate mapping. Since $2 \times 2$ matrices form a $4$-dimensional space, then this coordinate mapping turns a $2 \times 2$ matrix into an element of $\mathbb{R}^4$.

  • $\begingroup$ How does coordinate mapping work? The rows all extend downward to become one long column? Can all matrices with $n$ elements be multiplied by any matrix with $n$ columns? $\endgroup$ – CodyBugstein Oct 11 '12 at 23:35
  • $\begingroup$ What confused me is how $[T]_\alpha$ could be square. But one needs to remember that the columns are in vector notation too (not in matrix notation!), so they are actually not $k \times k$ but $k^2 \times 1$ and so in the end $[T]_\alpha$ is $k^2 \times k^2$. $\endgroup$ – philmcole Apr 10 '18 at 21:20

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